Physical (Year 1)

1. They are accelerated by an electric field.
2. They are detected by attracting and gaining an electron when they hit the negatively charged plate at the end of the chamber, which generates a current.
3. Their abundance is determined by the fact that it is relative to the size of the current produced.Note: the idea that the size of the current is proportional to the abundance is key.

1. A high energy electron is fired at the sample from an electron gun.Note: It is vital you say ‘high energy electron’.
2. This knocks off one outer shell electron from each atom, forming 1+ ions.Note: it is used for elements and low Mr compounds and the mass remains the same.

1. The sample is first dissolved in a volatile, polar solvent.
2. It is then injected through a fine hypodermic needle as a fine spray into a vacuum in the ionisation chamber.
3. A very high voltage is applied to the end of the needle where the spray emerges.
4. The particles gain a proton (H+) and become ions. The solvent then evaporates, leaving a fine mist of 1+ ions.Notes: this is used for high Mr compounds and the mass goes up by one due to the addition of the H⁺.

1. Finding relative atomic mass.
2. Finding relative molecular mass.
3. Finding relative isotopic abundances.

1. The average mass of an atom of an element.
2. Divided by 1/12th of the mass of an atom of 12C.

1. The sum of the average masses of atoms of each element in the molecular formula of a molecule.
2. Divided by 1/12 of the mass of an atom of 12C.

1. The elements gain more shells and get larger.
2. This means there is a weaker attraction of the ion/ nucleus to the lost, outer electron.Note: I’d also always mention the increase in shielding (the core electrons repelling the outer electrons) as being another reason for the decrease in ionisation energy. It appears on some mark schemes as an alternative thing you could mention, but isn’t usually required for full marks.

1. They have the largest number of protons of all the elements in a period.
2. They have the same amount of shielding and energy levels as the rest of the elements in their period.

1. The electron in oxygen is being removed from a doubly-occupied p orbital.
2. It therefore experiences repulsion from the other electron in the orbital, as the pair of electrons repel each other.
3. This means less energy is required to remove the electron from the p orbital in sulfur compared to phosphorus.

1. The electron is being removed from a positive 1+ ion, meaning it experiences a stronger force of attraction than if it were being removed from an atom.
2. The electron being removed is closer to the nucleus.Note: this is true even if the electron is being removed from the same subshell, as the ion will be smaller than the atom.

1. The group 2 ion loses an electron from a higher energy orbital, meaning it is further from the nucleus.Note: name the orbital it is removed from for both elements
2. There is more shielding in the group 2 ion.Note: name which orbital/ subshell shields the group 2 but not the group 1.

1. Contains protons and neutrons in a nucleus at the centre of the atom.
2. Contains electrons in energy levels around the nucleus.
3. Most of it is empty spaceNote: these three points should be enough for any ‘compare X model to the modern day model’ question.

1. The electron is removed from a lower energy orbital.Note: name the orbital it was removed from for the 2nd and compare this to the 3rd.
2. The electron being removed experiences less shielding and is closer to the nucleus.
3. This means there is a stronger attraction between the nucleus and the outer electron being removed.

1. Enthalpy needed to remove one mole of electrons from one mole of atoms.
2. In the gaseous state.

1. Molar ratios (big numbers) are used.
2. Mr is used.

1. Weigh out a known mass of the solid using weighing by difference.Note: or you could wash the solid from the weighing container into the beaker. I'd describe the process of weighing by difference if you go with that.
2. Add distilled water and stir until the solid has dissolved.
3. Transfer this to a volumetric flask, rinsing the beaker and glass rod into the flask using distilled water.
4. Add distilled water until the bottom of the meniscus reaches the graduation mark on the volumetric flask.
5. Shake the flask (or invert 10 times) to homogenise the solution.

1. Wash the beaker into the volumetric flask with distilled water.
2. Wash the glass stirring rod into the volumetric flask.
3. Use weighing by difference (describe if it’s a long mark question)
4. Ensure the bottom of the (liquid) meniscus is on the graduation mark.
5. Mix / shake the final solution in the flask (to homogenise the solution).

1. It would increase the titre value.
2. Because the solution would get diluted.

1. Not opening the tap before the titre to fill the jet.
2. There being an air bubble in the jet or other part of the burette.
3. Forgetting to remove the funnel from the top of the burette before beginning the titration.

