Grade 9 | AQA GCSE Chemistry Paper 1 | whole paper revision

Summary

In this video, Primrose Kitten Academy provides a comprehensive revision of the AQA GCSE Chemistry Paper 1. Tailored for students aiming for top grades like 9 or 7, the video covers the entire paper with a focus on essential concepts, common exam questions, and exam techniques. The creator, Charlotte, also delves into writing and structuring exam papers, predicting questions, and avoiding pitfalls during exams. Using detailed explanations and examples, she helps students understand atomic structures, chemical reactions, bond types, states of matter, electrolysis, and titration. The video serves as a robust guide for students aiming to excel in Chemistry, highlighting critical definitions, equations, and methods needed for the exam.

Highlights

• Comprehensive guide to AQA GCSE Chemistry Paper 1 revision 🌟.
• Explains complex topics in an easy-to-understand manner, using detailed examples and equations 📘.
• Covers essential chemistry areas like atomic structure, bonding, chemical reactions, and electrolysis 🔍.
• Helpful exam tips and techniques offered to avoid common pitfalls and enhance answer structuring ✏️.
• Interactive and engaging presentation style makes learning chemistry enjoyable and less daunting 🎢.

Key Takeaways

• The video targets students aiming for AQA GCSE Chemistry Paper 1 excellence with a focus on achieving grades 9 and 7 🎓.
• It emphasizes understanding key chemistry concepts such as atomic structure, chemical reactions, bonding, and states of matter 🧪.
• Charlotte provides techniques for structuring exam answers, avoiding common mistakes, and using predicted papers effectively 📚.
• Fundamental chemistry knowledge is discussed, from isotopes and ions to electrolysis and titration ⚗️.
• The video's engaging tone helps demystify complex topics and equip students with necessary exam strategies 💡.

Overview

Welcome, chemistry enthusiasts! In this AQA GCSE Chemistry Paper 1 revision session, we're diving into the depths of chemistry knowledge designed specifically to help students aiming for top grades like 9 and 7. With Charlotte at the helm, this video promises a journey through chemical concepts with clarity and insight. Whether you're perplexed by isotopes or tackling the intricacies of ionic bonds, this guide is your go-to resource before stepping into the exam hall!

The session embarks on explaining vital topics such as atoms, ions, isotopes, and chemical reactions. The presenter’s engaging style transforms daunting topics into digestible content. Expect a walkthrough of predicted exam papers and a comprehensive guide on how to decode exam questions effectively. Charlotte also shares how to avoid common traps and keep the nervous jitters at bay while maximizing your potential in the exam.

What makes this video indispensable is its wide-reaching focus on both fundamental and complex areas like titration, electrolysis, and the periodic table. Charlotte aligns her teachings with exam techniques, offering insights into writing compelling answers and utilizing key chemistry equations wisely. This isn’t just a revision video; it’s a toolkit for success in AQA GCSE Chemistry Paper 1, crafted to lighten your academic load and boost confidence.

Chapters

• 00:00 - 01:00: Introduction In this introductory chapter, aimed at students preparing for the AQA GCSE chemistry paper one at a Grade 9 standard, the video will guide them through the necessary content to achieve high grades. It includes a mention of sections specifically for separate science students, which will be indicated with a popup. Students can use timestamps or chapters to navigate to specific parts. Additionally, there are predicted papers and walkthroughs available to complement the video content.
• 01:00 - 20:30: Atomic Structure The chapter "Atomic Structure" provides insights into the expectations and strategies for answering exam questions on the topic. It features guidance from Charlotte, who has both created the video and authored related papers, offering a unique perspective on how to approach exam questions. The chapter highlights common question types, clues left by examiners in wording, and common pitfalls to avoid during exams. It advises students to digest the material thoroughly, suggesting that notes be taken while watching the accompanying video. This chapter aims to prepare students for exams by offering practical advice and insight into examiners' strategies.
• 20:30 - 30:00: The Periodic Table The chapter titled 'The Periodic Table' begins by addressing the fundamental question: 'What is an atom?' It explains that an atom is the smallest part of an element that can exist and is represented by a chemical symbol. These chemical symbols are standardized and always begin with a capital letter, consisting of one or two letters. For instance, the chemical symbol for oxygen is shown as an example.
• 30:00 - 70:00: Chemical Reactions The chapter titled 'Chemical Reactions' introduces basic atomic structure and measurements. It briefly explains how elements like oxygen and helium are denoted in chemical language, with oxygen as 'O' and helium as 'He'. The concept of atomic radius is addressed, stating that an atom has a radius of approximately 0.1 nanometers (nm) or 1 x 10^-10 meters (m). Furthermore, it highlights that the nucleus's radius is smaller than that of the atom, equating to about 1 x 10^-4 meters.
• 70:00 - 96:00: Acids and Alkalis The chapter introduces the concept of elements and provides a definition, highlighting that an element is a substance composed of only one type of atom. It mentions that the periodic table, which will be available during exams, contains over a hundred different elements, each represented by unique letters.
• 96:00 - 115:00: Energy Changes The chapter 'Energy Changes' begins with an explanation of the periodic table, where pairs of letters are used to represent different elements. Each element is distinguished by a unique set of properties. The chapter then revisits the concept of an atom, affirming it as the smallest unit of an element that can exist independently. An element is depicted as a substance composed solely of that type of atom. An example is given with an atom to illustrate this concept.
• 115:00 - 142:00: Chemical Cells and Fuel Cells The chapter begins with an exploration of the element carbon, highlighting its atomic structure and how it forms the basis for various compounds. A compound is defined as a substance containing two or more different elements chemically combined in fixed proportions. The chapter then introduces examples of chemical reactions that occur involving carbon compounds.

Grade 9 | AQA GCSE Chemistry Paper 1 | whole paper revision Transcription

• 00:00 - 00:30 hello lovelies in this video we're going through the whole of your AQA GCSE chemistry paper one and this is to a Grade 9 standard so if you're reaching for that grade seven that grade that grade nine this is what we are looking for any content that is separate science only is going to have a little popup to say that if you're doing F you don't need to watch it but you can use the time stamps or the chapters to jump to the bits that you're looking for now to go with this there are the predicted papers that we've written for this year and the walkthroughs of the predicted
• 00:30 - 01:00 papers so you can see Charlotte who's made this video has also written the papers and has done the walkthroughs for the papers so you can see how she would expect you to structure her answers if she was marking the exams what sort of questions we think come up this year what sort of Clues the exam that might be leaving you if the questions worded this way and how to avoid any mistakes in the exam take this calmly take this slowly can watch it on two times speed if you want to but I'd much rather you watched it made notes and tried to
• 01:00 - 01:30 questions as well good luck guys we're here with you every single step [Music] [Applause] away what is an atom an atom is the smallest part of an element that can exist atoms of an element are represented by a chemical symbol and these symbols will always start with a capital letter and consist of one or two letters as you can see here oxygen is
• 01:30 - 02:00 just represented with a capital O and helium is represented with a capital H followed by an e now we need to know how small an atom actually is atoms have a radius of about 0.1 Nom which is 1 * 10- 10 m and the radius of their nucleus is less than 1 10,000 of that of the atom so it's about 1 * 10-4 m
• 02:00 - 02:30 we've already mentioned this word element we need to know though what is an element what is its definition so we need to know going into your exam that an element is a substance made up of only one type of Aton if we look at our periodic table which we will be given in the exam you can see that we've got over a hundred different elements all of these different letters
• 02:30 - 03:00 or pairs of letter within the periodic table that you can see here is representing a different element and each of those different elements will have their own unique set of properties now if we link this back with what we said an atom was the smallest part of an element that can exist we can see that we can represent an element by showing a substance made of only that type of atom so here for inance instance I've got an atom of
• 03:00 - 03:30 carbon and the element of carbon is just shown as lots and lots of these atoms in a structure so now we're going to look at compounds what is a compound we should know that a compound is a substance which contains two or more different elements which are chemically combined in fixed proportions so an example of this could be the chemical reaction that occur
• 03:30 - 04:00 between ion and oxygen to form ion oxide here we should recognize that ion and oxygen are both elements because they're only made up of one type of atom whereas ion oxide is a compound you can see it's made up of two different elements the ion and the oxygen which have become chemically combined now it's important that we know that compound actually have different
• 04:00 - 04:30 properties from the elements that made them so the properties of ion oxide are actually very different to the properties of ion and of oxygen now we also need to know that compounds can only be separated into Elements by chemical reactions we aren't able to Simply separate them using filtration or distillation or some other physical process we have to use use a chemical
• 04:30 - 05:00 reaction now chemical reactions will always involve the formation of one or more new substances such as this example here where ion and oxygen made ion oxide and they will often involve an energy change so quite often we'll observe an increase or a decrease in temperature due to this energy change in chemistry we're often going to be asked about chemical reactions we might be asked write down the word
• 05:00 - 05:30 equation which would look like something like this water goes to hydrogen plus oxygen however sometimes we'll be asked to write down the symbol equation now this involves writing down the chemical names for each of these substances so we know that water is H2O hydrogen is H2 and oxygen is O2 now when we're asked to write down the symbol equation we will need to
• 05:30 - 06:00 balance it and this is a certain skill when we want to balance a symbol equation we need to make sure that we have the same number of each type of atom on each side of the arrow so on both the reactant side on the left and the product side on the right now let's use this as a perfect example to show you how to do this the first thing I like to do is to separate the reactants from the products by drawing a little line and then on the
• 06:00 - 06:30 left hand side on the reactant side let's count up how much of each type of atom we have I can see I have two h's and one oxygen if I do the same on the other side I can see that I have two hydrogens and I have two oxygens as well now looking at this I can see I have the same number of hes on the left and the right however I only have one
• 06:30 - 07:00 oxygen on the left and I have two on the right now the way we can balance this is by having more waters on our reactant side so let's write another H2O there now here I can clearly see that I have the two oxygens so I would cross out that one and replace it with a two however by introducing another water we now have more hydrogens as well so we cross out the two and we replace it this
• 07:00 - 07:30 time with four because I had two of these H2S so now although the oxygen are balanced the hydrogen's are now not balanced so what we need to do is we need to have more hydrogens on the right hand side so let's introduce another H2 now I can see H2 plus an H2 is going to give give me four
• 07:30 - 08:00 hydrogens so now the hydrogens will also be balanced I have four on the left and I have four on the right so this means that our equation is balanced now let's write this in the way that the examiner would want it I can see I have one two waters on my left hand side so I'm going to put a big number two in front of the H2O that means that we have two
• 08:00 - 08:30 waters so now let's look over on the right hand side here I can see that I have one and two hydrogens which means I want to put a big number two in front of the H2 and there we have it we have our balanced symbol equation now in order to write down all of these balanced equations there's some formula that we really need to know and
• 08:30 - 09:00 this is going to hold throughout the whole of chemistry we need to know that carbon dioxide is CO2 we need to know that water is H2O we need to know oxygen is O2 hydrogen is H2 nitrogen is N2 ammonia is NH3 hydrochloric acid is HCl and sulfuric acid is h2so4 if we learn these off by part we
• 09:00 - 09:30 will have such an easier time in our exam the atom is made up of three different types of particle so we have protons which are positively charged particles found within the central nucleus of the atom then we also have neutrons which are also found inside the nucleus of the atom neutrons however have no charge now we might see this referred to as being neut
• 09:30 - 10:00 neutal now that third particle found within an atom is the electron now electrons are negatively charged particles and we don't find them in the nucleus this time we find them in shells which orbit the nucleus so you can see all this in our little diagram here and we have to get very used to these diagrams and being able to label them accurately this table here summarizes all of the facts that we need to know about the protons neutrons and
• 10:00 - 10:30 electrons we need to know that the relative charge of a proton is+ one the relative charge of a neutron is zero and the relative charge of an electron is minus1 we also need to know that the relative mass of both the protons and the neutrons so both particles found within that nucleus is equal to one however the electron's relative mass is really really tiny compared to this
• 10:30 - 11:00 we might see it referred to in several different ways but essentially the relative mass of an electron is approximately 1 over 2,000 we may even sometimes see it referred to as just very small now it's important that we know a few key facts relating to this information so atoms of different elements will all have different numbers of protons this is part of what gives them different properties
• 11:00 - 11:30 however all atoms of a particular element actually have the same number of protons now finally if we have a think about the mass of an atom we can see that the relative mass of the protons and the neutrons is so much more than that of the electrons and this is why almost all of the mass of an atom is found within the nucleus in chemistry we will often see is opes coming up now we need to know
• 11:30 - 12:00 what is an isotope Isotopes are atoms of the same element that have the same number of protons but a different number of neutrons here you can see I've got two different isotopes of helium the helium 4 isotope has two protons two neutrons and two electrons however the helium 3 isotope has two proton
• 12:00 - 12:30 only one Neutron and two electrons so notice here the thing that makes these isotopes of each other is that they still have the same number of protons however they have a different number of neutrons now let's have a think about the ion what is an ion an ion is a charged particle which forms when an atom or a molecule gains or loses an electrons so here I got a diagram of a
• 12:30 - 13:00 burum atom I can see that it has four electrons now the burum atom can lose the two electrons within its outer shell to become a brillium ion now because it's lost two of these negative electrons it has become more positive think about it like if you lose a negative thought and you become more positive and happy that's exactly what this pilion has done so because it's lost two electrons it now has a 2+
• 13:00 - 13:30 charge and this is how we would represent this in an [Music] exam electrons are found within shells now shells can also be referred to as energy levels so use whichever one you prefer when we're adding our electrons to an atom we start by adding them to the innermost Shell First now we need to know that in the first shell that in a mo shell we can
• 13:30 - 14:00 only add up to two electrons then in the second shell we can only add up to eight electrons and the third shell we only add up to eight electrons again now this is going to be really really important when we're in the exam if we're asked to add the electrons to an atom so let's use carbon as an example here carbon has six electrons now if we want to add this to
• 14:00 - 14:30 the shells we can remember that we start with the innermost shell first so with the inim shell we know that we can add up to two electrons so we add our one two electrons to the innermost shell of this carbon atom then if you think we have six electrons in total that we need to add on if we do 6 minus that two we're left with four more
