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Hint 1: There should be an electronegativity difference of zero between the two.
Hint 1: Clue is in the shape of the molecule and cancelling out of bond polarities; correct answer is where no cancelling occurs
Hint 1: Reducing agents get oxidised
Hint 2: B (permanaganate) and D(dichromate) are oxidising agents. See them in the electrochemical series bottom left (here they gain electrons and are therefore oxidising agents)
Hint 3: hydrogen peroxide (response C) is used as an oxidising agent in Higher experiments. CO - remember role from blast furnace where it gains oxygen to form carbon dioxide.
Hint 1: Must manipulate equations to get same number of electrons featuring: here the common factor is 6
Hint 2: multiply first equation by 2, second by 3
Hint 3: then add equations and the electrons cancel
Hint 4: 3 moles zinc : 2 moles nitrate, then work out for 1 mole zinc
Hint 1: looking for the OH bearing carbon atom to have 3 atoms directly attached so not in a straight chain, eliminate respomse C
Hint 2: Looking for OH and carbon containing branch to be on same carbon atom within a chain. Sketch out if necessary
Hint 1: Amide functional groups break. The word partial implies that only one of these functional groups will break/hydrolyse.
Hint 2: Circle the amide links (CONH). Hydrolysis occurs at amide group and C-N bond breaks forming carboxylic acid (COOH) and amine group (NH2)
Hint 3: look at the connections in the parent molecule as the same connectivity patterns should be seen in products
Hint 4: Luckily response A is correct, but as a learning point for your revision rule out the others.
Hint 1: The formula shows two oxygen atoms so this implies ester, or carboxylic acid but not ketones/aldehydes which only have one oxygen.
Hint 2: Hint 1 rules out responses B and C so then consider response A. Pentylethanoate: pentyl means 5 carbons and ethanoate a further two. So the total number of carbon atoms would be seven so pentyl ethanoate cannot be the correct answer.
Hint 1: CuO - this will oxidise the reactant. Rule out pentan-3-one (response C) and acid (response D) - no oxidation
Hint 2: No reaction with Fehlings means the product is a ketone.
Hint 3: primary alcohols oxidise to form aldehydes initially: they WILL react with Fehlings.
Hint 1: Obviously not straight chain so rule out response A.
Hint 2: Look for longest chain of carbon atoms including carboxyl carbon.
Hint 3: Three carbons in the longest chain, so this is a branched propanoic acid.
Hint 4: two methyl branches on second carbon atom in the chain. Remember that carbon atom number 1 is the carbon in the carboxyl group (COOH)
Hint 1: Terpenes based on units of five carbons so C5, C10 etc.
Hint 1: looking for the opposite e.g, perhaps ketone back to secondary alcohol etc
Hint 2: Responses A, B and D are oxidation processes (check your notes) gain of oxygen or loss of H; increase in O:H ratio
Hint 1: Similar to primary/secondary alcohols description so just apply as stated in question
Hint 2: Response B: look at Nitrogen atom. It is directly attached to how many Carbon atoms?
Hint 1: Phosphate formula in data book p21 - valency 3. Work out chemical formula of the compound.
Hint 2: Cu3(PO4)2 Five ions in formula unit, so 5 moles.
Hint 3: If you were to work out GFM of, say, MgCl2 then you are effectively adding masses of one mole magnesium ion to two moles chloride ion. Same idea regarding moles of ions here.
Hint 1: Common mistake is to forget that some of the responses are diatomic so when calculating GFM you must multiply relative atomic mass by 2 before calculating moles present.
Hint 2: Hydrogen, nitrogen and chlorine are diatomic so GFM values are 2g ,28g and 71g respectively. 4g methane contain 0.25 moles as GFM 16g. Look for 0.25 moles in the four examples.
Hint 1: Is there an excess of one reactant? If so, which one? 0.05 mol of MgCO3 needs twice this number of moles of nitric acid. Is there enough nitric acid?
Hint 2: Not enough HNO3 so it is the limiting reactant. Use its amount in moles to work out moles of MgCO3 used.
Hint 1: The carbonyl functional group is polar. The carbon atom is slightly positive, and the oxygen atom slightly negative.
Hint 2: Interactions will be between carbonyl carbon atom in one molecule and the oxygen involved in carbonyl in other molecule.
