Chemistry: Post your doubts here!

Student12

Student12

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smzimran said:
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!
:)
Click to expand...
Thankk you :D How about Q30 and 39 ?
 
Romeo10

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Please i need someone to explain to me question no 8 in June 2010 paper 11 .. why it is magnesium??
and also i need the next question
thanks
 
G

geek101

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xhizors said:
Anyone Plz ?????
Click to expand...


Q1. From the table X is a substance with the mp less than zero and bp more than zero so its a liquid >> the options fr X are I2 and Br2. Iodine is a solid at rtp and Br2 is liquid so is Br2. Then Y and Z both have mpoints and bps above 0 so they are solids, BUT they have low densities >> from the options B is the only one with 2 metals with a low density. Na and Mg (1st and 2nd group so low density), tadaa the answer is B

Q2. this is simple calculation. write out the equation of the reaction. Using mole ratios or proportions find out the mass of bromine. the key here is to write the right equation, which is not hard...just balance it correctly.

Q4. Hmm...here the structure given is complicated, so just try to link it to the ones you know from the syllabus. Here ethene has a similar bond structure (just replace the N with a C). We know ethene has angles of approximately 120. So the answer is C.

Q6. First you divide the total enthalpy by 2. to get the enthalpy for one NF3. You get +834. Then to find the enthalpy for one N-F bond you divide by 3 since the compound has 3 N-F bonds. You get +278! C

Q8. here HSO3- changes to SO32- it does so by gaining a proton. see the reactants side only. To make sure look at the top reaction. HSO3- is becoming SO2 <<here it loses a OH- ion to become SO2, which is the property of bases.

Q10. The forward reaction in endothermic. An increase in temp favors the forward reaction and HI is formed, making the solution paler. So D.

excuse me to do only these. sorry. ill definitely post the rest when i have time :)
 
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maleeha123

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please someone explain me these questions Q:2,22,27
 

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smzimran

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Student12 said:
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 ??
Click to expand...
Q30:
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!
 
smzimran

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maleeha123 said:
please someone explain me these questions Q:2,22,27
Click to expand...
Aoa wr wb,
Q2:
In step A, o.s of nitrogen changes from +5 to +3
In step B, it changes from +3 to +2
In step C, it changes from +2 to +1
In step D, it changes from +1 to 0

In steps B,C and D oxidation number decreases by one unit
In step A, oxidation number decreases by 2 units

This is why answer is A.
:)

Q22:
CH3 as an electrophile means CH3+
It has 4 electrons of carbon + 3(1) electrons of hydrogen - 1 electron due to positive charge = 6 electrons

CH3 as free radical means .CH3
It has 4 carbon + 3(1) hydrogen = 7 electrons

CH3 as nucleophile means CH3-
It has 4 carbon + 3(1) hydrogen + 1 due to negative charge = 8 electrons

So, the correct answer is A.
P.s: carbon has 6 electrons in total but i have taken 4 because they are asking only for outer shell(valence) electrons in the question!

Q27:
Because they are talking about ethanoate means you have to look at the acid group to be ethanoic acid
A : propanoic acid (3 carbon)
B : propanoic acid (3 carbon)
C : methanoic acid (1 carbon)
D : ethanoic acid (2 carbon)

That is why D is correct.
:)
 
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trgirl

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9701_s11_ms_21.pdf
can anyone help me for question4 a)i and ii) i dont get why the answer is nucleophilic addition???
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_ms_21.pdf
 
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trgirl

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i am really confused what is the difference btw nucleophili addition and nucleophilic substution? can someone explain it to me properly?
 
Student12

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smzimran said:
Q30:
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!
Click to expand...
Thank you! May Allah bless you :)
 
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