• 14:30 - 15:00 electrons and we know that in the second shell we can add up to eight electrons so we can fit all of those four in that second shell and that's what I've done here I've got one 2 3 four electrons in its second shell now we can represent this using this diagram however we might also see it using this system here where in Brackets we do 2 comma 4 that represents 2 two in the first shell and four in the
• 15:00 - 15:30 second shell now let's have a look at how the atom is represented on the periodic table now the biggest number is known as the mass number the mass number is very important because it represents the sum of the number of protons and the number of neutrons now the smallest number of the two is the atomic number and the atomic number gives us the number of protons within that atom so so much information
• 15:30 - 16:00 given just from this periodic table so it's important that we know atoms have no overall charge now that then leads us on to the fact that in an atom the number of protons those positive particles must be equal to the number of electrons those negative particles and that's how they can have no overall charge
• 16:00 - 16:30 now we could be asked in the exam to work out the number of protons neutrons and electrons in a certain atom so if we look at Florine here as an example I know that that smallest number is the atomic number which gives me the number of protons therefore I know that the number of protons is equal to 9 now then we can think about the mass number which is the biggest number
• 16:30 - 17:00 and remember we said that the mass number was the sum of the protons and the neutrons so as we now know how many protons there are we can work out the number of neutrons by doing 19 - 9 and that ends up giving us 10 now finally we can work out the number of electrons by using the fact that we just said that in an atom the number of protons is equal to the number of
• 17:00 - 17:30 electrons so using that fact then we can say that we also have nine electrons and it's as simple as that so let's carry out the same process now for an argon atom I can see here that the mass number is 40 and I know that that mass number is equal to the sum of the protons and the neutrons I can also see here 18 the smallest number of the two is our atomic
• 17:30 - 18:00 number and I also of course remember that the atomic number is equal to the number of protons so now that I've looked at this I can start to work out how many protons neutrons and electrons we have so because the number of protons is equal to the atomic number I can say that I have 18 protons now I know that the number of neutrons is going to be equal to that mass number minus the number of
• 18:00 - 18:30 protons therefore to work out the number of neutrons I can do 40 minus 18 which gives me 22 so I know that I have 22 neutrons finally to work out the number of electrons that we have we use the fact that in an atom the number of electrons is equal to the number of protons so in this case I can say that I also have 18 electrons
• 18:30 - 19:00 sometimes we'll be asked a similar question but to work out the number of protons neutrons and electrons in an ion instead so here we're asked for br minus ion we should understand that in an ion like this which is negatively charged we must have more negative particles than we have positive particles now remember that electrons are the negative particles and protons are the positive particles so when in order for Bromine to have a negative charge in this case
• 19:00 - 19:30 we must have one more electron than proton we can only have more or less electrons not protons that's an important fact so if we look at this we can see that that smallest number the atomic number is equal to 35 so using our knowledge from before we know that the number of protons is equal to the atomic number so we have 35 protons
• 19:30 - 20:00 now we can see that the mass number here is 80 so to work out the number of neutrons we use the fact that the mass number is the protons plus the neutrons so to work out the neutrons I'm just going to do 80 minus the number of protons 35 which ends up giving me 45 so here I can see that I have 45 neutrons finally we need to write down the number of electrons and in order to do this because we have
• 20:00 - 20:30 an ion we have to use the fact that we said that in this case because it's BR minus we must have one more electron than the protons so as we have 35 of these positive protons we must have 36 of these negative electrons to get that overall negative charge what is relative atomic mass we need to know the definition of relative atomic mass is the average mass of all of the atoms of an element when compared
• 20:30 - 21:00 to 1 12th of the mass of a carbon 12 atom it's quite a technical definition so definitely one to add to your flashcards now the importance of relative atomic mass is that this definition explains why the relative atomic mass of some elements on the periodic table are not actually whole numbers what they actually are are averages of several different isotopes now on the periodic table we can look at
• 21:00 - 21:30 chlorine and notice that its mass number is actually 35.5 now that is its relative atomic mass that point five has come about because we've averaged all of the different isotopes that exist for chlorine it doesn't actually have 35.5 protons and neutrons it's different to what we're used to it's not quite the mass number in that case it's just the relative atomic mass now remember when we're talking
• 21:30 - 22:00 about Isotopes they have the same number of protons but a different number of neutrons so all of those different isotopes of chlorine are still going to have 17 protons however the number of neutrons that they have will vary a very common type of question that we will see in the exam is being asked to calculate relative atomic mass so here is the formula that we need to use relative atomic mass is equal to the
• 22:00 - 22:30 percentage of isotope 1 multili by the mass of isotope 1 plus the percentage of isotope 2 times by the mass of isotope 2 and we repeat this process for as many Isotopes as we have adding them all together and then divide through by 100 because all of these percentage abundances will of course total 100% so the three steps we're using are one multiply the percentage abundance of
• 22:30 - 23:00 each isotope by its mass two add all of these together and finally three divide by that total abundance so if we're using percentages that is of course going to be 100 let's have a look at an example here so lithium is found to have two isotopes lithium 6 has a 7% abundance and lithium 7 has a 93% abundance here we asked to calculate the relative atomic mass of
• 23:00 - 23:30 lithium so first things first let us remind ourselves of the equation so we know relative atomic mass is equal to the percentage of isotope 1 times the mass of isotope 1 plus the percentage of isotope 2 times the mass of isotope 2 and then we add those together divide by 100 so summarizing the information we have here we know the mass number of lithium 6 is six and its percentage abundance is 7% lithium 7 has a mass number of seven
• 23:30 - 24:00 and a percentage abundance of 93 so if we substitute these into our equation we find that the relative atomic mass is equal to 6 * 7 + 7 * 93 ID 100 and that gives us a relative atomic mass equal to 6.93 and so that is the relative atomic mass of lithium what is a mixture we need to know that a mixture consists
• 24:00 - 24:30 of two or more elements or compounds which are not chemically combined so this diagram here shows what a mixture looks like you can see we have a mixture here of these green molecules and these blue atoms they are not chemically combined and therefore we have a mixture so we need to know that mixtures can be separated by different physical processes these include filtration crystallization simple distillation
• 24:30 - 25:00 fractional distillation and chromatography and the method that we end up using of course depends on what makes up the mixture now these processes do not involve any chemical reactions and no new substances are made so let's start by looking at filtration filtration is used to separate mixtures of insoluble solids for example sand and liquids such as water as you can see from this diagram we will
• 25:00 - 25:30 pour a mixture of insoluble solid and liquid through some filter paper which will be sitting in a funnel the insoluble solid will end up staying in that filter paper because its particles are simply two large to pass through and the liquid will pass through the filter paper and ends up Gathering inside our conical flask like we said we would use this method whenever we want to separate a mixture of a solid which is insoluble and a
• 25:30 - 26:00 liquid now let's look at simple distillation now simple distillation is used to separate a solvent from a solution when we want to keep the liquid so an example of this is we want to produce water from a salt solution let's have a look at this apparatus here you can see that we have a condenser which is an important piece of apparatus which cools down our gas Vapors that will form to condense them
• 26:00 - 26:30 back down to a liquid so let's look at the method here so the first thing we're going to do is we're going to heat up our solution you can see we're doing that here using a buns and burner so the liquid part of our mixture will evaporate becoming a vapor and rising up now this vapor is going to pass through the condenser where it's going to get cooled down but when it gets cooled The Vapor will condense back down to a liquid this then separates the liquid
• 26:30 - 27:00 from the dissolved solid and we have successfully separated them fractional distillation is another method to separate substances now we use fractional distillation to separate two or more liquids which have different boiling points now this process is very very similar to that of simple distillation with one main difference to the apparatus and that is the presence of a fractionating p column it's that fractionating column that is key to
• 27:00 - 27:30 separating these two liquids so let's run through the steps here so the mixture is going to get heated to the temperature of the liquid with the lowest Boiling Point now the liquid with the lowest boiling point is going to evaporate first it's going to rise up through the fractionating column and pass through the condenser eventually Cooling and condensing now when all of this substance has seated and condensed we've separated the two
• 27:30 - 28:00 liquids now let's have a look at crystallization a crystalization is used to separate a dissolved solid a solute for example sodium chloride salt from a solution so we do this when the thing that we want is that dissolved solid so at the end of this process we want to end up with that solid let's look through the steps so the first thing we want to do is to gently heat the mix in an evaporating Basin now some of the solvent for
• 28:00 - 28:30 example water will evaporate away making the solution much more concentrated now we're going to remove this from the heat when the crystals begin to form we're going to leave it to cool and then we're going to filter it to remove any excess liquid that there may be finally we're going to leave those crystals to dry when they've dried we have successfully separated our solid from the
• 28:30 - 29:00 solution chromatography is used to separate mixtures of colored compounds for example inks or dyes and this can be great for telling us if certain inks or dyes are made up of several different components or if they're in fact pure so the way that we're going to do this is first we're going to start by putting a spot of the mixture along a pencil Line near the bottom of a piece of chromatography paper we place that paper vertically up up right in a solvent for example water the solvent
• 29:00 - 29:30 ends up getting soaked up the paper carrying all these different components with it now different components are going to move at different rates and this is what's going to separate them now if we had a pure substance of course we wouldn't have a mixture we'd only end up getting one spot on our chromatography P however if we had a mixture we would see the spot move up the chromatography paper and separate and give us two
• 29:30 - 30:00 different spots or more dependent on how many components there are to this mixture now an important thing to know about chromatography an extra little fact for you is that we always use a pencil line and they could ask you about why we do this the reason for this is that pencil is insoluble in water and many other solvents too and it doesn't move during the chromatography process imagine if it did it would completely interfere with with our results so we always use pencil
• 30:00 - 30:30 lines here we're going to have a look at the development of the model of the atom through time so we started off with Dalton's model Dalton's model told us that the atom was a solid sphere cannot be created destroyed or divided into any smaller parts now different types of spheres made up different elements but this sphere did not contain any protons neutrons or electrons now over time JJ Thompson came along and
• 30:30 - 31:00 proposed a new model called The Plum Pudding model here he said that the atom is not actually a solid sphere what it actually is is a cloud of positive charge with negative electrons embedded within it next came along the nuclear model and the experiment that determined this was Rutherford and his students Alpha scattering experiment now what Rutherford and his students did was he fired alpha
• 31:00 - 31:30 particles at a thin sheet of gold foil now if the plum pudding model was true we would expect all of these alpha particles just to bounce back however most of them passed straight through some of them were a little bit deflected and very very few came straight back now what this told us was that the mass of an atom was actually concentrated in a central nucleus which had to have been positively charged to have repelled this positive alpha particle and most of the atom actually was empty
• 31:30 - 32:00 space completely contradicting the plum hudding model and we know this because most of these atoms went straight through the gold foil with no deflection so over time or came up with the electron shell model and the electron shell model told us that rather than just randomly floating around the nucleus the electrons were actually found in fixed distances from the nucleus orbitting it in
• 32:00 - 32:30 shells later experiments found that the positive charge of any nucleus can actually be subdivided into a whole number of smaller particles of equal charge these were named protons finally James Chadwick carried out some experiments that provided evidence to show the existence of neutrons in the nucleus and this more or less brings us to our upto-date understanding of the atom here we're going to have a look at a vital part of chemistry the periodic
• 32:30 - 33:00 table we're going to use this so much throughout the course so periodic table why is it called a periodic table it's because similar properties occur at regular intervals or periodically now let's have a look at this periodic table then we should know that the columns are called groups and the periods are the rows so we start with period one which which is our first row then the next one
• 33:00 - 33:30 is our Period 2 period 3 four and so on now we need to know that all of the different groups have elements with similar properties within it now all of the elements within a group have the same number of electrons in their outer shell this is such useful information so for example all of the elements in group one have just one ele electron in that outer shell all of the elements in group
• 33:30 - 34:00 two have two electrons in the outer shell and so on another great fact to now is that elements in the same period all have the same number of electron shells so hydrogen and helium in Period one just have one shell all of the ones in Period two have two shells and so on early periodic tables like the ones here by Newland were incomplete as many of the elements were unknown before the protons neutrons and
• 34:00 - 34:30 electrons were discovered scientists arranged elements in order of their atomic weights however this meant that some elements were placed into incorrect groups as their chemical properties were ignored now Mel tried to solve this problem he arranged elements in columns based on their similar properties he arranged elements horizontally in order of atomic weight but left gaps to allow for elements that may have been undiscovered up to that point in some
• 34:30 - 35:00 places he changed the order of atomic weights to maintain the patterns in the columns now over time elements with the properties predicted by Mel were discovered filling the gaps now Isotopes explained why the order of atomic weight was sometimes Incorrect and that is how we got to the periodic table that we use today now we need to know that elements that react to form positive ions are metals and elements that react to form
• 35:00 - 35:30 negative ions are non-metals now on our periodic table we need to know where we find our metals and where we find our non-metals so let's have a look at it if we look at this little staircase that we have here where the staircase starts just above aluminium and goes all the way down we can see that the metals are on the left hand side of the periodic table and the non-metals are on the right hand side of the per iodic table let's look at group zero or the
• 35:30 - 36:00 noble gases now this is the column on the periodic table that is found furthest to the right hand side so the key facts we need to know about the group zero elements are that they have a full outer shell of electrons they are unreactive and they do not form molecules they are monatomic they exist as neon or helium not as molecules so why are the group zero elements