Hint 1: Use equation for atom economy on page 4 of Higher databook.
Hint 2: Remember to take into account the moles of iron and the reactants.
Hint 3: so ... 4 × 55.8/ [(2 × 159.6) + (3 × 12.0)] then × 100
Hint 1: 1:5 mole ratio. Check volumes present.
Hint 2: 100cm3 propane needs 500cm3 oxygen, so 100cm3 oxygen in excess
Hint 3: Use 100cm3 propane to work out volume CO2 produced (remember water a liquid from state symbol in equation)
Hint 4: total up 100 from oxygen with volume of carbon dioxide produced. (remember 1:3 ratio propane to carbon dioxide)
Hint 1: Imagine if we started with 100g. 60g would react and convert to product mass in step 1.
Hint 2: Only 60% conversion in step 1. Then only 90% of this amount converts in step 2. Overall yield equates to working out 90% of 60.
Hint 3: answer quick route = 100 × 0.6 × 0.9
Hint 1: excess zinc, so the moles of HCl control the total gas volume produced. We now have half the moles HCl (same concentration, but half volume)
Hint 2: Hint 1 rules out graphs A and D. Now focus on Rate. Acid concentration same. Powder means faster or slower?
Hint 3: faster means steeper initial slope to plateau
Hint 1: b is the subject of the equation so focus on the reaction W to Z. Instead of going that route directly we must go W to X to Y to Z. If we go this way we must ensure that the arrows all point that way too. This may require altering the direction of an arrow and then we must also change the sign of the enthalpy change. For example, we need to reverse the arrow to go from X to Y so the enthalpy change becomes -c in value.
Hint 2: we can get to X (so use value 'a') but from X to Y we need to reverse the arrow, so the value used is therefore -c. Apply same logic for step Y to Z.
Hint 1: Concentration: definitely does. Check notes for responses A and B, but think of energy profiles.
Hint 2: Activation energy is a barrier to reaction. Kinetic Energy is also crucial to the success of a reaction, so both will affect the rate.
Hint 3: Enthalpy change does not change. Think back to the use of a catalyst (lower energy pathway, but enthalpy change stays same)
Hint 1: lower pressure, lower numbers / moles of gas particles
Hint 2: If gas pressure is decreased then the equilibrium position moves to the side with a greater number of gas particles - we want this to be the product side.
Hint 1: Increasing Temperature increases the average Kinetic Energy of particles
Hint 2: Kinetic Energy values range from zero, but average position hovers around a higher value
Hint 3: Increasing temperature has NO effect on the activation energy value
Hint 1: Carefully look at reactant molecule and pick out its fragments in the product molecules. Imagine chopping through C=C and sticking oxygens on.
Hint 2: draw ethanal and propanone in the same way, as these produce molecules
1a)i) Hint 1: Start with Na. Look at RHS
1a)i) Hint 2: then balance Cl
1a)ii)A) Hint 3: total volume must be 50
1a)ii)B) Hint 4: 1/rate
1a)iii) Hint 5: Pick two rate values within the range plotted. Use graph to get T values for these.
1b) Hint 6: Sorry, but no hint is available - see the marking instructions.
1c)i) Hint 7: Activated complex is the high energy short-lived intermediate formed at the top of the energy curve
1c)ii) Hint 8: a catalyst provides a lower energy pathway from reactants to products
2a)i) Hint 1: Going across a period, what happens to nuclear charge?
2a)ii) Hint 2: Linked to atomic size. What are the factors here?.
2b)i) Hint 3: Check equation at very top of p11 of data book. Substitute symbol for nitrogen instead of E
2b)i) Hint 4: remember to use N not N2
2b)ii) Hint 5: remember that it is linked to the removal of electrons from the outer shell - look at electron arrangement of Nitrogen
2b)ii) Hint 6: N 2,5 so how many electrons require removal before outer shell 'empty'? What happens next?
2b)ii) Hint 7: energy required to remove 6th electron is linked to the 6th ionisation energy. This requires breaking into full energy level
2c) Hint 8: metals form positive ions by their atoms losing electrons wheras Phosphorus gains electrons
2c) Hint 9: Think about the electron arrangements of the respective ions
2d) Hint 10: Read carefully. Problem solving, so use the equation for radius ratio
2d) Hint 11: Use p17 of the data book and compare to the two values in the diagrams
Hint 1: You should focus on monatomics (group 0), discrete molecular (eg oxygen, sulfur, carbon - fullerene) and Covalent networks (carbon, silicon)
Hint 2: Look to talk about key forces of attraction. No polar bonds here as atoms are identical. No ions either.
