Chemistry: Post your doubts here!

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Help please!!!
Well you should always have tried something, i.e. outline your ideas then people can just help to guide you along. Remember, if you don't do that and just rely on someone explaining from zero, you risk losing it all big time in the real exam. By then it becomes too late.

Go for option 2 => since these involve specific isotopes, they both have mass of 34. Even if you remove some of the dioxygen, the average mass will remain the same.

Check number 1. Same deal with option 2. Average mass will just be 34 because all the molecules present have mass of 34.

Check number 3. H2S + 3/2 O2 --> SO2 + H2O (to balance the eqn - gcse/o level, no). I leave you to think about limiting and excess reactant based on the reacting mole ratio, and the information that you have equal number of these two molecules in the mixture. It all comes down is 3 correct or not.
 
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Since $$H_2SO_4$$ is a strong acid it will dissociate fully into $$H^+$$ ions, $$HSO_4^-$$ is the conjugate base and further dissociates into $$SO_4^{2-}$$ ions and protons therefore the solution wont contain an equal number of $$H^+$$ and $$HSO_4^-$$ ions so 2 is incorrect, hence since 2 is incorrect the only option available is D.

Also wait r u from sir shoukats class because i swear i just answered this question there rn
 
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chemistry past paper mayjune 2011 paper 22. (s11_22).
number 1. 1ci.
I still couldn't process why it is directly 0.04. I thought 0.04 reacted so why didn't it subtract 1-0.04 = 0.06 for ethanoic acid?
 
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i have attached two files here: a question paper and its markscheme. Please do help me with number 6b as soon as possible. Why 19.10 (which I think is got directly from the titre of titration 5). Why not sum up all titre values then divide by 5, I had got 19.23? How am I calculating wrong, or where am I not understanding the question?
 

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http://www.xtremepapers.com/CIE/Interna ... _qp_11.pdf

number 4. Why is the answer C and not D? With a catalyst, activation energy is lowered isnt it? :pardon:
in diagram 1, the higher temperature must be Q due to more molecules with high energy/ energy greater than ea, and less molecules with low energy.
in diagram 2, yes catalyst lowers the Ea, so only a small value of molecular speed is required for particles to make successful collisions. Less Ea, less speed necessary for particles. So it's X.
Thus C.
 
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http://www.xtremepapers.com/CIE/Interna ... _qp_11.pdf

number 4. Why is the answer C and not D? With a catalyst, activation energy is lowered isnt it? :pardon:
in diagram 1, the higher temperature must be Q due to more molecules with high energy/ energy greater than ea, and less molecules with low energy.
in diagram 2, yes catalyst lowers the Ea, so only a small value of molecular speed is required for particles to make successful collisions. Less Ea, less speed necessary for particles. So it's X.
Thus C.
 
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i have attached two files here: a question paper and its markscheme. Please do help me with number 6b as soon as possible. Why 19.10 (which I think is got directly from the titre of titration 5). Why not sum up all titre values then divide by 5, I had got 19.23? How am I calculating wrong, or where am I not understanding the question?
If you had done any A Level titrations, you should be aware of what concordant titres within +/-0.10 cm3 is the maximum acceptable. You are going to be marked within that margin relative to that of your teacher/centre supervisor who has to do the titration using the same batch of chemicals you did, and that is how titration accuracy marks are marked. Looking at the five readings, you should be able to see that reading 1 is comparatively different from the others. I didn't check all the titres, I just quickly glanced over them, so you get to them again.

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NOVEMBER 2015 P11 QUESTION 16

X is the oxide of a Period 3 element. X reacts with water to give an acidic solution.
A solution is prepared by reacting 0.100g of X with excess water. This solution was neutralised
by exactly 25.0cm3 of 0.100 moldm–3 sodium hydroxide solution.

What could be the identity of X?
A A l2O3
B MgO
C P 4O10
D SO3


Can anyone explain why answer is D?
 
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hi. so3 + 2naoh ->na2so4 + h2o
mol of NaOH = 2.5 x 10^-3 mol = 2 x mol of SO2
mol of SO2 = 1.25 x 10-3 mol
Mr of X oxide = mass / mol = 0.1 / 1.25x10^-3 = 80 g/mol

SO3 is (3 x 16) + 32 = 80 g/mol

i did this backwards, knowing the answer as you've told. u can do it forwards by writing down each balanced equations and see which one fits.
 
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here, can somebody explain me question 16 part (c) ??
the markscheme answer is pentane.
i wasn't able to find even a clue
I guess the question was wrong because as according to the above equation none of the options are correct. Is this a whole question from a paper? if yes please mention the year and variant.
if this is your school exam they might have mixed the question options
 
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