• 36:00 - 36:30 unreactive this is because they have full outershell of electrons so the next fact we need to know about the noble gases is how does the boiling point of the group zero elements change down the group we need to know that the boiling point increases down the group the group one metals or the alkaline metals as they're sometimes referred to have a few Key Properties first first one is that they all have one electron in their outer
• 36:30 - 37:00 shell and the second property we need to know is that they can react with oxygen chlorine and water now we need to know how does the reactivity of the group one elements change down the group it's important we know that the reactivity increases Down group one why does the reactivity of group one Metals increase down the group we need to know that the atoms increase inze as you go down the group there are also more shells and this
• 37:00 - 37:30 means that the outer shell is further away from the nucleus this means that the electrostatic attraction between the positive nucleus and the outer shell electron is weaker now this means it's easier to lose that one outer electron making those elements at the bottom more reactive let's have a look at how the group one metals react with water we need to know the general reaction is that metal plus water makes a metal hydroxide plus hydrogen now an example
• 37:30 - 38:00 of this of a group one metal reacting with water is the reaction of sodium with water to make sodium hydroxide and hydrogen the alkaline metals get their name from the fact that when they react with water they form this metal hydroxide which forms an Alkali solution with a pH greater than seven it's good to know that heat is also given in these reactions sometimes and this can even sometimes be seen with
• 38:00 - 38:30 a flame in these reactions so what is observed when alkali metals react with water let's think about our equation our general equation here we're going to observe fizzing due to the formation of hydrogen gas and we're also going to observe our solid metal disappearing over time and this is because the metal gets used up now let's have a look at the reactions
• 38:30 - 39:00 of those alkali metals with oxygen we need to know that a group one metal reacts with oxygen to make a metal oxide so an example of this is sodium plus oxygen makes sodium oxide now let's look at the reaction of group one Metals with chlorine so we need to know that a group one metal plus chlorine makes a metal chloride and an example of this is sod sodium plus chlorine makes sodium
• 39:00 - 39:30 chloride the halogens are found in group seven of the periodic table Group Seven Elements all have seven electrons in their outer shell they're all non-metals and they all consist of molecules made of pairs of atoms we call these diatomic molecules so they exist as for instance F2 cl2 br2 Etc now we need to know how does the reactivity of the group seven elements
• 39:30 - 40:00 change down the group we need to know that the reactivity decreases Down group seven so as well as knowing that the reactivity decreases down the group we also need to know why so we need to know the atoms increase in size as they go down the group we need to know that there are more shells and so the outer shell is further away from the nucleus and therefore the electrostatic attraction between the nucleus and the
• 40:00 - 40:30 outer shell is actually weaker this means it's going to be harder to gain an electron to fill that outer shell which in turn makes it less reactive as we go down group seven the melting and the boiling point of the halogens increases now this explains why the state changes from a gas to a liquid to a solid as we go down this group but why does the melting and boiling point increase we need to know these three points the molecules increase in size so the
• 40:30 - 41:00 intermolecular forces become stronger and therefore more energy is required to overcome the forces of attraction and that is why the melting and boiling point increases so here we have a table that summarizes all of the properties about the group seven elements that we need to know so the state of room temperature goes from a gas to a liquid for Bromine and a solid for iodine we can see that as we go down the group the relative
• 41:00 - 41:30 molecular mass increases as does the melting point and boiling point however the reactivity decreases down the group we also need to know about the appearance of group seven elements we need to know that at room temperature Florine is a gas that is yellow in color we need to know that chlorine at room temperature again is a yellow green gas however if we were to dissolve chlorine in water it would appear a pale
• 41:30 - 42:00 green solution groine now at room temperature is a red brown liquid however if we put that in water it would appear orange and finally iodine is a gray solid at room temperature however in solution it's actually dark brown in color when the group seven elements react with metals they produce salts for instance if we had sodium plus chlorine
• 42:00 - 42:30 we end up making the salt sodium chloride the halogens can also react with non-metals such as hydrogen now when halogens react with hydrogen they form hydrogen halides for example if we had hydrogen plus chlorine we end up making hydrogen chloride so H2 plus cl2 makes 2 HCL hydrogen halides are gases at room temperature and they dissolve in water
• 42:30 - 43:00 to form acidic solutions for example hydrogen chloride would dissolve in water to form hydrochloric acid now let's look at displacement reactions for the group seven elements so we know that reactivity decreases Down group seven now the more reactive group seven elements can take the place of less reative group seven elements in a compound this is called a displacement reaction so if we have chlorine reacting with
• 43:00 - 43:30 potassium bromide we will see that the chlorine which is more reactive will displace the bromine giving us potassium chloride and bromine however if we reacted chlorine with potassium fluoride because chlorine is less reactive than Florine we actually have no reaction take place because the chlorine cannot displace the more reactive Florine this table here summarizes the reactions
• 43:30 - 44:00 and the observations that we need to know so if we reacted potassium chloride with bromine or iodine you can see that we get no reaction each time that's because the chlorine is already more reactive than the bromine or the iodine so no displacement takes place however when we react potassium bromide with chlorine the chlorine will displace the bromide ions now as a result we'll end up forming
• 44:00 - 44:30 potassium chloride and bromine now that bromine will be ins solution so we will observe an orange color now if we reacted pottassium bromide with iodine we would see no reaction take place again and the reason for that is because bromine is more reactive than iodine finally here if we have potassium iodide reacting with chlorine because the chlorine is more reactive than the iodine we get a displacement reaction
• 44:30 - 45:00 take place we end up forming potassium chloride and iodine and because this is in solution a brown color is observed now the exact same thing happens with bromine react them with pottassium iodide again the bromine will displace the less reactive iodide ions and iodine will be formed and again we will observe that brown color of the iodine in solution the transition metals are found in the middle of the periodic table as you can
• 45:00 - 45:30 see here now the transition elements that you may get asked about are chromium manganese iron Cobalt nickel and copper now it's important that we recognize that although transition metals and the group one metals are both metals they actually have some very different properties if we compare their melting points their density their strength and their hardness we can see that these properties are all relatively high for
• 45:30 - 46:00 the transition metals however they're relatively low for the group one Metals likewise the reactivity of the transition metals with oxygen water and the halogens are very slow or maybe even not at all whereas we know that the group one metals react very easily with these things so what are the typical properties of transition elements the first property we need to know is that they have ions for different
• 46:00 - 46:30 charges for example we can get fe2+ and fe3+ they can also form colored compounds for example Ion 2 oxide forms a reddish brown solid and finally they can be used as catalysts for example manganese for oxide increases the rate of decomposition of hydrogen peroxide so there we have it those are the things things we need to know about transition metals
• 46:30 - 47:00 [Music] [Music]
• 47:00 - 47:30 [Music]
• 47:30 - 48:00 the three states of matter can be represented by a simple model which uses small solid spheres to represent the particles you can see that in the solid we've got this regular pattern whereas a a liquid although the particles are still touching they're in a random Arrangement whereas in a gas they are in a completely random Arrangement far apart and not touching so we know that going from a solid to a liquid is the process of melting a liquid to a solid is
• 48:00 - 48:30 freezing whereas going from a liquid to a gas is boiling and a gas to a liquid is condensing we've got three limitations with this simple model firstly there are no forces involved secondly all the particles are represented as spheres and finally all the Spheres are solid this table here nicely summarizes the difference between solids liquids and gases in a solid the particles are
• 48:30 - 49:00 in a regular pattern very close together and vibrating around fixed positions whereas in a liquid the particles are randomly arranged but close together and they move around each other and finally in a gas we have a random Arrangement particles are very far apart and they move quickly in all directions make sure you can draw all of these diagrams in chemical equations we represent the different states of matter using the
• 49:00 - 49:30 following symbols so a solid is represented with an s in Brackets a liquid is an L in Brackets gas is a g in Brackets and an aquous solution is represented with an AQ in Brackets there are three types of strong chemical bonds we have ionic bonds calent bonds and metallic bonds we need to know that we find ionic bonds
• 49:30 - 50:00 in substances that have metals and non-metals bonded together an example of that is sodium chloride calent bonding however is found when we have non-metal atoms bonded together an example of that is Diamond which is made up of just carbon atoms now finally we've got metallic bonding now that's found in metals an example of that is iron let's take a closer look at ionic bonding so we've said that ionic bonds
• 50:00 - 50:30 form between metals and non-metals when a metal atom reacts with a non-metal atom electrons from the outer shell of the metal atom are transferred the metal atoms lose electrons and become positively charged ions also known as cats because they're positive so here if we have a look at buril which has two electrons in its outer shell now burum wants to get a full
• 50:30 - 51:00 outer shell this means it needs to lose its two electrons by losing two electrons it ends up as a positive ion with a two plus charge and what we find is that all of the group one Metals form ions with a one plus charge the group two metals form ions with a two plus charge whereas the group six non-metals form ions with a two minus charge and group seven
• 51:00 - 51:30 nonmetals form ions with a one minus charge well we've looked at what happens to the methyl atom but what happens to the non-meal atom non-meal atoms are going to gain electrons and they're going to become negatively charged ions or anions let's have a look at oxygen as an example here I can see here that oxygen has six electrons in its outer shell now to get a full outer shell it needs to have eight electrons in its outer
• 51:30 - 52:00 shell and the easiest way that oxygen can do that is by gaining two electrons if it gains two electrons it becomes a negatively charged Ion with a two minus charge well all this leads us to the question what is an ionic bond what an ionic bond is is the strong electrostatic attraction between these oppositely charged ions of course we're going to end up with that positive methal ion and the
• 52:00 - 52:30 negative non-metal ion after reacting both that metal and non-metal ion are going to have a full outer shell and they now have the same electronic structure as a noble gas in group zero now we can show the electron transfer during the formation of an ionic bond using a DOT and cross diagram and when we draw a DOT and cross diagram we generally only draw the outermost shell of electrons well let's have a go at
• 52:30 - 53:00 looking at the formation of an ionic bond between Barum and oxygen so we know that burum has two electrons in its outer shell in fact its electronic configuration is 22 oxygen as well we just said earlier that we have six electrons in its outer shell so it has a 26 electronic configuration now what's going to happen is the buril is going to give its two electrons to the oxygen we end up with a burum 2 plus ion
• 53:00 - 53:30 and an oxygen 2 minus ion and we're going to write them just like this with square brackets around their full shells and the charges up on the top right hand side let's have a look at another example here draw a DOT and cross diagram to show the bonding in sodium chloride well sodium's in group one so we know it has one electron in its outer shell chlorine on the other hand I know is in group seven so it has seven
• 53:30 - 54:00 electrons in its outer shell I'm representing sodium's electrons with a cross and chlorines with a DOT you can do it the other way around and it still works so the sodium is going to give up its electron to get its F outter shell that leaves us with a one plus charge ion whereas the chlorine is going to gain that electron and it's going to therefore become a negative ion with the one minus charge so I just put a little minus in the top right hand corner for
• 54:00 - 54:30 chlorine what is an ionic compound an ionic compound is a giant ltis structure of ions held together by strong electrostatic forces of attraction between oppositely charged ions now these electrostatic forces of attraction act in all directions within the latis and they can be represented in several different ways they can be represented in a 3D structure or using a
• 54:30 - 55:00 ball and stick approach instead for ionic compounds we can work out the empirical formula from a diagram to do this all we need to do is to count the number of each type of ion and work out the lowest common ratio well in case you're wondering what an empirical formula actually is it's the simplest whole number ratio of atoms or ions within a structure let's have a look at this example here so I can see that the pink ions are sodium plus and the blue
• 55:00 - 55:30 ones are CL minus I can count in this diagram I have 6 na+ and six CL minus ions this is a 6:6 ratio which simplifies down to a 1 to1 ratio therefore we can say that the empirical formula of this compound is na1 cl1 or just NAC now we can represent ionic compounds in many different ways and there are limitations to these different types of diagrams that we used
• 55:30 - 56:00 to represent them for DOT and cross diagram it's very helpful because it shows the transfer of electrons the disadvantage is that it doesn't show how the ions are arranged in 3D space and it also doesn't show the relative sizes of the ions with the 3D ball and stick an advantage is that it shows the arrangement of irons in 3D space however the disadvantages includes the fact that it uses stick for bonds which is misleading because instead the forces of
• 56:00 - 56:30 attractions act in all directions in the 2D diagram we can see the arrangement of ions in one layer however it doesn't show the different layers of ions and it also doesn't show the 3D arrangement in space finally the 3D diagrams show the 3D arrangement in space which is an advantage however the disadvantages include the fact that it is not to scale and it gives no information about the forces of attraction between the
• 56:30 - 57:00 ions we need to know the properties of ionic compounds so ionic compounds have the following properties they have high melting points high boiling points they conduct electricity when melted molten or dissolved in water and they do not conduct electricity when in the solid state so why is it that ionic compounds have high melting or boiling points this is because there are strong electrostatic attractions between the
• 57:00 - 57:30 oppositely charged ions therefore we need a large amount of energy to overcome these forces of attraction we also need to know why ionic compounds conduct electricity when molten or dissolved in water now this is because the ions are free to move and the charge can flow freely and this is the same reason why a solid does not conduct electricity a solid ionic compound does not have ions which are free to move and
• 57:30 - 58:00 therefore the charge cannot flow freely let's have a look at calent bonding what is a calent bond we need to know that a calent bond is a bond formed when electrons are shared between non-metal atoms cently bonded substances can have different structures we can see them as small molecules for example H2O or O2 or CO2 or as giant calent structures for example diamond and silicon
• 58:00 - 58:30 dioxide we can use Dot and cross diagrams to show calent bonding when drawing Dot and cross diagrams remember we only draw the altero shell of electrons now the electrons of one atom are represented by dots and the electrons of the other atom are represented by crosses and when we show Dot and cross diagram with ve valent bonding we have overlapping regions and these overlapping regions show the sharing of
• 58:30 - 59:00 a pair of electrons this is a calent bond sometimes two pairs of electrons are shared and this would be a double calent bond and occasionally we even get three pairs of shared electrons which would be a triple bond and here you can see this example of chlorine Each of which have seven electrons in that outer shell they both need to gain one so what they do is they share one of their own electrons and therefore gain one from the other as
