Chemistry Qs
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Reply 20
Ariale
OP
14
Original post by TypicalNerd
The enthalpy of formation of glucose should be
6C(s) + 6H2(g) + 3O2(g) —> C6H12O6(s)
All you do is you make the top line the enthalpy of combustion of glucose and the bottom line the reactants in the equation above + 6 more molecules of O2(g)
6C(s) + 6H2(g) + 3O2(g) —> C6H12O6(s)
All you do is you make the top line the enthalpy of combustion of glucose and the bottom line the reactants in the equation above + 6 more molecules of O2(g)
Also for Q2a(ii) how am I meant to know which decomposes at the highest temperature?
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Periodic%20Table,%20Group%202%20and%20the%20Halogens%201%20QP.pdf
Reply 21
Ariale
OP
14
Original post by Ariale
Also for Q2a(ii) how am I meant to know which decomposes at the highest temperature?
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Periodic%20Table,%20Group%202%20and%20the%20Halogens%201%20QP.pdf
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Periodic%20Table,%20Group%202%20and%20the%20Halogens%201%20QP.pdf
@Pigster . Also for this question 2(b) why can’t I just find the mass of the solid divided that of the mr to find moles then compare molar ratios etc or not the mr divide by that solid plus water?
Reply 22
Original post by Ariale
Also for Q2a(ii) how am I meant to know which decomposes at the highest temperature?
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Periodic%20Table,%20Group%202%20and%20the%20Halogens%201%20QP.pdf
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Periodic%20Table,%20Group%202%20and%20the%20Halogens%201%20QP.pdf
I think that’s an old spec question.
The group 2 carbonates become increasingly thermally stable down the group, so the metals at the bottom of the group form the most stable carbonates (since the cations get larger and become less polarising, hence they become less able to weaken the C-O bonds in the carbonate ion thereby making the carbonate less prone to thermal decomposition).
Radium in theory should be the correct answer, but I’d be unsurprised if they wanted you to put barium.
Reply 23
Original post by Ariale
@Pigster . Also for this question 2(b) why can’t I just find the mass of the solid divided that of the mr to find moles then compare molar ratios etc or not the mr divide by that solid plus water?
You are given the means to calculate the moles of SrCl2.6H2O and the means to calculate the moles of water lost.
The moles of SrCl2.6H2O used are trivial to find as you have the mass used and relevant molar mass.
The mass of water lost is found by taking the difference in masses. From this you should be able to find the moles of water quite straightforwardly.
After calculating the moles of each, try working out what the molar ratio of SrCl2.6H2O used to H2O lost is. It should tell you how many of the 6 water molecules were ripped out of each formula unit and from that, you should be able to work out the formula of the final product.
Reply 24
Original post by Ariale
This is just a general enquiry. But when do you actually use molar ratios the big numbers. Do you use it in percentage yield , atom economy and titration calculations including molar rations when you times the Mr.
Or scenarios that you don’t use it like when you have the mass and mr do you times the mr by the big number to find the Mr then divided by mass to find the moles or don’t times anything and link with molar ratios ?
Or scenarios that you don’t use it like when you have the mass and mr do you times the mr by the big number to find the Mr then divided by mass to find the moles or don’t times anything and link with molar ratios ?
For reference, the “big numbers” are usually referred to as “stoichiometric coefficients” by chemists and so that is how I will refer to them in any subsequent posts.
Let’s run through some examples all based on the following reaction:
XeF6(s) —> Xe(g) + 3F2(g)
(I don’t know why this was the first reaction to come to mind - maybe I’ve been doing an unhealthy amount of main group chemistry lately LOL).
Let’s suppose in each example our desired product is fluorine, F2(g).
For atom economy, the big numbers do matter. Atom economy is the total Mr of all the desired product divided by the total Mr of reactants.
The total Mr of reactants in this case is just the Mr of XeF6(s) (Ar of Xe = 131.3, Ar of F = 19.0), which is 245.3.
The total Mr of desired product is 3 x the Mr of F2(g), or 3 x 38.0 = 114.0.
The atom economy is therefore 100% x 114.0/245.3 = 46.5% (3 sf)
For percentage yield, it is important to factor in via the molar ratio.