Hint 3: London dispersion forces, covalent bonds. Discuss fully and link to melting point.
4a)i) Hint 1: Sorry, but no hint is available - see the marking instructions.
4a)ii) Hint 2: calculate the mass of 50cm3 of cider using information provided then compute answer.
4b)i) Hint 3: Compare structures of malic and lactic acids. Difference?
4b)i) Hint 4: Have lost one carbon atom and two oxygen atoms and they must end up in the product.
4b)ii)A) Hint 5: look at cider 2 to find information about malic acid distance
4b)ii)B) Hint 6: there should be no spot which is aligned with the spot from acid 2
4c) Hint 7: No hint available
4d)i) Hint 8: always look firstly for OH groups
4d)i) Hint 9: why will OH groups help water solubility?
4d)ii) Hint 10: circle ester group (COO). Split molecule through the C-O bond. Fragment to the left is the acid. Alcohol was ethanol, so use the clue in the name of 'ester' to name the acid
4d)iii) Hint 11: Sorry, but no hint is available - see the marking instructions.
4e)i) Hint 12: ethanal is an aldehyde
4e)ii) Hint 13: aldehydes oxidise to their corresponding carboxylic acid
5a)i) Hint 1: Break all bonds (endothermic means energy in) and remember mole quantities. Then make all bonds (energy released)
5a)ii) Hint 2: Look at p10 of the data book e.g, H-H bond only one source where this bond is found, whereas many different molecules contain C-H in different environments.
5a)iii) Hint 3: Convert 200cm3 volume to moles by using the fact that 1 mole is 24000cm3 or 0.2/24
5a)iii) Hint 4: answer here will also be the same as the number of moles of carbon dioxide
5b)i) Hint 5: Sorry, but no hint is available - see the marking instructions.
5b)ii) Hint 6: Use cmΔT. Remember that water mass must be in kg so 0.4. Temperature change = 23° so 0.4 × 4.18 × 23 but this is just for 1.1g heptane
5b)ii) Hint 7: scale from 1.1g to GFM of heptane listed in table i.e., scaling up by 100/1.1
5b)iii) Hint 8: Sorry, but no hint is available - see the marking instructions.
6a)i)A) Hint 1: must have the same number of electrons if possible to make comparison
6a)i)B) Hint 2: When you see OH, think hydrogen bonding. S-H bond is polar by definition, but its electronegativity difference is only 0.3
6a)ii) Hint 3: use the clue from ethanethiol (contains 2 × C )and substitute name in accurately for 1 x C
6a)iii) Hint 4: multiply value by 1000 to get per litre, then by total volume in litres.
6b)i) Hint 5: link to the way alcohols are classified
6b)ii)A) Hint 6: Hydrogen sulfide when added is effectively two parts adding on: and HS and H.
6b)ii)A) Hint 7: Look at the product. SH group has added on to middle carbon and the other H atom has added on to the first carbon in the alkene double bond . The answer lies by adding the H and S atoms (-S-H) on to the first carbon in the double bond shown in the reactant alkene. The lone H atom adds on to the middle carbon in this case.
6b)ii)B) Hint 8: Calculate the number of moles of 2-methylpropane reacted, using 30.5g as mass
6b)ii)B) Hint 9: This answer is equal to the moles of product if yield was 100%. So work out this theoretical mass. Finally work out 84% of this mass
7a)i) Hint 1: new free radical formed
7a)ii) Hint 2: needs to break bonds
7a)iii) Hint 3: a substance that stops formation of, or disrupts process of, free radical formation
7b)i) Hint 4: work out molecular formulae of the two molecules shown, then work out difference in atoms. This gives clue to Y
7b)ii) Hint 5: look at the first example and trace the path of all the key atoms into the product. What new bonds are there? Then apply the same logic to the new starting molecule
7b)ii) Hint 6: notice how the four CH2 units are present in reactant and product. Look then at neighbouring groups. What happened to OH and COOH?