• 59:00 - 59:30 well resulting in our calent bond let's have a go at another example let's draw a DOT Andross diagram to show the bonding in CH4 we should know that carbon's in group four so it has four electrons in its outer shell and we should also know that hydrogen has one electron in its outer shell well if each of these hydrogens shares its electron with the carbon and that means each of the hydrogens end up with two electrons and get a full outer shell and the carbon
• 59:30 - 60:00 ends up with eight electrons and that also has a full outer shell so here we can see that there are four calent bonds in the bonding of C4 substances with giant calent structures are solids with very high melting points now all atoms in a giant coent structure are bonded to other atoms by these strong calent bonds now some examples of giant coent structures include Diamond graphite and
• 60:00 - 60:30 silicon dioxide we need to know that giant calance structures have very high melting points we need to know why this is we need to know that this is because the many strong coent bonds between atoms must be broken in order to melt or boil it this requires a lot of energy we also need to know why most giant calent structures do not conduct
• 60:30 - 61:00 electricity this is because there are no delocalized electrons or ions that are free to move to carry charge through the structure with the one exception of graphite which we will learn about in more detail later small molecules are usually gases or liquids and therefore they have relatively low melting and boiling points so some examples of small molecules include water cl2 oxygen anything that you're used to being a gas or a liquid at room
• 61:00 - 61:30 temperature so when boiling or melting substances made up of these small molecules the weak intermolecular forces between the molecules get overcome we're not actually breaking those strong calent bonds just these weak intermolecular forces now relatively speaking this doesn't require much energy at all and this explains the relatively low melting and boiling points now we need to know that as the molecules get bigger they have stronger intermolecular forces so as a result
• 61:30 - 62:00 bigger molecules have higher melting and boiling points now we should also know that small molecules do not conduct electricity and the reason for this is the fact that they have no charge what are polymers so we need to know that polymers are very large molecules made up of many repeating units joined by calent bonds now the polymer chains themselves are held together by intermolecular
• 62:00 - 62:30 forces and of course because polymers are such long molecules those intermolecular forces are relatively strong now the result of this is that polymers have higher melting points compared to smaller molecules and therefore they're generally solids at room temperature now a good use of polymers is to make Plastics now an example of a polymer is polyethene and we can represent this in different ways that we need to be able to recognize so it can be represented in
• 62:30 - 63:00 this very long way here which shows a short section of the polyethene polymer it actually contains thousands of these atoms but of course we're just showing a short section which still looks pretty long notice those open ends to show that it's continuing onwards or instead we can do it in this abbreviated form where the little n represents how many of these repeated units we have and N is going to be a very large
• 63:00 - 63:30 number we can represent calent compounds in many different WS however there are limitations to using different types of diagrams to represent them so dot and cross diagrams an advantage to this is that we can see the transfer of electrons and we can see which atom the bonding electrons come from however disadvantages include the fact that it doesn't show how the atoms are arranged in 3D space or show the relative sizes of the atom next up the 3D ball and stick this
• 63:30 - 64:00 is good because it shows the arrangement of atoms in space as well as the shape of the molecule however disadvantages include the fact that it uses sticks for the bonds and it doesn't show that the bonds are forces the atoms are placed far apart from each other but in reality the gaps are much smaller so now the 2D diagrams now some advantages to this is is that it shows what atoms are in a molecule and how they're connected
• 64:00 - 64:30 however disadvantages are that it doesn't show the relative size of the atoms and the bonds and it also doesn't show the 3D arrangement in space finally the 3D diagrams obviously an advantage is that it shows the 3D arrangement in space however disadvantages include the fact that the 3D diagrams aren't to scale and it gives no information about the forces of attraction between the atoms let's describe the structure and bonding in Diamond this is a very common
• 64:30 - 65:00 question we should know that diamant is a giant calent structure and every carbon atom forms four strong calent bonds with other carbon atoms make sure you include the number of calent bonds and the fact that they're strong so a very common question asks us why is Diamond so hard to answer this we're going to say that it's a giant coent structure that every carbon atom forms four strong calent bonds with other carbon atoms and
• 65:00 - 65:30 therefore it requires a lot of energy to break another very common question is why does diamond not conduct electricity and the answer to this is similar every carbon atom forms four strong coent bonds with other carbon atoms and so as a result there are no delocalized electrons or ions free to move and carry that charge we need to be able to describe the bonding and structure in
• 65:30 - 66:00 graphite so we should know that graphite is made of only carbon atoms we should know that it forms hexagonal Rings arranged in layers and we need to know that each carbon atom forms strong coent bonds with only three other carbon atoms not the four that it potentially could with its four electrons in its outer shell now this as a result leaves carbon with one spare electron and this electron gets delocalized and is free to move through the layers a very common question we're
• 66:00 - 66:30 going to get is why does breite conduct electricity so the answer we want to give is that each carbon atom has one spare electron which is delocalized and free to move and carry electrical charge through the layers another question we need to be able to answer is why is graphite slippery and so why can it be used in lubricants the reason for this is that it has layers which have weak intermolecular forces between them now
• 66:30 - 67:00 these weak intermolecular forces don't require much energy to overcome therefore making them slippery now let's have a look at graphine what is graphine graphine is a single layer of graphite and its properties make it useful in electronics and Composites so let's have a look at those properties then so it's got strong coent bonds between the carbon atoms a result of this is that graphine is going to be very strong and it's going to have high
• 67:00 - 67:30 melting and boiling points now this is because it's going to take a lot of energy to break those strong calent bonds if we look at the diagram we can see that each carbon atom in graphine is only bonded to three other carbon atoms we should know that as carbon has four electrons in its outer shell it can form four bonds and the fact that graphine only forms three bonds per carbon atom shows that we're going to
• 67:30 - 68:00 have a delocalized electron which is free to move and carry charge meaning that graphine is a conductor of electricity what are ferin ferin are molecules of carbon atoms which have hollow shapes the structure of ferin is based upon hexagonal Rings although we can sometimes have rings with five or seven carbon atoms at this level there'll generally be hexagonal rings so ferin can be arranged as a tube
• 68:00 - 68:30 for instance a nano tube or as a ball such as Buckminster ferine C60 carbon nanot tubes are cylindrical ferin with very high length to diameter ratios this means that they have a high tensile strength and so are difficult to break and they're useful for nanot technology electronics and materials Buckminster ferine however is a molecule made of 60 carbon atoms coal bonded into
• 68:30 - 69:00 a spherical shape they have weak intermolecular forces between the molecules which gives it a low melting point and making it slippery now that makes it perfect for use as a lubricant and for drug delivery let's have a look at Metals now metals have giant structures of atoms with strong metallic bonding now this means that most metals have high melting boiling points and this is because it takes a lot of energy to overcome these strong
• 69:00 - 69:30 bonds now let's have a think about pure Metals now in a pure metal the atoms are arranged in layers these layers are able to slide over each other and this means that pure metals can be bent and shaped they're malleable now a side effect of this though is that pure metals can be too soft now a way of remedying this is we can perhaps make an alloy an alloy is what we get if we mix
• 69:30 - 70:00 pure Metals with other elements now the way that this works is that atoms of different elements have different sizes Now by mixing these together this distorts the layers and the layers can't slide over each other as easily anymore we need much more Force for them to slide over each other now the result of this is that Alloys are hard harder and stronger than pure Metals making them much more useful for certain jobs than
• 70:00 - 70:30 pure Metals we need to know that metals are made up of giant structures of atoms arranged in a regular pattern the electrons in the outer shell of the metal atoms are delocalized forming metal ions now these delocalized electrons are now free to move through the layers of the metal ions so why is it that metals can conduct electricity the reason for this is that the delocalized electrons are free to move
• 70:30 - 71:00 between the layers and carry their charge why are metals good conductors of thermal energy this is because the delocalized electrons transfer energy and finally let's ask ourselves what causes metallic bonding this is due to the sharing of delocalized electrons we need to know about nanoparticles well what are nanoparticles nanoparticles are particles which are between 1 and 100
• 71:00 - 71:30 nanometers in size made up of a few hundred atoms we need to know the relative size of a few different particles we need to know that atoms are around the size of 1 * 10- 10 m nanop particles are between 1 * 10 - 9 and 1 * 10 - 7 m fine particles which we may see represented as PM 2.5 are between 1 * 10- 7 and 2.5 * 10-6
• 71:30 - 72:00 m and finally Co particles which are pm10 are between 2.5 * 10- 6 and 1 * 10 - 5 m now we need to know a few uses of nanop particles to we'll see that they can be used in medicine Electronics cosmetics suncream deodrant and catalysts now the main reason why
• 72:00 - 72:30 they're used in so many places is because they have a large surface area to volume ratio however as well as being useful for this fact there are also some risks associated with using nanoparticles firstly they can be breathed in because they're so small they can enter cells it can catalyze harmful reactions and also there's a chance that toxic substance es could bind to them due to their large surface area to volume ratios now let's explore this surface
• 72:30 - 73:00 area to volume ratio in a bit more detail let's work out the surface area of 1 cm Cube we know that there are six faces and each of those faces has an area of 1 * 1 so the surface area of this 1 cm cubed is 6 cm squared now if we work out its volume we do 1 * 1 * 1 which gives us 1 cm cubed so now if we need to get the
• 73:00 - 73:30 surface area to B INE ratio we simply divide the surface area by the volume so in this case that is 6 / 1 which gives us six now let's do exactly the same for 10 cm cubed six for the six faces * by 10 * by 10 which is the area of one face gives us 600 cm squ for the surface area now the volume will be 10 * 10 * 10 which gives us 1,000 cm
• 73:30 - 74:00 cubed now working out the surface area to volume ratio Again by doing surface area divided by volume we get 600 ided 1,000 which gives us 0.6 so what we can see here is that the smaller the particle the bigger the surface area to volume ratio so we can see that if we increase the size of a cube by a factor of 10 we actually decrease the surface area to volume ratio by a factor of 10 to [Music]
• 74:00 - 74:30 [Music]
• 74:30 - 75:00 [Music] the law of conservation of mass is an
• 75:00 - 75:30 important one in chemistry we need to know what it is so it states that no atoms are lost or made during a chemical reaction and that the mass of the product is equal to the mass of the reactants and if we had a balanced symbol equation we'll see that the number of atoms of each element is actually the same on both the reactant and the product side of this equation so let's have a look at this in practice we can have nitrogen and then we can add three
• 75:30 - 76:00 hydrogen molecules and we end up making two ammonia molecules see here that we have two of those green nitrogen atoms on both the left and the right hand side and you have six of those pink hydrogen atoms on again both the left and the right hand side now that is showing that no atoms are lost or made during that chemical reaction and we can also see this in
• 76:00 - 76:30 terms of mass so if we had 14 G of nitrogen and it reacted with 3 G of hydrogen we would end up making 17 G of ammonia so the mass of the products is equal to the mass of the reactants now we know that mass is always conserved however it might not always look like it is now this is going to depend on if we have an enclosed system like this one on the left hand side or a non-enclosed
• 76:30 - 77:00 system so in an enclosed system the mass will be observed to be conserved because everything is contained and nothing escapes whereas in a non-enclosed system gases can enter and leave the system now this may mean that the mass may appear to have changed where is actually all that's happened is some gases may have escaped from the system so that they can't be
• 77:00 - 77:30 measured however it's important to know that that mass is still there and the mass has been conserved relative formula mass is the sum of the relative atomic masses of the atoms in the numers shown in its formula now if we had a balanced chemical equation due to the conservation of mass we would expect the sum of the relative formula mes of all of the reactants to be equal to the sum of the relative formula masses of all of
• 77:30 - 78:00 the products now the easiest way to explain this is by looking at an example so let's have a look at working out the relative formula mass of sulfuric acid h2so4 so here I can see I have two hydrogen atoms and the relative atomic mass of hydrogen is 1 so I'm going to do 2 * 1 and then I'm going to do plus and I can see I have one sulfur atom so I'm going to do 1 * 32 which is the relative
• 78:00 - 78:30 atomic mass of sulfur and then I can see I have four oxygens so I'm going to add on 4 * 16 where 16 is the relative atomic mass of oxygen now popping that all into the calculator I find that sulfuric acid has a relative formula mass of 98 in the exam we're off asked how we calculate the percentage by mass now we need to know this
• 78:30 - 79:00 formula percentage by mass is equal to the total relative atomic mass of atoms of the element we're being asked about divided by the relative formula mass of the compound and then as always to turn it into a percentage we multiply it through by 100 so the easiest way to understand this is to have a look at an example so let's have a look at this question let's C calculate the percentage by mass of oxygen in calcium carbonate so the first step that we need
• 79:00 - 79:30 to do is firstly let's calculate the relative formula mass of calcium carbonate so to do this we can see that there's one calcium so we've got 40 plus one carbon that's + 12 and then plus the three oxygens so that is 3 * 16 and putting that all together we get 100 now we need to calculate the total mass of oxygen in the calcium carbonate so here because there's three oxygen
• 79:30 - 80:00 atoms in calcium carbonate we do 3 * 16 which gives us 48 now finally now that we've got all those parts we just substitute them into the equation above so we're going to do 48 / 100 times by 100 and that gives us 48% so the percentage by mass of oxygen in calcium carbonate is 48% and you can do that for anything okay we just repeat the same process and
• 80:00 - 80:30 we'll get our marks we need to be aware that whenever we make measurements in chemistry whether that's using a thermometer or a measuring cylinder or any other apparatus there's always going to be some uncertainty one example of this is when we have to judge the reading on a measuring instrument It's Not always completely obvious where your measurement is do we round up do we round down all of these things need to be taken into account another time when there might be
• 80:30 - 81:00 some uncertainty is judging whether or not your chemical reaction has actually ended so now let's have a look in a bit more detail but how we can estimate this uncertainty and there's two main methods that you're going to learn about first is how we can use the range of a set of repeated experiments to find that uncertainty and the second way is how we look at the resolution of the measuring instrument when we need to estimate
• 81:00 - 81:30 uncertainty from a set of repeat measurements there's two steps that we need to take first we need to work out the range of the results and then we need to half that range because the uncertainty will be plus or minus half the range that might seem a little confusing so let's have a look at putting this into practice so in this question here we want to work out the mean mass and estimate the uncertainty so let's look at these
