Higher Chemistry 2018-19 Watch

Pearls1
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#81
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No it’s okay don’t worry about it. It’s jusf, numeracy is probably my weak area. Don’t know how to study as it isn’t knowledge based
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Pearls1
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Size:  61.5 KBCould you please also help me with the trends queation? I don’t understand why the answer is around 99-124
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sbneelu
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#83
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After finishing writing this up I realised that most of this is unnecessary as I wrote out my entire thought process but I decided to just leave it in anyway because it would help with other questions like this. The important things for this question are in bold though so feel free to skip through the rest of it.

In general for questions like this, consider:
- Number of carbon atoms in the entire molecule
- Number of carbon atoms in main chain
- Branches
- Functional groups present (number, type and position)
- Molecule size (formula mass)

First consider the number of carbons in the entire molecule. All of the molecules given, as well as the molecule in the question, have 6 total carbons so that doesn't help us.


Some of the molecules are carboxylic acids and therefore have one straight chain but others are esters so they don't have a single main carbon chain so that doesn't help us.

Look at functional groups and branches together next. First compare 1 and 4 since they are both straight chains with no branches but 1 is a carboxylic acid while 4 is an ester. 1 has a greater boiling point than 4 which may indicate that carboxylic acids have greater boiling points than esters. Compare 2 and 5 next since they each have a methyl group on the same carbon, but 2 is a carboxylic acid and 5 is an ester. 2 has a greater boiling point than 5 so that supports our previous hypothesis that carboxylic acids have greater boiling points than esters. Comparing 3 and 6 shows the same thing.

Next look at the position of functional groups. Carboxyl groups can only go in one place (the first/last carbon) so the ester link is the only important one here. Compare 4 and 7 because they're both straight chains with no functional groups, and the only difference between them is where the ester link is. 4 has a greater boiling point than 7 so you could hypothesise that the further away from the first to second carbon the ester link is, the lower the boiling point of the ester. You can then support this hypothesis with 6 and 8.

The molecule in the question has one methyl group, and an ester link between the second and third carbons. Based on all the information before and this, the boiling point must be between an ester with ester link in the same position with no methyl groups, and one with two methyl groups (all attached to the same carbon). Therefore the boiling point must be between that of 7 (126ºC) and 8 (98ºC). I'm not exactly sure why they're also excluding 125ºC as an answer though.

(Original post by Pearls1)
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Size:  61.5 KBCould you please also help me with the trends queation? I don’t understand why the answer is around 99-124
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katherine340
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(Original post by Pearls1)
Also, is it true that see the cost numeracy questions are they taken out of the higher now?
No my teacher mentioned on the last week of school that we will have at least one numeracy question.
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Ali$hah
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Hey sorry for this difficult question due to wording But see if we were to find the how many moles of the product would be produced - we would be given the moles of reactants - we would find excess and limiting reacting using mole to mole ratio But would the excess convert into the product ? Or would the moles that have reacted convert to products ?Thank you in advance !
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sbneelu
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The excess would not be converted into product. All of the limiting reactant and the corresponding number of moles of any reactant in excess based on the molar ratio would be converted, but the rest of the reactant in excess would not be.
(Original post by Ali$hah)
Hey sorry for this difficult question due to wording But see if we were to find the how many moles of the product would be produced - we would be given the moles of reactants - we would find excess and limiting reacting using mole to mole ratio But would the excess convert into the product ? Or would the moles that have reacted convert to products ?Thank you in advance !
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Ali$hah
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Ok so the limiting reactant and the no. Of moles of the other reactant that can be reacted as well will become a product .. is that correct ?
(Original post by sbneelu)
The excess would not be converted into product. All of the limiting reactant and the corresponding number of moles of any reactant in excess based on the molar ratio would be converted, but the rest of the reactant in excess would not be.
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sbneelu
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Yep!
(Original post by Ali$hah)
Ok so the limiting reactant and the no. Of moles of the other reactant that can be reacted as well will become a product .. is that correct ?
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Ali$hah
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Thank you ! Xx
(Original post by sbneelu)
Yep!
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Pearls1
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Thanks a lot. The explanation was very very helpful!
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Pearls1
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Sorry I keep bothering you, but how would you do this question? I thought in this you look at Fehling’s solution as I thought thats being oxidised but you don’t you look at the reducing sugar. Could you please explain why? Name:  FB9484E6-6E21-4AFE-9752-B44C0E31AE06.jpg.jpeg
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sbneelu
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The Fehling's solution is being reduced, not oxidised. The charge on Copper goes from 2+ to + which means it has become less positive and therefore more negative, so it has gained an electron, and reduction is the gain of electrons. The reducing sugar has gone from having a 6:12 O : H ratio to having a 7:12 O : H ratio, so it has been oxidised. Another way to think about it is that reducing agents reduce other substances so they themselves are oxidised, and a reducing sugar is a reducing agent so it is oxidised.
(Original post by Pearls1)
Sorry I keep bothering you, but how would you do this question? I thought in this you look at Fehling’s solution as I thought thats being oxidised but you don’t you look at the reducing sugar. Could you please explain why? Name:  FB9484E6-6E21-4AFE-9752-B44C0E31AE06.jpg.jpeg
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Pearls1
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I thought if something is reduced then that means it’s oxidation and if something is oxidised it means it’s reduced?
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sbneelu
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(Original post by Pearls1)
I thought if something is reduced then that means it’s oxidation and if something is oxidised it means it’s reduced?
No, oxidation is when something is oxidised and reduction is when something is reduced, but a reducing agent is oxidised and an oxidising agent is reduced.
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Pearls1
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Okay I kind of understand
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Pearls1
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That makes sense now! Thanksss
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katherine340
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Hello.
On question 1 are we not supposed to be given a balanced equation to do that question? I am confused.
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sbneelu
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Not necessarily. You're expected to write the formulae for the two compounds using your own knowledge. Mg has a valency of 2, Br has a valency of 1 and SO4 has a charge of 2- so has a valency of 2, so magnesium bromide is MgBr2 and magnesium sulfate is MgSO4. Since there are 4 moles of bromide, there are 2 moles of magnesium in magnesium bromide. Since there are 3 moles of magnesium overall, the other 1 mole must be in magnesium sulfate, so since there is 1 mole of magnesium in magnesium sulfate, there must also be 1 mole of sulfate, so the answer is A.
(Original post by katherine340)
Hello.
On question 1 are we not supposed to be given a balanced equation to do that question? I am confused.
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536458
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(Original post by sbneelu)
Not necessarily. You're expected to write the formulae for the two compounds using your own knowledge. Mg has a valency of 2, Br has a valency of 1 and SO4 has a charge of 2- so has a valency of 2, so magnesium bromide is MgBr2 and magnesium sulfate is MgSO4. Since there are 4 moles of bromide, there are 2 moles of magnesium in magnesium bromide. Since there are 3 moles of magnesium overall, the other 1 mole must be in magnesium sulfate, so since there is 1 mole of magnesium in magnesium sulfate, there must also be 1 mole of sulfate, so the answer is A.
Great explanation, but I just noticed you put MgSO4 rather than Mg2SO4
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Pearls1
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Does anyone know why the sign is negative for this question? I thought it was because of enthalphy of combustion but I don’t think that’s right as there’s a similar question like this in the only difference is that they have Al+ on reactant side and Al3i + 2e- on product side in the other Past paper whereas this one is flipped so is that why it’s -ve? Name:  8EDE05D5-2157-4D28-B2F0-27C6842005CE.jpg.jpeg
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