Supposing I made 24.0 dm^3 of F2(g) under RTP from 98.12 g of XeF6(s), I would first work out the moles of each:
n(XeF6) = (98.12 g)/(245.3 g mol^-1) = 0.4 mol
n(F2) = (24.0 dm^3)/(24.0 dm^3 mol^-1) = 1.0 mol
The theoretical yield of F2(g) should be 3 x the number of moles of XeF6(s) as per the equation and so this is how we factor in the stoichiometric coefficients.
Theoretical n(F2) = 3 x 0.4 mol = 1.2 mol
% yield = 100% x (1.0 mol)/(1.2 mol) = 83.3% (3 sf)
For any other stoichiometry calculations, the equation gives the molar ratios in which your reactants react and the molar ratios in which your products form.
Under no circumstances, however, must you multiply an Mr by a stoichiometric coefficient when attempting to use the formula moles = mass/Mr.
Reply 25
20
Original post by Ariale
@Pigster . Also for this question 2(b) why can’t I just find the mass of the solid divided that of the mr to find moles then compare molar ratios etc or not the mr divide by that solid plus water?
Not really a lot left for me to do. If you want me to re-explain anything of TN's replies in my own way, I'll be happy to do so.
The relative thermal stabilities of G2 carbonates was on the old OCR A spec, but IS NOT on the current spec.
Reply 26
Ariale
OP
14
Original post by TypicalNerd
For reference, the “big numbers” are usually referred to as “stoichiometric coefficients” by chemists and so that is how I will refer to them in any subsequent posts.
Let’s run through some examples all based on the following reaction:
XeF6(s) —> Xe(g) + 3F2(g)
(I don’t know why this was the first reaction to come to mind - maybe I’ve been doing an unhealthy amount of main group chemistry lately LOL).
Let’s suppose in each example our desired product is fluorine, F2(g).
For atom economy, the big numbers do matter. Atom economy is the total Mr of all the desired product divided by the total Mr of reactants.
The total Mr of reactants in this case is just the Mr of XeF6(s) (Ar of Xe = 131.3, Ar of F = 19.0), which is 245.3.
The total Mr of desired product is 3 x the Mr of F2(g), or 3 x 38.0 = 114.0.
The atom economy is therefore 100% x 114.0/245.3 = 46.5% (3 sf)
For percentage yield, it is important to factor in via the molar ratio.
Supposing I made 24.0 dm^3 of F2(g) under RTP from 98.12 g of XeF6(s), I would first work out the moles of each:
n(XeF6) = (98.12 g)/(245.3 g mol^-1) = 0.4 mol
n(F2) = (24.0 dm^3)/(24.0 dm^3 mol^-1) = 1.0 mol
The theoretical yield of F2(g) should be 3 x the number of moles of XeF6(s) as per the equation and so this is how we factor in the stoichiometric coefficients.
Theoretical n(F2) = 3 x 0.4 mol = 1.2 mol
% yield = 100% x (1.0 mol)/(1.2 mol) = 83.3% (3 sf)
For any other stoichiometry calculations, the equation gives the molar ratios in which your reactants react and the molar ratios in which your products form.
Under no circumstances, however, must you multiply an Mr by a stoichiometric coefficient when attempting to use the formula moles = mass/Mr.
Let’s run through some examples all based on the following reaction:
XeF6(s) —> Xe(g) + 3F2(g)
(I don’t know why this was the first reaction to come to mind - maybe I’ve been doing an unhealthy amount of main group chemistry lately LOL).
Let’s suppose in each example our desired product is fluorine, F2(g).
For atom economy, the big numbers do matter. Atom economy is the total Mr of all the desired product divided by the total Mr of reactants.
The total Mr of reactants in this case is just the Mr of XeF6(s) (Ar of Xe = 131.3, Ar of F = 19.0), which is 245.3.
The total Mr of desired product is 3 x the Mr of F2(g), or 3 x 38.0 = 114.0.
The atom economy is therefore 100% x 114.0/245.3 = 46.5% (3 sf)
For percentage yield, it is important to factor in via the molar ratio.