• 81:30 - 82:00 results here I know that range is the biggest minus the smallest you should be familiar with this from maths so in this case the biggest value is 2.28 and the smallest is 2.21 so the range will be equal to 2.28 minus 2.21 and that gives me 0.07 so therefore the uncertainty is equal to plus or minus half the range harving this I get that the uncertainty
• 82:00 - 82:30 will be equal to plus or minus 0.035 G so the mean we find by adding all of these values together and dividing by four so in this case we're going to do 2.25 + 2.25 + 2.21 + 2.28 and then we're going to divide by four because we've got four different repeats
• 82:30 - 83:00 here so that gives me a mean equal to 22475 G and then we're going to do plus or minus that uncertainty so plus or minus 0.035 G now let's have a look at how we estimate uncertainty from measuring instruments I mentioned earlier that we do this by looking at the resolution of the measuring instrument but what is resolution the resolution of a measuring
• 83:00 - 83:30 instrument is the smallest division in that apparatus so if for instance you had a measuring cylinder and you can see every division represented 1 cm cubed we would say it has a resolution of 1 cm cubed now we estimate uncertainty slightly differently dependent on if we have an analog measuring cylinder such as a measuring cylinder or an analog thermometer or a digital measuring instrument and a digital measuring instrument gives you the numbers as a
• 83:30 - 84:00 decimal for instance it's going to have the digits actually written down there for you like your typical measuring balance if we have an analog measuring cylinder then the uncertainty will be estimated by taking plus or minus half of the smallest division on the scale so if we think back to that measuring cylinder that has a mark every 1 cm cub it's going to have an uncertainty of half of this of plus or minus 0.5 cm
• 84:00 - 84:30 cubed so now let's think about a digital measuring instrument so to estimate the uncertainty we're still going to do plus or minus half of the smallest division on the scale just that scale looks different because of course it's the digits on your reading so if for example we have a digital balance and it can take readings up to two decimal places the smallest division then would be 0.01 and so the uncertainty is going to be half of that smallest division so the
• 84:30 - 85:00 uncertainty in that case will be plus or minus 0.005 G so the mass of one mole of a substance in grounds is equal to its relative formula mass so if we had one mole of hydrogen atoms that would weigh 1 gr if we had one mole of oxygen molecules so O2 that would be equal to a mass 2 * 16
• 85:00 - 85:30 32 G finally if we think about having one mole of water the relative formula mass of water is 18 and so the mass of one mole of water is 18 G too now a lot of people get very confused about moles they think it's something really complicated and fancy but it's just a word to represent a really big number there's a word that we use all the time which has a similar job and that's dozen so we often say oh
• 85:30 - 86:00 we've got a dozen eggs or something else and everybody knows that means we have 12 eggs likewise if we had three dozen eggs we know that that's 3 * 12 36 eggs now moles do that same job except it's a really horrible number okay so if we have one mole of a substance we have 6.02 * 10^ 23 particles if we have 2 moles of a substance and we have 2 * 6.02 * 10 23
• 86:00 - 86:30 of those particles so just don't overthink this concept okay so when we talk about this 6.02 * 10 23 let's just talk about particles now that could be atoms it could be ions it could be electrons it could be anything okay it's just representing that much stuff and we call this number avagadro constant so putting this into practice quickly we may be asked how many
• 86:30 - 87:00 molecules of water are there in five moles of water well we should know that in one mole of water we have avagadro's number molecules of water we have 6.02 * 10 23 molecules of water in one mole so if we have 5 moles all we need to do is just times that by five and that ends up giving us 3.0 01 * 10 24 molecules we can calculate the mass of a
• 87:00 - 87:30 substance if we know how many moles there are so let's remember what we said the relative formula mass is equal to the mass of one mole of a substance so we can use this bag to turn it into a little equation we can say that the total mass is equal to the relative formula mass times by the number of moles and we can learn this in short hand relative formula mass is often referred to as m r and so we can say that mass equals Mr moles it's a short
• 87:30 - 88:00 Snappy way of remembering it so we might be asked to find the mass of 0.5 moles of ammonia with a relative formula mass of 17 so in this case we can use that same formula mass equals relative formula mass time moles or mass equals Mr moles so the mass will equal 7 for that relative formula mass times by the 0.5 for the moles and that gives us
• 88:00 - 88:30 8.5 G now when we have a balanced equation it shows us the number of moles that react together and the number of moles of products formed so the balanced equation gives us the ratio of moles in a reaction so this reaction here I can see that I have one mole of nitrogen reacting with three moles of hydrogen and making 2 moles of ammonia so we can use these balanced equations to work out the mass of one of
• 88:30 - 89:00 the reactants or products so long as we know the mass of one of the others let's take a look at this question here how much ammonia can be made if you have 42 gram of nitrogen and an excess of hydrogen so there's three steps to take here first let us work out the number of moles of nitrogen so we should now know that when working out the number of moles moles is
• 89:00 - 89:30 the mass divided by the relative formula mass so let's work out the relative formula mass of the nitrogen then N2 is going to be 2 * 14 because the relative atomic mass of nitrogen is 14 so the Mr of nitrogen is 28 so the moles are equal to 42 because that's the mass of nitrogen ID by 28 and that gives me 1.5 moles so now we know this we're on to
• 89:30 - 90:00 step two here we want to use the ratio to work out the number of moles of ammonia so here I can see I have one mole of nitrogen for every two moles of ammonia so this is a 1:2 ratio so to work out the number of moles of ammonia we simply need to do 2 * 1.5 and that gives us 3 moles of ammonia so now on to that last step now
• 90:00 - 90:30 we need to work out the mass of ammonia let's use the equation that states that mass equals Mr moles so we need to get that Mr or relative formula mass of ammonia of course there's one nitrogen so we've got 14 plus 3 * 1 for the three hydrogens and that gives us an MR equal to 177 so in this case here we're going to say that the mass is equal to that relative formula mass 17 * by 3 that
• 90:30 - 91:00 gives us a mass of 51 G so here the answer is 51 G of ammonia can be made if you have 42 G of nitrogen now a really useful thing we can do is we can use masses to work out the balancing numbers in an equation so to do this first work out the relative formula mass of all the substances and calculate the number of moles of each substance next work out the simplest
• 91:00 - 91:30 whole number ratio by dividing by the smallest and finally use these numbers to balance the equation now let's put this into practice we've got seven gram of nitrogen reacting with 1.5 gr of hydrogen to produce ammonia deduce the balanced equation so here I've got a little table first let's work out the Mr of nitrogen so 2 * 14 gives us 28 now let's work out the Mr of hydrogen
• 91:30 - 92:00 that's 2 * 1 which gives us two so now let's work out the number of moles and remember we do that by saying moles equals mass over the relative formula mass the Mr so for nitrogen we have 7 G / 28 which gives us 0.25 and for hydrogen we have 1.5 G / 2 which gives us 0. .75 so finally we divide by the smallest amount the smallest of those two moles
• 92:00 - 92:30 is 0.25 so we're going to divide both of these by 0.25 so 0.25 over 0.25 gives us 1 and 0.75 / 0.25 gives us 3 therefore we know that one mole of nitrogen reacts with three moles of hydrogen and we can finally use this to balance in the normal way 1 N2 means that we have two n's and 3 H2S means we have six hydrogens
• 92:30 - 93:00 therefore we must have two nh3s made to balance this equation so in a chemical reaction that has two or more reactants in all likelihood one of the reactants is going to run out before the others and we call this a limiting reactant and the remaining reactant that hasn't run out yet yet is said to be in excess now the amount of limiting reactant that we have is going to affect
• 93:00 - 93:30 the maximum mass of products that are made because once that limiting reactant runs out the reaction stops let's look at this example here 15 G of nitrogen is reacted with an excess of hydrogen calculate the maximum mass of ammonia made so the first thing we need to do here is to write a balanced equation and they may well give this to you in the exam anyway so here we have N2 + 3 H2
• 93:30 - 94:00 goes to 2 NH3 but like I say they may well give this to you in the exam so we know we have 15 G of nitrogen and we're looking at the mass of ammonium mate that means that those are the only two bits we care about in this equation I'm going to use my grid method here I've got three rows I've got Mass the relative formula mass or the Mr and the moles and I'm filling in the information I know I know that we have a mass of 15 G of nitrogen to begin with
• 94:00 - 94:30 and you may find in the exam they give you the relative formula mass of all of these substances here let's work it out though so nitrogen is 2 * 14 which is 28 so to calculate the moles we've got moles is mass divided by the relative formula mass and I like this grid method because it's just the 15 over the 28 now that gives me 0.535 7 that's the number of moles of
• 94:30 - 95:00 nitrogen we have but in this question we're not being asked about nitrogen we're being asked about ammonia and this is where that balanced equation that they will probably give you will help you here we can see we have a one to2 ratios so if we had one mole of nitrogen we have 2 moles of ammonia if we have 0.535 7 m of nitrogen we have 2 * 0.535 moles of
• 95:00 - 95:30 ammonia I'm putting this into my calculator I find that I have 1714 moles of ammonia so we're nearly there now okay just following through our grid method we need to have the relative formula mass of ammonia now again they may well give this to you in an exam question but let's work it out 14 for the nitrogen plus three * 1 for the hydrogen gives us 17 so let's fill that in here so
• 95:30 - 96:00 17 and of course we need to find the maximum Mass made and that's assuming that 1:2 ratio so if we use that number that we found and the fact that mass equals Mr moles or the relative formula masstimes moles we end up getting a maximum mass of ammonium made equal to 18.2 G and that's our answer so the method that we used there is we worked out moles of one of the things that we needed used
• 96:00 - 96:30 molar ratios and then used the new moles to calculate the mass of that final substance many chemical reactions actually take place in Solutions and therefore concentration is an important measurement because it can really represent the amount of stuff in your solution so the concentration of a solution can be measured in mass per unit volume so how many grams are in
• 96:30 - 97:00 that volume that you're looking at maybe GRS per decimeter cubed or ctim cubed now that's where this equation comes in we can calculate the concentration of a solution by taking the mass of a substance and dividing it by the volume of that solution now let's have a look at that in practice let's calculate the mass of 25 decim Cub of sodium chloride solution with a concentration of 0.1 G per decim
• 97:00 - 97:30 cubed now to solve this we'd need to rearrange the equation above and we'd find that the mass is equal to the concentration times by the volume now if we substitute these numbers in then we find that mass is equal to 0.1 * by 25 which gives us a mass equal to 2.5 g now an important tip here is is that we might find that we need to convert between ctime cubed and decimet cubed in this equation above we always want to
• 97:30 - 98:00 make sure that the concentration and volume units are consistent so they both use decimeter cubed or they both use centimet cubed and that's where this conversion can come in handy so when we want to go from centim cubed to decimet cubed we simply divide by 1,00 however when we want to go the other way when we want to go from decimeter cubed to centimeters cubed we multiply through by 100 so make sure you learn that little fact there so as well as thinking of
• 98:00 - 98:30 concentration as the mass per unit volume we can also think of it in a different way we can instead think of it as the moles per unit volume and commonly we'll be asked calculate the concentration by dividing the moles by the volume so concentration in moles per decimeter cubed equals the moles divided by the volume in decim cubed and we might even need to convert between moles per decim cubed and G per decim cubed
• 98:30 - 99:00 the way that we do that is if we have grams per decim cubed we divide by the relative formula mass or the Mr to get moles per decim cubed and if we want to go in the other direction we simply do the opposite so the moles per desm cubed times by that relative formula mass gives us the concentration in GRS per Destin cubed now if the volume of two solutions react together completely and we know the concentration of one of the two solutions we can calculate the
• 99:00 - 99:30 concentration of the other solution let's have a look at this in a calculation in a reaction 50 cm cubed of 0.2 moles per desm cubed of sodium hydroxide reacts completely with 25 cm cubed of hydrochloric acid here we asked work out the concentration of hydrochloric acid so we know two volume and one concentration and we need to find the second concentration so what we need here is a
• 99:30 - 100:00 balanced equation now they may give this to us in the exam here it's going to be NaOH plus HCL makes NAC plus H2O so I'm going to use the grid method here I know that we have concentrations volumes and we can use those to calculate the moles so starting off here looking at the sodium hydroxide side we're told we have 50 cm cubed now because concentrations in moles per decim cubed we need to convert this to
• 100:00 - 100:30 decimet cubed and we do that simply by dividing through by 1,000 so 50 divided 1,000 gives us a volume of 0.05 decim cubed we're told that the concentration is 0.2 m per cubed so we fill that into our little grid and here we can see we've got got two out of three things in our grid so we can use that to calculate the third thing the moles now we know that concentration is
• 100:30 - 101:00 moles over volume so therefore moles is equal to concentration times the volume and so 0.2 * by 0.05 gives us 0.01 moles of sodium hydroxide now the question wants the concentration of hydrochloric acid so let's put a little question mark there in our Rd so we can see here that between that sodium hydroxide and the HCL we have a
• 101:00 - 101:30 1: one ratio means that we have the same number of moles of HCL as we do NaOH so we've got 0.01 moles of HCL now we're told that we have 25 cm cubed of hydrochloric acid we've got the same problem here we need to convert that to decimeter cubed and we do that again just by dividing by 1,000 so that gives us 0.025 decim cubed now finally we're just going to
• 101:30 - 102:00 use the equation that we know for concentration concentration equals the number of moles divided by the volume in this case that's going to be 0.01 / 0.025 and that gives us a concentration for the HCL equal to 0.4 moles per decim cubed and that there is our answer to this question the percentage yield is calculated by taking the mass of the
• 102:00 - 102:30 product actually made and dividing it by the maximum theoretical mass of the product and timesing it by 100 and when we say maximum theoretical Mass that's if the reaction went perfectly and all of the atoms from the reactants went on into making our product so in reality as you can imagine most reaction don't actually have 100% percentage yield now we need to know why this is the case and
• 102:30 - 103:00 this is a very common type of question you may see so one of the reasons is that the reaction may actually be reversible and so it's never actually going to completion another reason is that some of the products may be lost and it's separated from the reaction mixture imagine you're pouring it from one container to another there'll always be some residue left it's it's never a perfect process and every time that happens you're losing some of the mixture and therefore your
• 103:00 - 103:30 yield the final reason you need to know about is that some of the reactants May react in different ways than you expected so other reactions might be taking place at the same time so let's have a look at what this might look like in practice they may tell us that in an experiment the theoretical yield is 8 G only 6.2 G is act made so they may ask us to calculate the percentage yield in this scenario so using our equation the
• 103:30 - 104:00 percentage yield will be equal to 6.2 divided by 8.0 and timesing it by 100 and that gives us a percentage yield equal to 77.5% occasionally in a more challenging percentage yield question they won't tell you what that theoretical mass is instead they will want you to work out yourself so let's have a look at how we'll do that so in this question we're asked to calculate theoretical mass of
