• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

dumb human

dumb human

Messages
338
Reaction score
868
Points
93
Suchal Riaz said:
i recently scored 40/40 in 31 minutes(physics paper) and then 39/40 in 41 minutes(chemistry)
when i started doing papers i was at 27. the marks increased with practice till maximum of 36/40. with more and more practice and if only i am totally awake when doing paper i score 38+ otherwise 37 is an average.
Click to expand...
Chemistry June 2013, Paper 1_2 Question 31?? y A?
 
ZaqZainab

ZaqZainab

Messages
1,764
Reaction score
3,472
Points
273
K

kingo44

Messages
164
Reaction score
226
Points
53
3 In aqueous solution, the acid HIO disproportionates according to the following equation where m,
n, p and q are simple whole numbers in their lowest ratios.

mHIO → nI2 + pHIO3 + qH2O

This equation can be balanced using oxidation numbers.

What are the values for n and p?

n p
A 1 2
B 2 1
C 4 1
D 4 2
 
U

unique111

Messages
420
Reaction score
1,426
Points
153
kingo44 said:
3 In aqueous solution, the acid HIO disproportionates according to the following equation where m,
n, p and q are simple whole numbers in their lowest ratios.

mHIO → nI2 + pHIO3 + qH2O

This equation can be balanced using oxidation numbers.

What are the values for n and p?

n p
A 1 2
B 2 1
C 4 1
D 4 2
Click to expand...
Here's the link to the solution. But i don't get it. Hope you do. Plz, explain to me too. :)
https://www.xtremepapers.com/commun...-for-chemistry-only.19324/page-23#post-609519
 
S

sagar65265

Messages
227
Reaction score
571
Points
103
unique111 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf
Q.4 C
Q.6 B (why not C?)
Q.10 A
Q.11 C ( why not A? Doesn't Br go from -1 to 0?)
Q.17 c
Q.18 D ( but question says made from water? :confused:
Q.26 A
Q. 29 D
Q. 35 B
Q.37 D
Q. 40 A
Please, somebody clear these doubts?
Click to expand...

Q4)

Alright, so if 120 kilograms of hydrogen are used initially and at equilibrium, 96 kilograms of hydrogen remain, we can say that 120 - 96 = 24 kilograms of Hydrogen have been used up. (Note, that here we are concerned with H₂ molecules).

Since the mass of 1 mole of H₂ molecules is 2 grams, and 24,000 grams of Hydrogen have been used, we can say that (24,000)/(2) = 12,000 moles of H₂ molecules have been used in the reaction.

This reaction is as follows : N2 + 3 H2 → 2 NH3

So, we note that for every 3 moles of H2 used in the reaction, we get 2 moles of Ammonia. Therefore, for 1 mole of H2, we should get 2/3 moles of Ammonia. Since we have used up 12,000 moles of H2, we should get (2/3) * (12,000) = 8,000 moles of Ammonia in the final reaction.

We also know that 1 mole of Ammonia has a mass of (14 + 3 * 1) = 17 grams = 0.017 kilograms. Therefore, the mass of 8,000 moles of Ammonia should be equal to
0.017 * 8000 = 136 kilograms = C.

Q6)

From the last column alone we can make out quite a lot - Ammonia and HCl are soluble in water, and so A and D can be eliminated (Ammonia because it can form co-ordinate bonds with water - I think - to form NH₄OH, and HCl since we all know it's an acid and it wouldn't be an acid without being able to dissolve in water).

We also know that MgO forms ions when molten - this means that it's liquid conductivity should be good. Also, since it's lattice energy (A2, i think, but if you've gone through this it might hopefully make sense) is rather high due to it's doubly charged ions, it would be difficult for hydrogen bonding alone to cut through that bonding and let the ions in the lattice dissolve, so MgO is insoluble.

Silicon Dioxide, however, is mostly covalent, made up of Silicon and Oxygen atoms bonded together by Covalent bonds. Therefore, when it melts, it will not split up into ions and so will not respond critically to an electric field. Therefore, pure liquid Silicon Dioxide should not be a good conductor.

This is most likely why the answer is B instead of C - i wasn't able to draw much from the second column of the answers, NaF and KCl look to be good candidates for the characteristics shown in the options, but there was a clear difference in the first column, which I think is what decides the answer.

I'll post the rest whenever I get time.

Hope this helped!
Good Luck for all your exams!
 