1. Only a few drops.
2. Only a small amount should be added as the indicators are weak acids, meaning more alkali would be required to neutralise the solution if too much is added.

1. Drop sizes varying.Note: no more explanation was given on mark scheme - I assume it means that there will be inconsistencies in the size of the last drop before the colour change.
2. The percentage error in the equipment being large enough to cause inconsistencies in results. This would occur if the volume of either solution used is too small.

1. Reading the burette at the start of the reaction.
2. Reading the burette at the end of the reaction.
3. Judging the end point to within one drop.

Note: this means the total uncertainty is three times the uncertainty of a single reading.

You should use a larger volume by:

1. Dissolving the current solution in a solvent (water in most cases, but an organic solvent in some cases).
2. Then transferring this solution to a volumetric flask and making it up to 250cm3 with more solvent.
3. Then titrating 25cm3 samples from the flask.

1. There are strong electrostatic forces of attraction.
2. Between oppositely charged (positive and negative) ions.
3. They form a giant ionic lattice, meaning there are many of these electrostatic attractions/ ionic bonds to overcome. Always talk about the ‘giant ionic lattice’ and how this causes there to be ‘many’ ionic bonds to overcome.

1. Al₂O₃ forms an ionic lattice so is held together by strong ionic attractions.
2. Sr has a metallic structure with relatively weaker bonding.Note: It seems that in these questions where you’re comparing boiling points of ionic vs. metallic structures, you should just say the metal’s metallic bonding is weaker than the ionic compound’s ionic bonding. You should also then describe the structure and bonding of each substance to make sure all possible marking points are covered.

1. Nitrogen and oxygen are very electronegative.Note: clearly you should mention the electronegativity of every atom involved, even if it isn’t bonded to an H.
2. Therefore, C=O and N-H are very polar.Note: again mention all of the bonds being polar and name the bonds with chemical symbols.
3. Which results in the formation of a hydrogen bond between the O on C=O and H on N-H.Note: if describing a specific hydrogen bond, always mention that it is a hydrogen bond and say what it forms between.
4. In which the δ+H is strongly attracted to the lone pair on the O.Note: of course I’d also mention the H nucleus being deshielded, and this causing it to have the delta positive. However, it seems you should always say the delta positive on the H is what’s attracted to the lone pair, not the deshielded nucleus.

1. They form a giant lattice of positive metal ions.
2. Where there are strong electrostatic attractions between the ions and the sea of delocalised electronsNote: remember to always pluralise ‘attractions’

1. Because they have a giant molecular structure.
2. And a lot of energy is required to break the many covalent bonds in the structure.

1. They contain delocalised electrons/ mobile ions that are free to move and are charged.
2. These are able to flow through the structure.Note: probably don’t say ‘carry charge’, just say the particles ‘are charged’

1. Chlorine has a higher electronegativity than carbon/ they have different electronegativities.
2. This means the electron density is unsymmetrical/ chlorine becomes delta (δ) negative and the carbon becomes δ positive.Note: I’d always mention the delta negative and positives.

1. Differences in electronegativities.Note: this only causes a polar molecule if the molecule is asymmetrical.
2. Lone pairs.

1. It’s symmetrical.
2. This means the dipoles cancel out.Note: always say this, don’t talk about polar bonds cancelling out.

1. Random electron movement in the first molecule creates a temporary dipole.
2. This induces a dipole in a neighbouring molecule.
3. The slight positive (δ+) dipole in one molecule attracts the slight negative (δ-) molecule in the adjacent molecule, forming a temporary attraction.

1. They are single atoms, meaning the electrons are closer to the nucleus and they have fewer electrons.
2. This means they cannot be polarised easily as the electron cloud cannot be easily distorted.Note: I believe this explanation can also be applied to small molecules.

1. Straight chain molecules have more surface contact with each other and pack closer together.
2. They therefore have stronger van der Waals’ forces between molecules.

1. Difference in electronegativities (Cl more electronegative than H) causes partial negative (on chlorine) and positive (on hydrogen) dipoles to form in each molecule.
2. The ∂+ on the H of one molecule attracts the ∂- on the Cl of a neighbouring molecule.

1. The Van der Waals forces in the non-polar molecules can be stronger than the permanent dipole-dipole forces in the polar molecules.
2. If the non-polar molecules have many more electrons.