• 104:00 - 104:30 magnesium oxide made when 12 G of magnesium reacts with an excess of oxygen so here we've been given the balanced equation and we're going to need to use our molar ratios and our mole equation to work this out so we know that we have 12 G of magnesium and we need to find the theoretical mass of magnesium oxide so those are the only two parts of this balanced equation that we care about let's use our grid method so here I've got my mass my relative formula mass and
• 104:30 - 105:00 my moles I know that I can fill in that I have 12 G of magnesium and for the Mr or relative formula mass of the Magnesium well it's just going to be 24 because it's all on its own here now we can work out the number of moles of magnesium so moles is mass over the Mr so that's going to be 12 over 24 and that ends up giving us 0.5 so we know that we have 0.5 moles of
• 105:00 - 105:30 magnesium here you can see we have a 2:2 or a 1:1 ratio between the magnesium and the magnesium oxide so we also have 0.5 moles of magnesium oxide so we need to get the Mr of magnesium oxide next it's going to be the 24 for the magnesium plus the 16 for the oxygen which gives us 40 so let's fill that into our grid
• 105:30 - 106:00 now and now we have enough information we've got two out of three to work out what that final number is the mass in this case we need so the mass is going to equal Mr moles the Mr * moles which will be 40 * 0.5 which gives us 20 so in this case the theoretical mass of magnesium oxide is 20 grand and potentially this could then flow on into a percentage yield question where they
• 106:00 - 106:30 tell you how much is actually made and you use that number together with the answer for our theoretical Mass to calculate percentage yield atom economy is a measure of the amount of starting materials that end up as useful products we've got this formula here that we can use to calculate it in an exam the atom economy equals the relative formula mass of the desired product divided by the sum of the relative formula masses of all of the
• 106:30 - 107:00 products or you might have learned it as reactants but it works out exactly the same and then we times it by 100 to make this a percentage in chemistry we'll see that high atom economy is very important and we need to know why so there's two reasons you need to be able to talk about the first one is sustainable development now what this means is if we have a High atom economy we're going to have less waste being produced and fewer natural resources being needed another reason is economic
• 107:00 - 107:30 reasons imagine the more things you make that you don't want to make you've got to spend money getting rid of disposing of you might even need more reactants in order to make stuff you don't even need so it makes good economic sense to have a high atom economy now top tip here if we have a reaction that only has one product we should know that the atom economy is 100% we don't even need to do a
• 107:30 - 108:00 calculation now let's put this into practice with this example so this following reaction is used to make sodium chloride calculate the atom economy of the reaction the first thing we always need to do is to identify what was the useful product and here the question has clearly stated that we're making sodium chloride so that is our useful product let's remind ourselves of the formula for atom economy so atom economy is the rfm of
• 108:00 - 108:30 our useful or our desired product divided by the rfm of all of the products or you may use the reactants they work out the same either way and then finally we times it by 100 to change it into a percentage so in this question we have said that the useful product the desired product is sodium Chlor right so in this case you can see that we're given all of the information that we need to be able to work out that relative formula mass
• 108:30 - 109:00 of sodium chloride so that's going to be 23 + 35.5 which gives us 58.5 now that of course is our desired product but we want all of the products here we can see we've also got H2O as well so let's work out the relative formula mass of that as well that's going to be 2 * 1 for the hydrogen and then plus the 16 for the oxygen and that gives us 18 for the relative formula
• 109:00 - 109:30 mass of water so let's substitute these numbers into our equation here then so atom economy is going to be equal to that 58.5 of that desired product divided by 58.5 + 18 and then we Times by 100 100 to change it to a percentage so putting that into the calculator I end up getting
• 109:30 - 110:00 76.47% and that is our atom economy of this reaction in chemistry when we talk about gases there's a really interesting fact it doesn't matter what gas you have if you have the same amount of it so you have one mole of it it will occupy the same volume under room temperature temp and pressure now that has a very specific value that you need to learn so at room temperature and pressure one mole of any
• 110:00 - 110:30 gas whatever the gas occupies 24 decim cubed or 24,000 cm cubed now room temperature we say is 20° C and room pressure is one atmosphere pressure now under the same temperature and pressure say one mole of carbon dioxide gas will occupy the exact same volume as one mole of chlorine gas or oxygen gas or ammonia gas or any
• 110:30 - 111:00 gas let's look at this example let's calculate the volume of gas produced in decim Cube when 10 G of sodium reacts with an excess of HCL at room temperature and pressure so we're given this balanced equation here we know two sodiums plus two hcl's makes two nac's plus H2 now I like to use the grid method here I'm going to draw on three rows I'm going to have the mass the relative
• 111:00 - 111:30 formula mass and the moles and the point of this if I know the mass of sodium 10 G I can work out the number of moles of sodium I have so I know that I have 10 gram of sodium I know that it's relative atomic mass in this case is just 23 and I know the formula that says moles is mass over the relative formula mass so I can calculate the number of moles of sodium by doing 10 /
• 111:30 - 112:00 23 that gives me 0.435 mol of sodium so now in this question we're being asked to find the volume of gas now the only gas that's produced here is the hydrogen gas and if I look at the relationship between the sodium and the hydrogen I can see I have a 2:1 ratio that means I have half as many moles of hydrogen than I did sodium so the moles of hydrogen will
• 112:00 - 112:30 simply be that 0.435 we just calculated divided by 2 now that gives me for the moles of hydrogen 02175 moles now that we know how many moles we can use the fact that one mole occupies 24 decim cubed so in this case 02175 * 24 is going to be the volume
• 112:30 - 113:00 occupied in this case that gives me a volume of 5.22 decim cubed so always make sure that you know that you take the number of moles you have and times it by that 24 decim cubed or 24,000 cm cubed if that's what the question wanted [Music]
• 113:00 - 113:30 [Music] [Music] [Music]
• 113:30 - 114:00 [Music] what when metals react with oxygen they form metal oxides this is called an oxidation reaction so an example here is
• 114:00 - 114:30 when magnesium reacts with oxygen forming magnesium oxide now reduction on the other hand happens when we remove oxygen from a metal oxide for example the reaction between copper oxide with carbon to form copper and carbon dioxide now we can summarize this so oxidation will involve a gain of oxygen and reduction will involve a loss of oxygen now Metals will generally react
• 114:30 - 115:00 with water to form a metal hydroxide and hydrogen now this General reaction is one that we need to learn so metal plus water makes a metal hydroxide and hydrogen so if we have a look at an example we could see sodium plus water makes sodium hydroxide and hydrogen it's quite common we be asked about the observations of this reaction and I always find the state symbols of the key to this we will notice the solid metal
• 115:00 - 115:30 is going to dissolve and we're going to see a fizzing or effervescence as that hydrogen gas is formed also if we were to add Universal indicators to the solution we will notice it will turn purple and the reason for this is that hydroxides are alkaly and alkaly of course course turn Universal indicator purple now generally speaking as the metals get more reactive they will react more quickly and
• 115:30 - 116:00 vigorously an example of this is potassium which is a very reactive metal and when we add it to water we're going to see it Fizz move quickly around the surface propelled by all those bubbles and it's going to heat up to the point where we see a lilac flame and sometimes even a small explosion now of course we've already said haven't we that when a metal does react with water we form a metal hydroxide and hydrogen however we also want to know
• 116:00 - 116:30 what that reaction is going to look like so we should recognize that the reaction of potassium and sodium with water is going to be a violent reaction lithium and calcium is not going to be violent but it's still going to be very quick magnesium on the other hand is very slowly zinc you'll usually see no reaction happen iron is going to rust very slowly and for copper we will see no reaction so what we can deduce from
• 116:30 - 117:00 those results is that those violent quick reactions show us the most reactive metals and where we have no reaction we have the least reative Metals now let's look at the reactions of metals with dilute acids so generally Metals will react with a dilute acid to form a salt and hydrogen I really like the acronym mash for this so metal and acid make salt and hydrogen just remember Mash so here's an example if we have
• 117:00 - 117:30 sodium plus hydrochloric acid we'll make sodium chloride and hydrogen gas now again we might be asked about our observations here again look at those state symbols we can see that the metal is going to disappear or dissolve and over on the other side you can see the formation of this gas so we're going to see a fizzing or an effervescence observed as that gas is formed again the same rule applies and
• 117:30 - 118:00 more reactive metals will react more quickly and vigorously however we've got this extra little Factor know here if a metal is below hydrogen on the reactivity series it will not react with dilute acids and the reason for this is that the metal is going to displace the hydrogen in the acid normally so if the hydrogen is at more reactive then the metal can't displace it the hydrogen wall stay where it is so let's compare the speed of the
• 118:00 - 118:30 reaction for these different Metals for potassium sodium and lithium we're going to get a violent reaction with the acid for calcium and magnesium we're going to see a vigorous reaction for both zinc and ion we're going to see a slower reaction take place and for copper we're going to observe no reaction now again the scale of these reactions can tell us the order of reactivity and so potassium is going to be the most reactive of those metals and
• 118:30 - 119:00 copper will be the least reactive because it's the only one where there's no reaction a common question we'll see in the exam is we may be asked to place Metals in order of their reactivity based on experimental results so this could be either looking at the reaction with water or acid so here I'm going to show you the observations when four different metals a b c and d react with water so see here C has no bubbles D has the most Bubbles and a and b have
• 119:00 - 119:30 somewhere in the middle so we can use this information to see that D must be the most reactive as the most bubbles are released a is going to be the second most reactive as the second most bubbles are released and C must be the least reactive as no bubbles are released at all there must be no reaction taking place so we could use those observations to conclude the order of reactivity so D is
• 119:30 - 120:00 the most reactive followed by a b and then C metals can be arranged in order of their reactivity in the reactivity series we'll often find included in these tables the elements carbon and hydrogen although they're not Metals it can be very helpful to compare the reactivity of the metals to these two ele elements now the reactivity of a metal is related to how easily it's going to form a positive ion so the more easily
• 120:00 - 120:30 it's going to form that positive metal ion the more reactive our metal so here you can see potassium is the most reactive metal followed by sodium lithium calcium magnesium then we have carbon the non-metal then zinc oxygen then hydrogen another nonmetal and then finally copper and gold we can use what we know about the reactivity series to look at
• 120:30 - 121:00 displacement reactions so in a displacement reaction a more reactive element can displace a less reactive element in a compound for example we know from the reactivity theories that magnesium is more reactive than ion so as a result magnesium is able to displace ion from an ion compound an example of this is if we had magnesium reacting with Ion 2 sulfate so an ion compound what's going to happen
• 121:00 - 121:30 is that magnesium will displace the ion to form ion plus magnesium sulfate so there we have it we have our displacement reaction a common question in the exam is about Metal extraction so we should know that unreactive Metals For example gold or silver will be found in the earth as simply pure Metals however it's not normally that easy most metals aren't just found on
• 121:30 - 122:00 their own as pure Metals instead they're found as compounds in rock and we're going to need to extract them from the rock using different chemical reactions and this is where the reactivity series and including something like carbon can be so helpful and this is because Metals less reactive than carbon can be extracted from their oxides by reducing them with carbon we already know about oxidation and reduction we know that oxidation
• 122:00 - 122:30 involves the gain of oxygen and reduction is the loss of oxygen however we also need to know about them in terms of electrons we need to know that oxidation is the loss of electrons and reduction is the gain of electrons we've got a great little pneumonic here to help us out so so oil rake oxidation is loss of electrons and reduction is gain of electrons so let's have a look at this
• 122:30 - 123:00 little summary here to help us know the difference between oxidation and reduction so oxidation is the gain of oxygen I always think that kind of makes sense oxidation you're gaining oxygen reduction on the other hand means getting rid of something you're getting rid of that oxygen however in terms of electrons there's nothing about the name oxidation that links with electrons really does it and that's where oil rig comes in oxidation is the loss of electrons
• 123:00 - 123:30 reduction is the gain of electrons when we have a displacement reaction we can instead write them in terms of ions let's have a look at this example where the more reactive ion reacts with copper sulfate a displacement reaction takes place and copper and iron sulfate are formed well this can be written as an ionic equation you might be wondering how we can do this well what we do is if we look at all of the aquous substances
• 123:30 - 124:00 in our earlier equation if we break those down into their respective ions that's going to be our first step here so we've got the Fe which was a solid we left that as it was the copper sulfate which was aquous we've broken down into cu2 plus and S so42 minus it's important we know the formula for that sulfate ion then on the other side copper which is as solid we just leave as it is but the aquous ion sulfate we're going to break down again into its respective ions in
• 124:00 - 124:30 this case into fe2+ and S so42 minus so this is where the clever bit comes in this is our next step when an ion is on both sides of an equation we have a spectator ion and spectator ions are doing absolutely nothing so we can just count cancel them so we're just going to strike through the sulfate on each side leaving us with our ionic equation which is Fe solid
• 124:30 - 125:00 plus cu2 plus aquous makes c s plus fe2 plus aquous and that is it that's how we get our ionic equations so when we have an ionic equation like this we can actually split it into two half equ equations let's start off by just looking at ion so here the Fe solid has lost two
• 125:00 - 125:30 electrons to become the fe2+ ion so we can write this as Fe goes to fe2+ plus 2 e minus and because the ion has lost two electrons remember oil rig we could say oxidation is the loss of electrons so we know that the ion has oxidized on the other side if we have a look at what's left here so the copper part of the
• 125:30 - 126:00 equation we can see that those copper 2+ ions must have gained two electrons to just become copper so our half equation will be cu2+ plus two electrons makes copper so here we have another half equation and we can see that the cu2+ ion has gained two electrons now this means that the cu2+ has been reduced again using our oil rig reduction is gain of
• 126:00 - 126:30 electrons in chemistry it's very important that we know about three main acids hydrochloric acid sulfuric acid and nitric acid we need to know that the formula for hydrochloric acid is HCl and we need to know that when that is put into water into solution it will dissociate to form a H+ and cl minus ions with sulfuric acid we should recognize that that has the formula of
• 126:30 - 127:00 h2so4 and again when that's in solution it forms H+ and S so42 minus ions nitric acid on the other hand has the formula H3 and that will dissociate to give the ions of H+ and NO3 minus now you might be wondering why this is so important whenever we need to write four formulas for certain things we need to know these ions to get the formula correct we also need to know the acids and metals can
• 127:00 - 127:30 react so Metals will generally react with dilute acids to form a salt and hydrogen that's the mash acronym so metal and acid makes a salt and hydrogen now we've looked at how metals react with acids we also need to know what type of reaction this is we need to know that this is an example of a Redux reaction but what is a Redux reaction we
