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chemistry atp

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Mostly Easy.
Titration ques seemed quite normal to me.
Final answer i.e. the percentage of FeSO4 was 8.79%
The very last answer was 8 something, right? (Grams of CaCO3 remaining in the reaction flask)
All in all, alhamdolillah, it was a decent exam.
 
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I actually got 8.81% FeSO4 because I used 3 significant figures
 
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I made a mistake on the first part of the titration, I calculated the average volume wrong, I lost one mark there. Because of that my whole titration could be wrong, so will there be an Error Carried forward(e.c.f)?
 
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I made a mistake on the first part of the titration, I calculated the average volume wrong, I lost one mark there. Because of that my whole titration could be wrong, so will there be an Error Carried forward(e.c.f)?
the first and third reading were to be ticked right?
 
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Variant 42. Can anyone tell me what alcohol X and Z were?
Yes, the first and third reading were the best results. You should get around 1.42g FeSO4 at the end.
 
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Variant 42. Can anyone tell me what alcohol X and Z were?
Yes, the first and third reading were the best results. You should get around 1.42g FeSO4 at the end.
Do you remember titration answers? Was one of the answer was 21.21% ..?do you remember anything else?
 
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Do you remember titration answers? Was one of the answer was 21.21% ..?do you remember anything else?
Percentage of N in (NH4)2SO4 was 21.2%.
The titration answers were something like these:
25.3 cm^3
0.000506 mol
0.00253 mol
0.0253 mol
1.4168 g (I wrote this as 1.42 g because of the Mark Scheme's strict adherence to the 3 s.f. thing)
8.79% (I got 8.81%)

I am confused about the alcohol thing. What'd you put for X and Z?
 
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Alcohol X and Z,don't know if I am
Percentage of N in (NH4)2SO4 was 21.2%.
The titration answers were something like these:
25.3 cm^3
0.000506 mol
0.00253 mol
0.0253 mol
1.4168 g (I wrote this as 1.42 g because of the Mark Scheme's strict adherence to the 3 s.f. thing)
8.79% (I got 8.81%)

I am confused about the alcohol thing. What'd you put for X and Z?

i dont know if i am right but i wrote
C2H7O for X and C4H9O for Z...as far as i can remember
 
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Do you mean C2H5OH for X and C4H9OH for Z? I wrote the same thing.

dont remember but yeah something like that!i think the one with lesser temp should had lesser carbons and hydrogen while one with higher temp had more carbons and hydrogen?
 
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anyways do u remember what was the mass of l which had to calculated at the start of titration by subtracting the mass of container? i may have done it wrong :(
 
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It's definitely 8.2g. They asked for what mass remained. So you had to subtract 1.8g from 10g.
 
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