OCR B Salter's Chemistry by Design F335 - 15th June 2015 Watch

greencat55
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(Original post by strugglingstdnt)
!!!! can anyone explain question 2 e (ii) on Jan 2011 paper! its so hard and confusing! please)
This question? Is it hard?
Basicaly phenol and sulfur group (acids) donate a proton because it is an alkali solution. Amine group (base) does not change, it only acepts a proton in acidic conditions. NO2 won't be affected.

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Priya265
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#222
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Can someone please explain how you do this Q?

SO2 + 2H2S 3S + 2H2O

(ii) 44.3 g of SO2 are mixed with 44.3 g of H2S.Calculate the maximum mass of sulfur that could be formed.Show your working.


Mark scheme:

1.BOTH moles SO2 = 44.3/64(.1) or 0.69AND moles H2S = 44.3/34(.1) or 1.3 
2.SO2 in excess/ H2S is limiting (AW)
3.mass S formed (= 1.3 x 1.5 x 32.1) = 62.6 g[62.4 if 32 used as Ar](Allow any number between 62.4 and 63)

I dont understand what they did in step 3, or how they got step 2.
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Diamond Crafter
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(Original post by Priya265)
Can someone please explain how you do this Q?

SO2 + 2H2S 3S + 2H2O

(ii) 44.3 g of SO2 are mixed with 44.3 g of H2S.Calculate the maximum mass of sulfur that could be formed.Show your working.


Mark scheme:

1.BOTH moles SO2 = 44.3/64(.1) or 0.69AND moles H2S = 44.3/34(.1) or 1.3 
2.SO2 in excess/ H2S is limiting (AW)
3.mass S formed (= 1.3 x 1.5 x 32.1) = 62.6 g[62.4 if 32 used as Ar](Allow any number between 62.4 and 63)

I dont understand what they did in step 3, or how they got step 2.
Step 2. You need 2 moles of H2S and only 1 mole SO2 to make 3 moles of S thus H2S is the limiting factor NOT SO2.

Step 3. Using the equation: for every 2 moles of H2S you get 3 moles of S thus it is essentially the moles of H2S x (3/2 -> which is 1.5) and then multiply this by the Mr of S to get the mass.

Hope this helps.
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MrChemKid
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#224
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(Original post by Priya265)
Can someone please explain how you do this Q?

SO2 + 2H2S 3S + 2H2O

(ii) 44.3 g of SO2 are mixed with 44.3 g of H2S.Calculate the maximum mass of sulfur that could be formed.Show your working.


Mark scheme:

1.BOTH moles SO2 = 44.3/64(.1) or 0.69AND moles H2S = 44.3/34(.1) or 1.3 
2.SO2 in excess/ H2S is limiting (AW)
3.mass S formed (= 1.3 x 1.5 x 32.1) = 62.6 g[62.4 if 32 used as Ar](Allow any number between 62.4 and 63)

I dont understand what they did in step 3, or how they got step 2.
Step 3 - 1.3 *32 ( moles times Molar mass =mass)

1.5 is the ration ( aka 3 to 2 , in the equation))

I dont get step 2 either lmao
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TheGingerBrit
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(Original post by sunshine12321)
Hi, can anyone help with Jan 2011, question 5f part ii and iii please. I haven't got a clue how they got the answer they did. It's about equilibrium constants and shiz.

Thanks x
Hey Sunshine

I did this paper as practice yesterday! It sure was a mean one :')

I also got these wrong, but on reflection:

ii) We know that, at equilibrium, 0.7mol of the acid has been used up (1.1 - 0.4), which means it must've become the ester. I'm not sure what if any relevance the 1.10 mols of alcohol has, apart from part iii), but it's the best I've got for you. Couple that with the fact it's only a one mark question, and you know they must be looking for something fairly simple.

iii) Draw out the Kc value --> [RCOOR] X [H2O] / [ROH] X [RCOOH]
Take the value you got from ii), and assume RCOOR and H2O both have that value (Carboxylic Acids are strong acids, so will dissociate fully into A- and H+, and we assume they have used 1dm3 of all the substances (I think?))

