- Messages
- 121
- Reaction score
- 32
- Points
- 38
But the mark scheme of this paper( O/N 03) says the answer is B...
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Oh yes, a little mistake by me, check i corrected the first part!But the mark scheme of this paper( O/N 03) says the answer is B...
Thankk you How about Q30 and 39 ?Aoa,
Q1:
Mr (TiO2) = 47.9 + 16 + 16 = 79.9
Mr (FeTiO2) = 55.8 + 47.9 + 3(16) = 151.7
FeTiO3 ---> TiO2 + FeO
molar ratio b/w FeTiO3 and TiO2 is 1:1
so
151.7 g of ore gives 79.9 g of TiO2
1 g of ore gives (79.9 / 151.7 g) of TiO2
19 tonnes of ore will give
= 19 * (79.9 / 151.7 g)
= 10 tonnes
So, A is the answer!
Q4:
The hydrogencarbonate anion is (HCO3)-1
Total electrons = 1 of H + 6 of C + 3(8) of O + 1 extra electron due to -1 charge
= 1 + 6 + 24 + 1
= 32
So, C is the answer!
Q22:
B and C have two similar groups (methyl) attached to one side of carbon having double bond, that means there cannot be cis-trans isomers for these
A will form cis-trans isomers if 1 H is substituted on the R.H.S but it will not have a chiral centre
D already has cis-trans isomers, and it will have a chiral centre at the RED carbo atom if one of the BLUE hydrogen is replaced by halogen!
CH3CH=CHCH2CH3
So, D is correct!
Anyone Plz ?????AsSalamalikum wr wb, i have some question from w11 qp 11 Please help, May Allah(swt) help you and make your way easy for Distinctions Ameen.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf
Q4
Q6
Q2
Q1
Q8 why ? HSO3- is donating H means proton donor means acid
Q10
Q14
Q15
Q18
Q16
Q20
Q3
Q9
Q22
Q23
Anyone Plz ?????
What is the colour of hydrogen bromide gas?
Q30:http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
Can anyone explain 1,4,22,30 and 39 from this paper plz ??
Aoa wr wb,please someone explain me these questions Q:2,22,27
Thank you! May Allah bless youQ30:
Mr(Ethanol) = 46
Mr(Ethanoic acid) = 60
Mr(Ester) = 88
n(Ethanol) = 30 / 46 = 0.652
n(Ethanoic acid) = 30 / 60 = 0.50
That means ethanoic acid is the limiting reactant, so ester's yield should be checked by ethanoic acid and NOT by ethanol
n(Ester) = 22 / 88 =0.25
yield = (0.25 / 0.50) * 100% = 50%
So, C is the answer!
Q39:
During bromination of propane (C3H8),
the free radical will be .C3H7
It can be of two types:
(i) CH3 - .CH - CH3
(ii) .CH2 - CH2 - CH3
In option 1, one radical of type(i) and another radical of type(ii) are combining, so this is correct
In option 2, two radicals of type(ii) are combining, so this is also correct
However, in option 3, one butyl radical is combining with one ethyl radical, so this cannot be formed in bromination of propane!
This is why B is the correct answer!
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