1. It is the attraction between the lone pair on the O of one molecule.
2. And the delta positive H on another molecule.Note: obviously talk about the 'deshielded nucleus' and 'differences in electronegativity' too, but always remember that the H’s attraction is to the lone pair of the NOF element.

1. Lone pairs on the NOF atoms involved in the hydrogen bond.
2. Partial charges (O δ−, H δ+) on the atoms involved in the hydrogen bond.
3. A dotted line between the lone pair on the NOF atom and the correct H.
4. The O–H⋅⋅⋅⋅⋅O being linear.Note: it is vital you draw it this way.

1. The atom the hydrogen is bonding to is not as electronegative as nitrogen, oxygen or fluorine.
2. The difference in electronegativity between the two atoms is not large enough.

1. How many electrons the central atom has in its outer shell.
2. How many electrons are given by the surrounding elements, and how this adds up to having 8 electrons in the outer shell in total.
3. How many bonding and lone pairs there are.
4. The fact that bonding pairs repel equally and spread as far apart as possible. Mention this on every shapes question, even small 2 markers.
5. The fact that lone pairs repel more than bonding pairs (if there are lone pairs). The new vocab is ‘lone-pair bonding-pair repulsion is stronger than bonding-pair bonding-pair repulsion.’
6. The shape of the molecule.
7. The bond angle of the molecule AND the bond angle it is ‘reduced’ from due to the lone pairs. E.g. Angle is reduced from 109.5 to 107.

1. It is the enthalpy required to break 1 mol of bonds.
2. Averaged over a range of compounds.

1. Mean bond enthalpies are calculated using enthalpies from a wide range of compounds.
2. Calculations using mean bond enthalpies don’t take into account the enthalpy changes that occur when substances change state, meaning they don’t account for the vaporisation of the fuel or the condensing of water.Note: this is the weaker of the two - I’d stick with the first.

1. The density of the solutions are 1.00 g cm-3.
2. The specific heat capacity of the solutions are 4.18 J g⁻¹ K⁻¹Note: basically you’re acting like the solutions are water, even when they’ll have solids dissolved in themNote 2: this hasn’t been tested before - usually they’re just assumptions the question will tell you to make.

1. Use a wind shield to insulate the entire experiment.Note: This is the ‘best way’ as it was the only answer option on one mark scheme.
2. Reduce the distance between the flame and the beaker.
3. Add a lid.

Stage 1:

1. Measure water out with a named apparatus (e.g. measuring cylinder)
2. Use a suitable mass of water (in the range of 10 - 200g).
3. Add the water to an insulated container/ polystyrene cup.Note: it can be easy to forget to do this for calorimetry questions as usually you have to use a copper beaker, but remember that if heat is being released by the reactant IN the water, you need to insulate the reaction.
4. Add a known mass of MgCl₂.
5. Using weighing by differenceNote: remember that any time you are adding a known but unspecified mass of something you need to mention using weighing by difference. Adding it ‘with washings’ wouldn’t work in this scenario as the MgCl₂ would react with the distilled water. Stage 2:
6. Record the initial temperature every 30 seconds for 2 minutesNote: the mark scheme specified a ‘min of 2 measurements’.
7. Record the temperature every 30 seconds for a minimum of 10 minutes after adding the MgCl₂ or until a trend is seen.Note: I’d also add ‘stir the solution before each reading to disperse the heat throughout the solution, ensuring the temperature is the same throughout the solution. Of course also mention the whole ‘don’t record the temperature when you add the reactant’ thing in these questions, too.
8. Plot temperature against time.Note: you could've received full marks for this section by simply drawing the graph and showcasing all of this stuff in the graph. Stage 3:
9. Extrapolate the lines to when the solid was added.
10. Find the temperature change at the point of addition by taking the initial temperature away from the temperature found at the point of addition on the extrapolated line.
11. Use the formula q = mc∆T to find the energy released by the reaction.
12. Find the moles of the reactant using Moles = Mass/Mr.
13. The enthalpy change is -q/ molesNote: make sure you show you understand you have to make q negative since the reaction is exothermic.

1. Measure and record the initial temperature of the solution for a few minutes before addition.
2. Add the reactant but do NOT record the temperature at the time of addition.Note: you don’t have to usually specify this, but do just in case.
3. Measure and record the temperature after the addition at regular intervals (e.g. each minute) for 8+ minutes.
4. Plot a graph of temperature against time.Note: it's important you specify what the graph actually shows, not just ‘a graph of the results’.
5. Extrapolate to the point of addition.
6. Determine ΔT at the point of addition by taking initial T away from final T.