--> [0.7]^2 / [1.1]^sq = 3.1

Hope I have helped at least somewhat :/

TheGingerBrit
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Priya265
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(Original post by Diamond Crafter)
Step 2. You need 2 moles of H2S and only 1 mole SO2 to make 3 moles of S thus H2S is the limiting factor NOT SO2.

Step 3. Using the equation: for every 2 moles of H2S you get 3 moles of S thus it is essentially the moles of H2S x (3/2 -> which is 1.5) and then multiply this by the Mr of S to get the mass.

Hope this helps.
(Original post by MrChemKid)
Step 3 - 1.3 *32 ( moles times Molar mass =mass)

1.5 is the ration ( aka 3 to 2 , in the equation))

I dont get step 2 either lmao

I get it now, thank you both so much!!
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Logger
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Good luck all!
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hannnahcox
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So... how'd everyone find it? I thought it was ok, gas volume question was weird and not sure about the pH value when you added alkali, but it was ok!

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Millie0118
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Yeah what did you put for the gas volume question? It was the % of ammonia converted to NO. I worked out the % of Nintrogen in Ammonia as only the nitrogen was used in NO.



ALSO what about the bond angle question of C2H4? I put 120 for between the hydrogens but people put 180 for between the carbons???
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username1082804
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What was the H-S-H bond?
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hannnahcox
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(Original post by Millie0118)
Yeah what did you put for the gas volume question? It was the % of ammonia converted to NO. I worked out the % of Nintrogen in Ammonia as only the nitrogen was used in NO.



ALSO what about the bond angle question of C2H4? I put 120 for between the hydrogens but people put 180 for between the carbons???
Ohh right yeah I see what you did, I put some random numbers together and got 34% ish bond angle I got 180 with a triple bond between the carbons!

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Millie0118
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Actually it was C2H2 for that molecule wasn't it :/ damn I drew C2H4 D:
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LordGaben
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Was a nice exam. Made the odd mistake but its out of 120 haha
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BrokenS0ulz
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Wtf at the Kw question (got 12 somehow). Wtf at the 3 peaks in nmr? Thought there was 2 proton environments. Other than that it was fine.
I got propagation x 3 then initiation x 2 at end what did everyone else get?
And what was the linescale forming when that thing was boiled on the first question. Did everyone get -1162 for enthalpy of hydration for co32-?
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Charletitia
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Found that okay?? Know I've dropped marks left right and centre but it wasn't horrific?


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Beasttty
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(Original post by BrokenS0ulz)
Wtf at the Kw question (got 12 somehow). Wtf at the 3 peaks in nmr? Thought there was 2 proton environments. Other than that it was fine.
I got propagation x 3 then initiation x 2 at end what did everyone else get?
And what was the linescale forming when that thing was boiled on the first question. Did everyone get -1162 for enthalpy of hydration for co32-?
I got propagation 4 times and initiation once. because water evaporated so less h2o so equilibrium to the left so caco3 formed more. something like that
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Anon12345678
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#237
What was the long question about benzene?
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emliche
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#238
That exam was so much better than f334 im so glad
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LordGaben
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(Original post by BrokenS0ulz)
Wtf at the Kw question (got 12 somehow). Wtf at the 3 peaks in nmr? Thought there was 2 proton environments. Other than that it was fine.
I got propagation x 3 then initiation x 2 at end what did everyone else get?
And what was the linescale forming when that thing was boiled on the first question. Did everyone get -1162 for enthalpy of hydration for co32-?
When you boil the concentration of water in liquid state decreases. Equilibrium position shifts to left. Yh I got that for hydration. Kw I had no idea just plugged some numbers and hoped for the best. I think I got only 1 as initiation.
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JakeDaw
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Salters had to go throw what I said last night about being able to draw hexagons well in my face by making us draw an octagon. They're just purposely trying to mess with us


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