• 127:30 - 128:00 need to know the definition that a Redux reaction is a reaction where both reduction and oxidation occur at the same time so let's look at an example to put this into action the ionic equation for the reaction of zinc with hydrochic acid is 2 H+ plus ZN for zinc goes to zn2+ plus H2 now if we split this into two half equations let's start by having a look
• 128:00 - 128:30 at the 2 H+ goes to H2 now in order for that to happen it must have gained two of these negative electrons using oil rig we know the oxidation is lost reduction is gain of electrons so here we can see that the H+ has gained electrons and been reduced now on the other hand if we look at the zinc we know zinc goes to zinc 2+ in order to form the 2 plus ion we
• 128:30 - 129:00 must have lost two electrons and so those zinc atoms have lost electrons and using oil rig again oxidation is lost we can say that the zinc atoms have been oxidized so we can see here that in this reaction between zinc and hydrochloric acid both reduction and oxidation have a occurred in that same reaction and so it's a Redux reaction we also need to learn about how acids can be neutralized and produce
• 129:00 - 129:30 salts so there's a few different reactions that can take place which are neutralization reactions so acids can be neutralized by alkaly which are soluble metal hydroxides like sodium hydroxide by bases such as insoluble metal hydroxides and methal oxides like calcium oxide and metal carbonates like calcium carbonate so it's very important we get memorizing these two reactions so we need to know acid plus a
• 129:30 - 130:00 metal carbonate makes a salt plus water plus carbon dioxide and we also need to know that an acid plus a metal hydroxide or an oxide makes a salt and water now different acids are going to produce different types of salt this is really important we know this so that we name correct salt in a question so if we ever have hydrochloric acid in our reaction we're going to make a chloride salt sulfuric acid is going to form a
• 130:00 - 130:30 sulfate salt and nitric acid is going to form a nitrate salt so let's have a look at what this is like in practice so if we had hydrochloric acid and it reacted with calcium carbonate now we just said that when an acid reacts with a carbonate we make a salt plus water plus carbon dioxide but what salt is it to do that we need to look at what the metal ion is so in this
• 130:30 - 131:00 case that's ca2+ the salt ion in this case is going to be CL minus because we have hydrochloric acid and so therefore the salt will be calcium chloride now we can work out the formula of this by doing our crisscross method okay so I'm going to crisscross these charges and that is going to give me CA cl2 and the whole reason for this is
• 131:00 - 131:30 that we need to have the same charge of both the positive and the negative so we need two of the negative CLS to balance with the ca2+ so now if we have sulfuric acid reacting with sodium hydroxide the methyl ion is na+ the salt ion is going to come from the sulfuric acid that's the s so42 minus the sulfate ion and that's going to make us sodium sulfate let's use the crisscross method again and here I can see I have
• 131:30 - 132:00 na2 so4 now finally if we have nitric acid with magnesium hydroxide the methyl ion is mg2+ the salt ion is NO3 minus and the salt in this case is going to give us magnesium nitrate because we have that nitrate ion from the nitric acid let's use our crisscross method again and here I can see the formula will be mg NO3 and the whole thing is
• 132:00 - 132:30 times two so it's mg3 in Brackets with a little two down below a common question is them asking us how we make a soluble salt so generally speaking we can make a soluble salt from an acid by reacting them with a solid in soluble substance such as a metal a metal oxide or carbonate so what we do is we add this solid in soluble substance to the acid
• 132:30 - 133:00 and then we should observe a reaction taking place when this reaction stops we're going to take this solution and we're going to filter it and any of that excess solid will be filtered off staying in the filter paper and going through the filter paper into our beer or our flask is going to be our salt solution now that we've removed the solid and we just have that salt solution we can do crystallization to
• 133:00 - 133:30 remove the water in that solution leaving us just without our salt now let's look at our experiment this is what we want to say if we're asked to describe how to prepare a pure dry sample of a soluble salt from an insoluble base so the first thing we need to do is to choose the correct acid and base required to make the desired salt so maybe the question has asked you to make a pure dry sample of copper sulfate
• 133:30 - 134:00 perhaps then because of the sulfate you're going to use sulfuric acid and because we need the copper perhaps we could use copper oxide so now you're going to add some of that dilute acid to a flask we're going to heat it gently using a bunson burner you may have used a water bath to do this and then we're going to add a small amount of the base and stir it we're going to continue to add the base until the reaction is going to stop
• 134:00 - 134:30 and the importance of this is that this means the base must be in excess at this point now when the base is in excess because it's a solid we can very easily remove it by using filtration so we filter it and then we're going to pour the remaining Solution that's gone through the filter paper into an evaporating Basin now this solution is going to contain the salt as well as water now we're going to use an electric
• 134:30 - 135:00 heater or a water bath to gently evaporate this water away this is going to leave us with crystals of the salt and so this is our method make sure that you're familiar with this method you can describe it and most importantly make sure you know that you need to choose the correct acid and base dependent on the salt that the question's asking you about we need to know what ions the acid produce in aquous solution so this is
• 135:00 - 135:30 such an easy Mark if you just learn this one fat so acids produce hydrogen ions we also need to know what ions alkalides produce in aquous solution this is again such an easy Mark so please do learn it it's hydroxide ions we also need to know how we can measure the pH of the solution so we could use one of two methods the first method is universal indicator we should know that when we add Universal indicated to a solution of
• 135:30 - 136:00 acid or Alkali or even a neutral solution we will observe a color change however this is quite an approximate answer you know how you're not going to get a number or anything are you you're just going to get a rough idea if it's an acid or an Alkali or neutral now a pH probe is the other method we could use and this is great because this is an exact method this will tell us the pH of our solution so the pH scale ranges from zero all the way up to 14 and if we were
• 136:00 - 136:30 to add Universal indicator we can see this color scale we can see that when we have a strong acid so somewhere around the range of zero or one the universal indicator will change to red when it's neutral the universal indicator will show you a green color and of course when it's very alkaline say 14 pH we will see a dark purple color so we should know of course that
• 136:30 - 137:00 in terms of pH a neutral solution has a pH of seven an acid has a pH less than seven and an Alkali will be more than seven so neutralization is the reaction that takes place between an acid and an Alkali and we need to know the ionic equation that takes place when we have a neutralization reaction this is such a simple one when you think about it we've got the H+ from the acid plus the O minus from The Alkali which go together
• 137:00 - 137:30 and make water so make sure you learn this titration are carried out to work out the concentration of an unknown solution in a reaction between an acid and an Alkali so we need to know how do we carry out titration where an acid is added to an alkali and vice versa as well let's look at it this way first so first off let's use a pet to add an amount of alkali to a conical flask in
• 137:30 - 138:00 this case let's say 25 cm cubed now let's add a few drops of indicator to this alkaline let's put our flask on a white tile the reason for this is when we get a color change we can see it much more clearly up against this white tile so now we're going to get a buet and we're going to fill that buet with acid we're going to look at that buet and we're going to record the starting volume of acid on that
• 138:00 - 138:30 buet then we're going to turn the tap and we're going to add the acid drop drop drop to that Alkali it's very important that we just do it drop by drop because then we're going to get the closest volume as possible it's needed to neutralize that alkaline so after every drop we're going to SWR the solution to ensure that it mixes and we're going to keep doing those drops until we observe a color change and then we're going to stop we're going to record the final
• 138:30 - 139:00 volume and then we're going to repeat until we get concordant tighter so that brings us to a question really doesn't it what is the Tighter and the tighter is the difference between the final and the initial volume reading on the buet and that's why you need to take that reading before before you do any drops and after we've had that color change and what are concordant results as well those are tighs within
• 139:00 - 139:30 0.10 cm cubed of each other so for example let's find the mean titer of the following results here we can see in this table we need to work out the tighter we do this by subtracting the initial volume from the final volume and then we fill them in the table now we need to look for which of these results are concordant so within that 0.10 cm cubed so the first thing we need to know is that we ignored the
• 139:30 - 140:00 rough Tighter and the reason for this is that just use that to get a ballp part as to how much we're going to need to add we don't actually use it in our measurements so then if we look at 1 2 and three I can see that one and two are concordant they're within that 0.10 so the mean tighter is going to be 1935 + 19.45 / 2 and that gives us 19.4 cm cubed now these readings should
• 140:00 - 140:30 always be to two decimal places and they must always end in a zero or a five so make sure that all of your results are filled in in that way so we can use the results of a titration to work out the concentration of the unknown solution so the way that we do this is we write a balanced equation then we calculate how many moles we have of the known substance the
• 140:30 - 141:00 substance whose volume and concentration we know and then we use a balanced equation and the ratios between all of the different substances to work out the moles of the unknown solution finally we use those moles that we just worked out and the mean tighter so that mean volume to work out the concentration of the unknown solution so let's have a look at this example so in a titration 50 cm cubed of
• 141:00 - 141:30 0.1 mole per decim cubed of potassium hydroxide is exactly neutralized by 25 cm cubed of hydrochloric acid here we need to work out the concentration of hydrochloric acid so the very first thing we need to do is to write a balanced equation so potassium hydroxide is K plus hydrochloric acid which is HCl and they'll react together to make potassium chloride KCl plus H2O and that
• 141:30 - 142:00 is already balanced actually so we're in Lu so I like to use this grid method so I'm going to do three rows one for concentration one for volume and one for moles now looking at the information for potassium hydroxide I know both the volume and the concentration but notice the units are different we've got cm cubed for the volume let's divide that by 1000 to get the volume also in
• 142:00 - 142:30 terms of decimet cubed so here I'm going to write it all in into this little grid that I've got the volume of pottassium hydroxide is 0.05 decim cubed and the concentration is 0.1 mol per desm cubed so I need to work out the moles here I know that moles is equal to the concentration times by the volume so here I'm going to do 0.1 * by
• 142:30 - 143:00 0.05 and that ends up giving me 0.05 moles so now I've got the number of moles of pottassium hydroxide I can use my balanced equation and the molar ratio between potassium hydroxide and hydrochloric acid to work out how many moles there are of hydrochloric acid so here I can see that there's no balancing numbers in front of them which means there's one of each so I've got a 1: one ratio between the
• 143:00 - 143:30 two so we have 0.5 moles of HCL so looking back at the information in the question I also know I have 25 cm cubed of hydrochic acid now let's get that into decim cubed again as well by dividing through by 1,000 so this means I now also know the volume of hydrochloric acid and I can fill it into my grid again I've got
• 143:30 - 144:00 0.025 decim cubed of HCL so now I can see I've got two things that I know about HCL and I'm being asked to find the third thing that concentration so how do we do that we just use the same equation we did earlier that moles equals concentration time volume but we rearranged this equation to find that concentration equals moles over volume so in this case then substituting
• 144:00 - 144:30 in the numbers I have 0.05 / 0.025 and popping that into the calculator I find that I have a concentration equal to 0.2 and that's m per decim cubed and that there is our answer so really the key thing in a question like this is to lay out all of your
• 144:30 - 145:00 information in this grid method is always my favorite way of doing it so that you've got all of the information you need to answer the question we need to know the difference between strong acids and weak acids so a strong acid is an acid that is completely ionized in aquous solution well alternatively we may say that it's an acid fully dissociates in aquous solution so examples of strong acids are the classic hydrochloric acid nitric
• 145:00 - 145:30 acid and sulfuric acid and we can show the dissociation this ionization using an equation so hydrochloric acid for instance becomes H+ and cl minus now on the other hand a weak acid is an acid that's only partially ionized or dissociated in aquous solution examples of weak acids include ethanolic acid citric acid which we find in oranges and other citrus fruit and
• 145:30 - 146:00 carbonic acid so we can show this using this equation so here we've got methanolic acid HC and then that becomes H+ and HC minus however we didn't just do our normal Arrow instead we've done this special Arrow which shows that this is a reversible reaction concentration tells us how much of a substance is dissolved in water so we need to know the difference between a concentrated acid and a dilute acid what
• 146:00 - 146:30 if we have a concentrated acid we've got a lot of acid in a small volume of water so so we could say that we've got a lot of acid per unit volume now a dilute acid on the other hand has a small amount of acid in that same amount of water so we could say that a dilute acid has a small amount of acid per unit volume and these diagrams demonstrate this you can see that there are fewer acid molecules
• 146:30 - 147:00 in that dilute acid in that same space compared to the concentrated acid now at any given concentration a stronger acid is going to result in a lower pH now we need to know that as the hydrogen ion concentration increases by by a factor of 10 the pH will actually decrease by one unit now this can sound a bit confusing so the best way to see this is to have a look at an
• 147:00 - 147:30 example so if we're told that the hydrogen ion concentration of an acid of ph3 increases from 0.01 moles per decim cubed to 0.1 moles per decim cubed and we're asked what is the new pH of the acid here we need to recognize that the concentration is inreased by factor of 10 so using our fact above we can say that the pH will decrease by one unit now if it started at three if we
• 147:30 - 148:00 subtract one from that we get two and so therefore the pH is now two and this is what we mean when we make that statement as the hydrogen ion concentration increases by a factor of 10 the pH decreases by one unit electrolysis is the process of splitting up an ionic compound using electricity when an iic compound such as sodium chloride is melted or dissolved in water its ions are free to move
• 148:00 - 148:30 through the liquid or solution now this now this liquid or solution is able to conduct electricity and we call this substance an electrolyte now if we place a negatively charged cathode and a positively charged anode into this electrolyte we're going to see an electric current flow the reason for this is the positive ions from our ionic compound are going to be attracted to
• 148:30 - 149:00 the negative electrode the cathode and the negative ions from that solution or molten liquid are going to be attracted to the positive electrode the anode and this therefore completes the circuit now this process of separating the positive and the negative ions is called electrolysis positive ions end up gaining electrons from the negative cathode now remember oxidation is lost reduction is gained so we can say that
• 149:00 - 149:30 those positive ions have been reduced at the cathode now the negative ions lose electrons at the anode again remembering oil rig oxidation is loss of electrons we can see that those negative ions get oxidized at the anode now if we have a look at molten lead bromide as an electrolyte we can see that at that negative cathode the lead ions gain electrons forming lead we can see this
• 149:30 - 150:00 using the ionic equation of pb2+ plus two electrons makes PB so we've got a reduction taking place there whereas at the anode the BR minus loses electrons producing bromine again we can see this using our ionic equation so 2 BR minus makes br2 plus 2 e minus so here we can see because the BR minus has lost two electrons it has been