7b)iii) Hint 7: Look at the name for the hydroxyacid in part (b)(i) (5-hydroxypentanoic acid) and apply to this new molecule.
8a)i) Hint 1: Circle the CONH links, then look at how many amino acid molecules are shown
8a)ii) Hint 2: Sorry, but no hint is available - see the marking instructions.
8b)i) Hint 3: Sorry, but no hint is available - see the marking instructions.
8b)ii) Hint 4: look at differences
8b)ii) Hint 5: difference of 1.0, 2.0, 3.0 so next should be 4.0
8c)i) Hint 6: Sorry, but no hint is available - see the marking instructions.
8c)ii) Hint 7: 1 gram of pineapple supplies 13.2mg. If you divide 500 by 13.2, then you will get mass of pineapple
9a)i) Hint 1: it is the amount of energy required to go from reactant level (at 0) to top of energy curve
9a)ii) Hint 2: raising the Temperature does not favour forward reaction i.e., production of chlorine
9a)ii) Hint 3: the energy profile shows an exothermic forward reaction
9b) Hint 4: Sorry, but no hint is available - see the marking instructions.
9c) Hint 5: Try to make the top equation your target equation
9c) Hint 6: Add the three equations together. You may have to alter the direction and mole quantities
9c) Hint 7: You must reverse the equation which involves methane. We need 1 moles of CCl4 so keep second equation as it is.
9c) Hint 8: We also need 4 moles HCl on RHS, so third equation needs to maintain direction but multiply it by 4. Add the three equations together.
10a) Hint 1: Sorry, but no hint is available - see the marking instructions.
10b)i) Hint 2: Sorry, but no hint is available - see the marking instructions.
10b)ii) Hint 3: Sorry, but no hint is available - see the marking instructions.
10b)iii) Hint 4: Calculate the number of moles of silver chloride. Then, link this to moles magnesium chloride using the equation
10b)iii) Hint 5: Divide moles AgCl by 2 to get moles of Magnesium chloride, then link to mass
10c) Hint 6: (mass of impure)/(mass of pure) × 100
Hint 1: Physical Properties, such as Melting Point/Boiling Point, solubility and link to functional groups. Chemical properties such as link to chemical behaviour, such as reactions.
Hint 2: All are molecular so they have low melting points. London Dispersion Forces: limonene, beta-carotene. Hydrogen bonding: fructose, vitamin C, citric acid, ethyl butanoate (with water interaction), limited with octanal
Hint 3: octanal is insoluble in water (large hydrophobic chain). OH groups give water soluble possibility. Limonen and beta-carotene are insoluble in water but soluble in, for example, hexane. Why?
Hint 4: reactions ... lots! Limonene and beta-carotene have addition reactions. Fructose has possible oxidation. Citric acid is acidic(!). Ethyl butanoate is an ester: hydrolysis. Think about their combustion, etc. Have an emphasis on their differences.
12a)i) Hint 1: response B - is this structure ionic?
12a)i) Hint 2: Head group: look for clues with compound C as it is ionic too. Watch charge on head.
12a)ii)A) Hint 3: Sorry, but no hint is available - see the marking instructions.
12a)ii)B) Hint 4: Sorry, but no hint is available - see the marking instructions.
12a)iii) Hint 5: Sorry, but no hint is available - see the marking instructions.
12b)i) Hint 6: think electronegativity of bonding atoms
12b)ii) Hint 7: Sorry, but no hint is available - see the marking instructions.
12c) Hint 8: Sorry, but no hint is available - see the marking instructions.
12d)i) Hint 9: Look at the I- ions. All they can do is lose electrons to form iodine molecules . This is therefore oxidation and rules them out. So the key lies with hypochlorite ion. Remember electrons must feature in your answer.
12d)i) Hint 10: Use data booklet p12 to see iodide/iodine involves 2 electrons so hypochlorite must also.
12d)i) Hint 11: Alternative is to take equation in step one and score out the iodine species. What is left is the core of the answer. Jjust add 2 electrons on left hand side.
12d)ii) Hint 12: calculate moles of thiosulfate used, then dividie by 2 to get moles of iodine (see mole ratio in step 2 equation)
12d)ii) Hint 13: Now link iodine and hypochlorite - see step 1 equation
12d)ii) Hint 14: 1 mole iodine means 1 mole hypochlorite