S

sagar65265

Messages
227
Reaction score
571
Points
103
unique111 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf
Q.4 C
Q.6 B (why not C?)
Q.10 A
Q.11 C ( why not A? Doesn't Br go from -1 to 0?)
Q.17 c
Q.18 D ( but question says made from water? :confused:
Q.26 A
Q. 29 D
Q. 35 B
Q.37 D
Q. 40 A
Please, somebody clear these doubts?
Click to expand...

Q10) I'll try to do a proper explanation later, i'm a little fuzzy on this one still.

Q11)

When two different atoms are bonded together, from what I recall (well, this page confirms it, either ways) the more electronegative kind of atom receives the negative value of the oxidation state. So, since Fluorine is the most electronegative atom, in any molecule that contains Fluorine, it has a negative oxidation state and number.

So, in Trifluorobromine (I think that's the name, i'm not sure) BrF3, we can see that Fluorine is present and thus gets the negative oxidation number. So, the oxidation state of Fluorine is -3 in BrF3( i'm just assuming that you know where to get the 3 from - the negative sign is the reason for this explanation). To ensure that the molecule has no net charge, the other atoms in the molecule should have oxidation states adding up to +3. Since the only atom in the molecule is a single Bromine atom, it has to have an oxidation number of +3.

After the reaction is complete, we can see that Bromine forms it's molecular state. In a single-atom-type molecule (Like O2, F2, etc) the oxidation state of any of the atoms in that molecule is zero. As ChemGuide puts it,
    • The oxidation state of an uncombined element is zero. That's obviously so, because it hasn't been either oxidised or reduced yet! This applies whatever the structure of the element - whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon.
Therefore we can say that Bromine goes from an oxidation state of +3 to 0. This is, clearly, a decrease in the oxidation state, and is therefore reduction.
In case you aren't sure why C is the answer (rather than why A isn't the answer) just let me know as soon as possible.

Q17) I'll post this later, it's going to take me some time to figure this one out.

Q18)

The first part of the question speaks only about the formation of fresh lime mortar. This formation requires water, but that is not the focus of the question.
The focus of the question is finding out how "Fresh" lime mortar turns into "Old" lime mortar. So the equations given are in no way related to the first line, as far as I know. That line is just to provide some sort of connection between the question material and the syllabus, I guess.

Having established that, we notice one thing - "fresh" lime is made using quicklime - having looked Calcium Oxide on Wikipedia, it turns out this is called quicklime. Therefore, "Fresh" lime is something formed from CaO reacting with H2O. This seems most likely to be Ca(OH)2.
Whatever it is, we know that almost by definition, "fresh" lime is not Calcium Oxide, since it is actually the result of a reaction between CaO and something else.
This pretty much straight away eliminates A and B, since they show "fresh" lime turning to "old" lime with CaO turning into something else.

(Personally I can't think of any other way to eliminate the first two options, really - hopefully the above paragraph makes sense).

Either ways, the most important part of this question is noting that "Both fresh and old lime mortar react with aqueous hydrochloric acid but only the old lime mortar effervesces during the reaction." This basically tells us that we have to pick out the equation that has some compound on the "reactants" side that does not effervesce during reaction with HCl, and some compound on the "products" side that does effervesce during reaction with HCl.

Let's see which option meets these criteria - since we have eliminated A and B, let's start with option C.

On the left side, we have Ca(OH)2 - on reaction with HCl, we obtain Calcium Chloride CaCl as one product, and water as the other, H2O. Okay, no effervescence.
On the right side, we have CaO - on reaction with HCl, this produces Calcium Chloride and water, again. No effervescence?! This is supposed to be effervescent, unless it isn't the answer - which, really, is the only workaround here as far as I can see.

So, if we have eliminated A and B, and C seems to be wrong by the brief of the question, then we can say that D is the only option left, and so is the right time.
But let's just check to make sure;)
On the left side, we have Ca(OH)2 - on reaction with HCl, we obtain Calcium Chloride CaCl as one product, and water as the other, H2O. No effervescence. Okay.
On the right side, we have CaCO3 - on reaction with HCl, we obtain Calcium Chloride, water and Carbon Dioxide. Effervescence! So this is clearly right.

Hope this helped!
Good Luck for all your exams!
 
A

Abdel Moniem

Messages
162
Reaction score
249
Points
53
Count the number of hydrogen and carbon present, make
CH3CH2CHO, CH3COCH3
CH3CH2COOH, CH3COOCH3--> when ethanoic acid reacts with alcohol
CH3CH=CH2, CH3COCH3
Ratio in aldehydes: 3C:6H
Ratio in Ketone : 3C:6H
Ratio in carboxylic acid : 3C:6H
Ratio in ester: 3C:6H
Ratio in Alkene: 3C:6H
 
You must log in or register to reply here.
Top