• 150:00 - 150:30 oxidized now when we look at extracting Metals some metals can be extracted from molten compounds using the process of electrolysis now we generally use electrolysis if the metal is more reactive than carbon or if it's likely to react with the carbon so the classic example that you need to know about is the fact that aluminium is extracted from aluminium oxide by this process of
• 150:30 - 151:00 electrolysis so we take the following steps we take out aluminium oxide and we mix it with a substance called prolite we heat up this mixture until it melts and this mixture that we get is electrolyzed using graphite electrodes now we're going to see alumin inium forming at the negative cathode and oxygen forming at the anode as you can see from the diagram
• 151:00 - 151:30 when that aluminium gets made at the cathode it's molten still and what happens is we can simply drain it away via a little tap and remove it from this system and use it for whatever purposes we need it for we need to be able to explain what's happening in terms of the ionic equations so we know that at the cathode aluminium ions gain electrons forming aluminium atoms therefore due to oil rig reduction is gain we can say that the aluminium
• 151:30 - 152:00 ions get reduced and the ionic equation here is a3+ plus 3 e minus makes a l so over at the cathode the negative electrode the oxide ions will lose electrons and they're going to form oxygen gas now of course using oil rig oxidation is loss of electrons we can say that those oxide ions have been oxidized so the ionic equation we need to know here is
• 152:00 - 152:30 that 202 minus makes O2 plus 4 electrons the overall reaction taking place here is aluminium oxide becoming aluminium plus oxygen now there are a couple of very important questions they can ask us about this electrolysis of aluminium ox side the first question that we'll see come up a lot is why is a mixture used for the electrolyte why do we mix together the aluminium oxide with the calite so the reason for this is that
• 152:30 - 153:00 this process lowers the melting temperature and reduces the cost as less energy is needed to melt this compared to aluminium oxide on its own the second question we'll see a lot is why must the positive electrode the anode be continually replaced so the reason for this is oxygen is formed at the anodes and this will react with those carbon graphite anodes and make carbon dioxide now this means that the
• 153:00 - 153:30 anodes are oxidized and will need replacing so we need to know about electrolysis of aquous solutions too so electrolyzing aquous Solutions can be beneficial over molten substances because less energy is required you don't need to melt down the substance simply dissolve it however working out the products is a bit harder because the water in that solution is also going to break down so that means as well as
• 153:30 - 154:00 having the ions from your ionic substance you're also going to have the H+ and the O minus from the water now the products formed is going to depend on the relative reactivity of the elements so over at that negative cathode hydrogen is going to be produced if the is more reactive than hydrogen over at the anod though a little bit trickier here if haly ions are present like chloride or bromide the hallogen will be
• 154:00 - 154:30 produced however if no halide ions are present then oxygen will be produced we need to know the products of electrolysis of the following Solutions so sodium chloride sodium sulfate copper chloride and copper sulfate so let's have a look at how we work out what the products are so with sodium chloride the positive ions in a solution of that substance will be na+ and H+ from the water now
• 154:30 - 155:00 the negative ions will be CL minus and O minus from the water now in terms of the products because sodium is more reactive than hydrogen hydrogen will be formed at the cathode whereas over at the anode because halide ions are present here chlorine is going to be produced now in this case we also see the na+ coming together with the O minus to also form sodium hydroxide now let's look at sodium
• 155:00 - 155:30 sulfate now again sodium is more reactive than hydrogen so hydrogen will be formed at cathode now in terms of the negative ions there isn't a hogen present as a result because there's no halide ion oxygen is going to form at the anode now for copper chloride we have the positive ions of cu2+ and H+ now copper is actually less reactive than hydrogen therefore we're going to form
• 155:30 - 156:00 Copper at the cathode now for the negative ions you can see we have a halide ion present so we're going to make a hogen at the anode we make chlorine now for copper sulfate again we've got cu2+ versus H+ hydrogen is more reative than copper so copper is formed and out of the negative ions we have the sulfate I and the hydroxide ion neither of those are the halides so
• 156:00 - 156:30 therefore we're going to form oxygen at the anode and there we have it make sure that you give those a good practice and that you can recognize what's made in those cases now when doing the electrolysis of acous solutions we also need to know the ionic equations that happen at the cathode when forming hydrogen and at the anode when forming oxygen so at the cathode when we form the hydrogen this one's quite simple really we take the
• 156:30 - 157:00 two H+ ions plus two electrons and you make H2 now here because the H+ ions have gained electrons we have reduction taking place again we're using our oil rig now over at the anode though is a little bit trickier because this is all about hydroxide ions so you can learn one of these two reactions you can either learn that 4 oh minus goes to O2 plus 2 H2O plus four electrons or you
• 157:00 - 157:30 could simply have those electrons on the other side and instead think of it as 4 o minus lose four electrons so minus four electrons to make O2 plus 2 H2O and notice these are both the same and here you can see the hydroxide has lost electrons therefore it has been oxidized so make sure that you get learning these ionic equations [Music]
• 157:30 - 158:00 [Music] [Music]
• 158:00 - 158:30 [Music] energy is always conserved in chemical
• 158:30 - 159:00 reactions what this means is we can never create energy or destroy it in fact the total energy in the universe before a reaction takes place is exactly the same as the total energy in the universe after the reaction takes place so if we have a chemical reaction which is transferring energy to the surroundings such as a combustion reaction then the molecules of the products produced must have less energy
• 159:00 - 159:30 than the reactants did before that reaction now this deficit of energy between the products and the reactants should equal the energy that's been given out to the surroundings now when we talking about energy transfers we've got two different types of chemical reactions there's exothermic reaction and endothermic reactions let's start off by having a look in a bit more detail at exothermic reactions so we need to know what is an
• 159:30 - 160:00 exothermic reaction this is a reaction which transfers energy to the surroundings and we see the temperature of the surroundings increase if we break down that word exothermic thermic is referring to heat energy just like the word thermometer and EXO is referring to energy exiting and that's one of my top tips as you can see here so we've got some examples of exothermic reactions that we should be
• 160:00 - 160:30 familiar with combustion is an example oxidation reactions and also neutralization as well as just having everyday uses such as hand warmers that we might use in our coats in the winter and also self-heating cans which would be great if we're going out camping and we just want a way to get hot meal in us so for exothermic reactions make sure you know those examples and make sure you understand that definition of an exothermic
• 160:30 - 161:00 reaction now let's have a look at endothermic reactions we need to know that endothermic reactions are reactions which take in energy from the surroundings and so the temperature of the surroundings decreases again if we deconstruct that word we've got thermic which is refer referring to that heat energy and the Endo is referring to the energy entering again little top tip below here to help you
• 161:00 - 161:30 remember just like with exothermic we need to know a couple of examples of endothermic reactions so these are a little bit more Technical and less everyday we've got thermal decomposition that's one example and another reaction is the reaction between sodium hydrogen carbonate and citric acid so some everyday examples of an endothermic reaction include some sports injury packs so we can use these if we
• 161:30 - 162:00 have an injury and we want to cool down the area we can use one of these Sports Injury packs and it would cool down we need to make sure that we're really familiar with reaction profiles what they look like and what they represent so reaction profiles can be used to show the relative energies of reactants and products the overall energy change of a reaction and the activation energy if we look at this exothermic reaction we can see that the
• 162:00 - 162:30 reactants have more energy than the products and energy is released as we go from our reactants to our products our activation energy is represented by the difference in energy between the reactants and the peak of this curve now endothermic reactions on the other hand we know take in energy and so here we can see that the reactants have less energy than the products and that's because energy gets absorbed to give us
• 162:30 - 163:00 the energy of the products the activation energy again is represented from the difference in energy between the reactants and the peak of this curve what is activation energy we need to know that activation energy is the minimum amount of energy that particles must have to react during chemical reactions energy gets taken in to break the bonds in the reactants and energy is released to form
• 163:00 - 163:30 the bonds in the products therefore we can work out the overall energy change of a reaction to do that we just need to find the sum of the energy needed to break all the bonds in the reactants and subtract the sum of the energy released when forming all of the bonds in the product as a result of this we can say that during exothermic reactions the energy released when forming new bonds is greater than the energy needed to break
• 163:30 - 164:00 existing bonds and this of course explains why energy is released to the surroundings during an exothermic reaction well at the same time we can say for an endothermic reaction The energy needed to break the existing bonds is greater than the energy released when forming new bonds and that's why endothermic reactions they take in energy from the surroundings now we might find that we need to do some calculations here let's look at this
• 164:00 - 164:30 example so methane combusts in oxygen to form carbon dioxide and water the reaction is shown below calculate the energy change for the reaction so notice here that all of the bonds are shown If instead you were just given a balanced equation you would need to write this out just like I have here and if there's any balancing numbers like a big two in front of the oxygen make sure you write those down twice just like I have here notice also on the right hand side we've got this little
• 164:30 - 165:00 table which has the bond energies of all of the different bonds that we're going to see in this question well of course we know that to calculate the overall energy change of a reaction we need to work out the sum of the energy needed to break the bonds and the reactants and then we need to take away the sum of the energy released least when forming Bonds in the products so let's have a look at how we calculate this let's start by just looking at the reactants so starting off here I can see
• 165:00 - 165:30 that in the methane I have four CH bonds so looking at the table I can see that each bond is 412 K per mole so I'm going to do 4 * 412 now I can see I have two o double bond O's and so again looking at the table I want to do 2 * 498 now let's put that into the calculator to simplify so that ends up equaling
• 165:30 - 166:00 2,644 K per mole so now let us have a look at the products so here I can see I have two cble Bond O's in the carbon dioxide so from our table here I can see I need to do 2 * 743 now onto the waters I can see that I have 1 2 3 four o bonds so here I'm
• 166:00 - 166:30 doing four * 463 now if I pop that into the calculator now I end up getting the energy released when forming Bonds in the products and that is 3,338 K per mole so let's have a look here then I know that to get the overall energy change of the reaction I need to take the energy from those reactants and take
• 166:30 - 167:00 away the products so breaking minus making is how I always think about it so 2,644 minus 3338 gives me - 694 K per mole and that is how we would do one of these questions in the exam so let's summarize here what we know about breaking bonds and making bonds so when breaking bonds we know that heat energy is taken in we know that the
• 167:00 - 167:30 temperature of the surroundings will decrease and we know that this is an endothermic reaction forming Bonds on the other hand we know that heat energy is given out we know that there's going to be an increase in the temperature of the surroundings and we know that this is an exothermic reaction in chemistry we learn about cells but they're not the cells that we learn about in biology no instead cells in chemistry are devices which contain
• 167:30 - 168:00 chemicals which will react to produce electricity now we may be asked how can we make a simple cell and simply all we need to do is connect two different Metals in contact with an electrolyte just like this diagram here now the voltage that we end up getting from this cell is going to be dependent on two different things the first one is the type of electrode so what metals or
• 168:00 - 168:30 materials we're using for the two different electrodes and also what electrolyte we have as well so what solution do we have connecting these two electrodes so we might hear the term battery as well and a lot of people can get conf conf used about the difference between a battery and a cell and actually the batteries that you know in day-to-day life are actually cells but a battery would consist of two or more cells connected in series and the whole
• 168:30 - 169:00 purpose of this is that we're going to get a greater voltage from a battery rather than just a single cell we may well be asked to describe the difference between non-rechargeable and rechargeable cells and batteries we need to know that non-rechargeable cell or battery is one where if the reactants get used up the chemical reactions stop and the battery just stops working and an example of
• 169:00 - 169:30 that is the typical alkaline battery that we add to many of our devices around the home now a rechargeable battery is one where the chemical reactions are reversible when an external electric current is applied and we have an example here that we're all using all the time and that that is our mobile phone batteries our laptop batteries examples similar to that where we can just plug it in and recharge that battery fuel cells are cells which are
• 169:30 - 170:00 supplied with an external source of fuel for example hydrogen and oxygen we need to know how fuel cells work we should know that in a fuel cell the fuel is oxidized electrochemically within the fuel cell to produce a potential difference now hydrogen fuel cells generate this electricity by reacting together hydrogen and oxygen and forming water now the type of reaction that we have in this case is an oxidation
• 170:00 - 170:30 reaction and the reason why this is oxidation is because the hydrogen gas gains oxygen to produce water therefore we can say that the hydrogen is oxidized now we know that hydrogen fuel cells generate electricity using the fuel of hydrogen and oxygen and they provide a really really valuable alternative to rechargeable cells and batteries however for your exam you need
• 170:30 - 171:00 to know the half equations that are taking place so we need to know that at the negative electrode at the cathode 2 H2 Gus that hydrogen fuel is reacting with four o minus ions to make four H2O Plus plus four electrons then we need to know that over on the positive electrode the anode we have our oxygen gas and it reacts with 2 H2O picks up four electrons and it makes
• 171:00 - 171:30 four hydroxide so four o minus ions and in this reaction due to oil rig remember oxidation is loss reduction is gain we can see here that the hydrogen has lost four electrons and therefore we can say that the hydrogen is oxidized to produce water because it has lost electrons and likewise we can say that
• 171:30 - 172:00 the oxygen has been reduced because it has gained electrons finally here let's summarize the advantages and disadvantages of using hydrogen fuel cells so a great Advantage is that the only waste product produced is water we're not having to find suitable places to dispose of these batteries which have dangerous chemicals in them another great Advantage is that they don't need to be recharged however the disadvantages
• 172:00 - 172:30 include the fact that hydrogen can be very difficult to store particularly due to the fact that it's highly flammable another disadvantage of using hydrogen is that it's often produced from nonrenewable resources particularly fossil fuels such as natural gas and so although when we use use the fuel cell we only make water great in order to get the fuels that we needed for this fuel cell actually other products were probably formed along the way and that's all while also depleting those
• 172:30 - 173:00 non-renewable resources so make sure you know these advantages and disadvantages of hydrogen fuel cells ouch this is why in some videos I explain scratches [Music]