Supposing I made 24.0 dm^3 of F2(g) under RTP from 98.12 g of XeF6(s), I would first work out the moles of each:
n(XeF6) = (98.12 g)/(245.3 g mol^-1) = 0.4 mol
n(F2) = (24.0 dm^3)/(24.0 dm^3 mol^-1) = 1.0 mol
The theoretical yield of F2(g) should be 3 x the number of moles of XeF6(s) as per the equation and so this is how we factor in the stoichiometric coefficients.
Theoretical n(F2) = 3 x 0.4 mol = 1.2 mol
% yield = 100% x (1.0 mol)/(1.2 mol) = 83.3% (3 sf)
For any other stoichiometry calculations, the equation gives the molar ratios in which your reactants react and the molar ratios in which your products form.
Under no circumstances, however, must you multiply an Mr by a stoichiometric coefficient when attempting to use the formula moles = mass/Mr.
https://pmt.physicsandmathstutor.com/download/Additional-Assessment-Material/Chemistry/AS-level/OCR-A/Component-2/Component%202%20MA%2014.pdf.
1c mark scheme says less effervescence and ph would be lower. Now I don’t understand going down the group 2 metals reactivity suppose to increase. So why is calcium suppose to have lots of effervescence than strontium?
Also how do you distinguishing between acidity and alkalinity in the periodic table ?
Reply 27
Ariale
OP
14
Original post by Ariale
https://pmt.physicsandmathstutor.com/download/Additional-Assessment-Material/Chemistry/AS-level/OCR-A/Component-2/Component%202%20MA%2014.pdf.
1c mark scheme says less effervescence and ph would be lower. Now I don’t understand going down the group 2 metals reactivity suppose to increase. So why is calcium suppose to have lots of effervescence than strontium?
Also how do you distinguishing between acidity and alkalinity in the periodic table ?
1c mark scheme says less effervescence and ph would be lower. Now I don’t understand going down the group 2 metals reactivity suppose to increase. So why is calcium suppose to have lots of effervescence than strontium?
Also how do you distinguishing between acidity and alkalinity in the periodic table ?
This was 1cii
Reply 28
Original post by Ariale
https://pmt.physicsandmathstutor.com/download/Additional-Assessment-Material/Chemistry/AS-level/OCR-A/Component-2/Component%202%20MA%2014.pdf.
1c mark scheme says less effervescence and ph would be lower. Now I don’t understand going down the group 2 metals reactivity suppose to increase. So why is calcium suppose to have lots of effervescence than strontium?
Also how do you distinguishing between acidity and alkalinity in the periodic table ?
1c mark scheme says less effervescence and ph would be lower. Now I don’t understand going down the group 2 metals reactivity suppose to increase. So why is calcium suppose to have lots of effervescence than strontium?
Also how do you distinguishing between acidity and alkalinity in the periodic table ?
The question asks what if you were to replace the strontium with calcium (e.g it’s asking you to describe the reaction of calcium).
As you’ve said, calcium is less reactive than strontium (which you can justify based on an ionisation energy argument) and so the problems are actually agreeing with you.
The pH comes from the relative degrees of solubility of the products formed. Down the group, the hydroxides of the group 2 metals become more soluble in water. That is to say that Ca(OH)2 will be less soluble than Sr(OH)2 and so [OH^-] will be reduced when strontium is swapped out for calcium and therefore the pH at the end of the reaction will be lower.
Reply 29
Ariale
OP
14
Original post by TypicalNerd
The question asks what if you were to replace the strontium with calcium (e.g it’s asking you to describe the reaction of calcium).
As you’ve said, calcium is less reactive than strontium (which you can justify based on an ionisation energy argument) and so the problems are actually agreeing with you.
The pH comes from the relative degrees of solubility of the products formed. Down the group, the hydroxides of the group 2 metals become more soluble in water. That is to say that Ca(OH)2 will be less soluble than Sr(OH)2 and so [OH^-] will be reduced when strontium is swapped out for calcium and therefore the pH at the end of the reaction will be lower.
As you’ve said, calcium is less reactive than strontium (which you can justify based on an ionisation energy argument) and so the problems are actually agreeing with you.
The pH comes from the relative degrees of solubility of the products formed. Down the group, the hydroxides of the group 2 metals become more soluble in water. That is to say that Ca(OH)2 will be less soluble than Sr(OH)2 and so [OH^-] will be reduced when strontium is swapped out for calcium and therefore the pH at the end of the reaction will be lower.
