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Last minute preparation on MCQS 9701 and 9702. Discussion here.

Hows the preparation?

  • Awesome. (Y)

  • Yet not started. :P

  • Subtle. :)


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Recently some friends texted me those. :) IDK from which paper it is. :)
Any one please make me know the answers.
B?
 

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Again answer to this question is simple :p That is IDK :mad:
D?
Reading through the question, I assumed W is Magnesium compound because usually Magnesium compounds are usually used in kilns. I assumed X to be a calcium compound because usually calcium carbonate is used as building material. and yes I read in the book (AS and A level chemistry by Roger Norris pg: 181) that Calcium carbonate is heated in kiln to produce calcium oxide, which is compound Y here.
Compound Y, calcium oxide can be hydrated to produce compound Z (which is Calcium Hydroxide). Calcium hydroxide is used to treat soils but to "increase" the pH of soil, hence #3 is wrong.
#1 is right because Carbonates get less soluble going down the group (hence less alkaline) but here we are comparing Calcium Carbonate (Y) to Mg Oxide. (X). And Oxides are more soluble than Carbonates. Carbonates are always sparingly soluble.
#2 is wrong because Going down the group II, reactivity and solubility increases down the group so Y is more reactive with water than W
 
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Recently some friends texted me those. :) IDK from which paper it is. :)
Any one please make me know the answers.
B?
First of all we'll make an ionic equation after balancing oxidation states.
5Sn^+2 +2MnO4^-1 + 16H^+1 ----> 2Mn^+2 + 5Sn^+4 +8H2O
n(SnCl) = 9.5/190 = 0.05
using proportion:
5 moles of Sn : 2 moles of Mn
0.05 moles of Sn: x moles of Mn
cross multiply and x= 0.02
 
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Can someone explain how this type of Qs are solved?
View attachment 57443
Any oxidising agent gets reduced. So just look at which equation the oxidation no. of the given element gets reduced and you have the answer.
For example, in the above question, the answer is D. H2 oxdn no. is 0 and in NaH it becomes -1. Therefore, it gets reduced and is an oxidising agent.
 
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yup, it's A.

12188756_1645968388999907_2047147922_n.jpg
 
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A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm. This measurement is made using a digital caliper with a zero error of (−0.20 ± 0.02) mm.
What is the percentage uncertainty in the calculated thickness of a single sheet?
A 1.0% B 2.0% C 4.0% D 6.7%
 
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A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm. This measurement is made using a digital caliper with a zero error of (−0.20 ± 0.02) mm.
What is the percentage uncertainty in the calculated thickness of a single sheet?
A 1.0% B 2.0% C 4.0% D 6.7%
Total UC in VC and sheet is 0.04
total thickness = 1
So %UC = 0.04/1 * 100 = 4%

Physicist is it correct way?
 
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