1. Temperature (initial and final, so you can calculate the change to use in the calculation for q)
2. Mass of the reactant that enthalpy change is being calculated for (initial and final, so you can find the moles used in the reaction)

1. Heat loss to the surroundings.
2. Incomplete combustionNote: remember that these two are the go-tos, but some fuel being lost by evaporation is a third option.

1. Heat loss to the surroundings.Note: you should say 'heat loss' here, not just 'heat exchange'.
2. Incomplete combustion.
3. Evaporation of fuel.
4. Experiment not done under standard conditions.

1. Reduce the distance between the flame and the container of water.
2. Put a sleeve around the flame to protect it from drafts.
3. Add a lid to the top of the container.Note: any insulation must be at the top only - we want conduction at the bottom.
4. Use a copper calorimeter instead of a beaker.

1. Replace the glass beaker with a polystyrene cup.Note: don’t just say use a polystyrene cup - say what you are using it instead of.
2. Insulate the glass beaker.Note: probably don’t use this one.
3. Use a lid.

1. Heating up the copper container/ thermometer of the calorimeter.Note: this is heat transfer to the surroundings, but not the ‘atmosphere’.
2. Evaporation of the alcohol.Note: this will affect the change in mass of the alcohol, changing the calculated moles of alcohol that were combusted.
3. The experiment was not done under standard conditions.

1. If the temperature change is much bigger than the uncertainty this will mean that the percentage uncertainty is still fairly low.
2. Heat loss is more significant of an issue than uncertainty caused by equipment.

1. Since ethanedioic acid is a weak acid (unlike sulfuric), energy will be needed to break its bonds during the reaction to cause it to fully dissociate and react.
2. Therefore less energy will be released in total as some will be taken in for the endothermic process of dissociation of the weak acid.

1. The molecules have greater kinetic energy.
2. Many more molecules have energy greater than activation energy.Note: ‘many’ was its own mark.
3. This leads to more frequent successful collisions.

1. It doubles it, because there are twice as many particles per unit volume.
2. This means the frequency of successful collisions (with E ≥ Ea) doubles.

1. Increased pressure means there are more particles in a given volume.
2. This causes more frequent successful collisions above the activation energy.

1. A substance that increases the rate of reaction.
2. And remains chemically unchanged at the end of the reaction.

1. They provide an alternative pathway.
2. With a lower activation energy.

1. Use a water bath to maintain a constant temperature so the results aren’t affected by the temperature being different in different experiments.
2. Use a colorimeter to monitor the colour change to eliminate human error in judging the colour change.

1. The forward and reverse reactions proceed at equal rates.
2. The concentrations of reactants and products remain constant.Note: 'the system being closed' doesn’t appear on mark schemes.

1. The equilibrium shifts in the endothermic direction.Note: also say whether it goes to the right or left.
2. To decrease the temperature, or to oppose the increase in temperature.Note: don’t talk about ‘minimising’ or just a ‘change’ in temperature.

1. The position of equilibria shifts in the endothermic direction.Note: state whether this is to the left or right.
2. This decreases the temperature.
3. This will decrease/ increase the yield of the products/ reactants, causing Kc to increase/ decrease.

1. There is an equal number of moles of gas on each side of the reaction.
2. This means increasing the pressure has an equal effect on the forwards and backwards reactions.

1. The catalyst increases the rate of reaction of the forwards and backwards reactions.
2. Equally.

1. It creates lower electrical pumping costs.
2. It allows the use of lower pressure, less expensive equipment.

1. The equilibrium will shift to the side with the most (aqueous) moles.
2. To oppose the decrease in concentration of all reactants and products.

1. Mention the change in its oxidation state.
2. State that this means electrons are lost/ gained.

1. Balance the electrons on each side by multiplying one/both of the equations so the electrons cancel out when you combine the equations.Note: your overall equation should have no electrons in it.
2. Cancel out H⁺s and H₂Os that appear on both sides, so you only have them on one side.

1. Bromine is not a strong enough oxidising agent to displace chloride ions.
2. This is because it is a weaker oxidising agent than Cl₂.Note: Never talk about halide ions having oxidising ability.