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/4-Core-Organic-Chemistry/Set-H/QP/Organic%20Synthesis%20QP.pdf
For Q1b. I don’t really get it. I can argue that simple distillation prevents any gases from escaping so it can’t be B. So what is the key factor of the differences of reflux and distillation?
For Q3b I got everything right apart from suggest a formula. I despise these questions as they give little context and suppose you to trial and error. Is there a faster way or maybe any hints for the previous question linking to this one?
Reply 30
Ariale
OP
14
Original post by TypicalNerd
The question asks what if you were to replace the strontium with calcium (e.g it’s asking you to describe the reaction of calcium).
As you’ve said, calcium is less reactive than strontium (which you can justify based on an ionisation energy argument) and so the problems are actually agreeing with you.
The pH comes from the relative degrees of solubility of the products formed. Down the group, the hydroxides of the group 2 metals become more soluble in water. That is to say that Ca(OH)2 will be less soluble than Sr(OH)2 and so [OH^-] will be reduced when strontium is swapped out for calcium and therefore the pH at the end of the reaction will be lower.
As you’ve said, calcium is less reactive than strontium (which you can justify based on an ionisation energy argument) and so the problems are actually agreeing with you.
The pH comes from the relative degrees of solubility of the products formed. Down the group, the hydroxides of the group 2 metals become more soluble in water. That is to say that Ca(OH)2 will be less soluble than Sr(OH)2 and so [OH^-] will be reduced when strontium is swapped out for calcium and therefore the pH at the end of the reaction will be lower.
Also how would you simplify the ratio 1.33:2.66:1. With these questions what is the simplest way instead of times by 2 or 100? Examples would be grateful as I’m stuck with numbers like this for my empirical formula ?
Reply 31
Original post by Ariale
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/4-Core-Organic-Chemistry/Set-H/QP/Organic%20Synthesis%20QP.pdf
For Q1b. I don’t really get it. I can argue that simple distillation prevents any gases from escaping so it can’t be B. So what is the key factor of the differences of reflux and distillation?
For Q3b I got everything right apart from suggest a formula. I despise these questions as they give little context and suppose you to trial and error. Is there a faster way or maybe any hints for the previous question linking to this one?
For Q1b. I don’t really get it. I can argue that simple distillation prevents any gases from escaping so it can’t be B. So what is the key factor of the differences of reflux and distillation?
For Q3b I got everything right apart from suggest a formula. I despise these questions as they give little context and suppose you to trial and error. Is there a faster way or maybe any hints for the previous question linking to this one?
Q1: The condenser is present to liquefy any vapours in order to ensure you don’t lose any solvent and any partially reacted reagent goes back into the solution so it can react further. As such, B seems to be the most appropriate answer.
Q3: Using the ideal gas law, you should be able to calculate the moles of gas. In this case, n ≈ 0.0362 mol. We also know that 1.321 g of gas was used, so we can find its molar mass:
(1.321 g)/(0.0362 mol) = 36.5 g/mol (3 sf)
You may simply recognise that this is the Mr of HCl, but it’s more helpful to recall that when chlorinated organic compounds combust, HCl is usually one of the products. If you look at the structure of MAC given to you, you’ll see there is a chlorine atom in its structure and so you’d expect it to form HCl upon combustion.
Reply 32
Original post by Ariale
Also how would you simplify the ratio 1.33:2.66:1. With these questions what is the simplest way instead of times by 2 or 100? Examples would be grateful as I’m stuck with numbers like this for my empirical formula ?
It might be helpful to memorise a few reciprocals:
1/2 = 0.5
1/3 = 0.333…
1/4 = 0.25
1/5 = 0.2
1/6 = 0.166…
1/7 = 0.142857…
1/8 = 0.125
1/9 = 0.111…
1/10 = 0.1
If you see anything that looks like one of these sequences of decimal places, multiply the ratio through by the corresponding number.
The ratio 1.33 : 2.66 : 1 has a sequence of decimals that looks close in value to 1/3, so try multiplying it through by 3:
4.99 : 7.98 : 3
This ratio is very close to 5 : 8 : 3, so this is what you take it as.
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