Chemistry: Post your doubts here!

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sagar65265

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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!
 
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sagar65265

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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!
 
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Abdel Moniem

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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
 
GCE As and a level

GCE As and a level

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GCE As and a level

GCE As and a level

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GCE As and a level said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9 , Q26 . Q37
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Q21 , 23 , 32
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q20 , Q25
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
Q25,26,38
PLZ GUYS
Click to expand...
GUYS PLZ even if only 1 Q :D
 
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