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Chemistry MCQ thread...

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meoooow said:
24 Tartaric acid is present in some wines. It may be synthesised in the laboratory in two steps.
OHCCHO intermediate HO2CCH(OH)CH(OH)CO2H
step 1 step 2
tartaric acid
Which reagents could be used for this synthesis?
step 1 step 2
A HCl (aq) HCN(g)
B HCN, NaCN(aq/alcoholic) H2SO4(aq)
C H2SO4(aq) K2Cr2O7 / H2SO4(aq)
D KCN(aq/alcoholic) K2Cr2O7 / H2SO4(aq

I know its B, but i want to know why its not D!
can someone pleease explain?
KCN(aq/alcoholic) is used only for nucleophilic substitution, there are no groups here to be substituted. but CHO can undergo nucleophilic addition thats why we r using HCN in the first step and then aqueous acid to hydrolys each of the CN groups tp COOH
 
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guyz plz...
10 The value of the equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol
and ethanoic acid is 4.0 at 60 °C.
C2H5OH + CH3CO2H CH3CO2C2H5 + H2O
When 1.0 mol of ethanol and 1.0 mol of ethanoic acid are allowed to reach equilibrium at 60 °C,
what is the number of moles of ethyl ethanoate formed?
A 1/3, b)2/3 c)1/4 ,d)3/4
 
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xHazeMx said:
meoooow said:
24 Tartaric acid is present in some wines. It may be synthesised in the laboratory in two steps.
OHCCHO intermediate HO2CCH(OH)CH(OH)CO2H
step 1 step 2
tartaric acid
Which reagents could be used for this synthesis?
step 1 step 2
A HCl (aq) HCN(g)
B HCN, NaCN(aq/alcoholic) H2SO4(aq)
C H2SO4(aq) K2Cr2O7 / H2SO4(aq)
D KCN(aq/alcoholic) K2Cr2O7 / H2SO4(aq

I know its B, but i want to know why its not D!
can someone pleease explain?
KCN(aq/alcoholic) is used only for nucleophilic substitution, there are no groups here to be substituted. but CHO can undergo nucleophilic addition thats why we r using HCN in the first step and then aqueous acid to hydrolys each of the CN groups tp COOH

ohh, okay i get it! thanks :D
 
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sparten said:
guyz plz...
10 The value of the equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol
and ethanoic acid is 4.0 at 60 °C.
C2H5OH + CH3CO2H CH3CO2C2H5 + H2O
When 1.0 mol of ethanol and 1.0 mol of ethanoic acid are allowed to reach equilibrium at 60 °C,
what is the number of moles of ethyl ethanoate formed?
A 1/3, b)2/3 c)1/4 ,d)3/4
C2H5OH + CH3CO2H CH3CO2C2H5 + H2O

initial moles: 1 ...... 1 ...... 0 ....... 0
At equilibrium: 1 - x ..... 1 - x ...... x ....... x ( as x mole is dissociated )
so, 4 = (x) (x) / (1 - x) (1 - x)
4 = x^2 / (1 - 2x + x^2)
4 - 8x + 4x^2 = x^2
3x^2 - 8x + 4 = 0
solve the quadratic formula and u will get the value of x as 2/3 .. so number of moles of ethyl ethanoate x is equal to 2/3 which is B
 
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its nt a mathematics ppr that u resorted to use QUADRATIC formula.....just take the sqrt on both sides and it makes it hell simple
 
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ShootingStar said:
Question!
Please explain

for question 9,

the answer is D

if in SO3 (Charge : -2) the oxidation number is +3.... when it changes into SO4 (still -2) there is an extra -2 charge due to the additional oxygen ion, but the final charge still remains (-2). So the Sulphur ion must've balanced the extra (-2) . So it must have +5 charge
 
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Can someone answer these please?

Q9 w10_11 In a calorimetric experiment 1.60 g of a fuel is burnt. 45 % of the energy released is absorbed by
200 g of water whose temperature rises from 18 ° C to 66 °C. The specific heat capacity of water is
4.2 J g–1K–1
.
What is the total energy released per gram of fuel burnt?
A 25 200 J B 56 000 J C 89 600 J D 143 360 J

Q 35 w10_11 Disproportionation is the term used to describe a reaction in which a reactant is simultaneously
both oxidised and reduced.

To which incomplete equations does the term disproportionation apply?
1 Cl2(g) + 2OH–
(aq) → H2O(l) + Cl –
(aq) + ……
2 3Cl 2(g) + 6OH–
(aq) → 3H2O(l) + ClO3

(aq) + ……
3 2NO2(g) + H2O(l) → HNO3(aq) + ……


Q 32 S10_11 Which reactions are redox reactions?
1 CaBr2 + 2H2SO4 → CaSO4 + Br2 + SO2 + 2H2O
2 CaBr2 + 2H3PO4 → Ca(H2PO4)2 + 2HBr
3 CaBr2 + 2AgNO3 → Ca(NO3)2 + 2AgBr
 
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the equation is NaCl + H2SO4 -------> NaHSO4 + HCl
if u look to each oxidation number of each element u will find out that there is no change in any of the oxidation numbers so thats why H2SO4 is not acting as an oxidising agent. but it acts as an acid as it is an acid/base reaction which forms a salt
 
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isin't the answer A?

cause from what i know, chlorine is a stronger oxidising agent than sulphuric acid.
 
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hassam said:
its nt a mathematics ppr that u resorted to use QUADRATIC formula.....just take the sqrt on both sides and it makes it hell simple
its a one-step job in the calculator, it can be solved in ur way also
 
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ideggkr said:
hassam said:
welll y it shud nt be B but A???

I think you've discussed this enough with Zishi..

Well

delta H = bonds broken - bonds formed
bonds broken : c-c in graphite (let it be x)
bonds formed : c-c in diamond (let it be y)

x-y = +3
so x is greater than y
Meh, I've explained this question to many people at XPF like 100 times. -_- Answer is A, anyway.
 
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i agree zzishi...bt see it was more -ve than that for graphite..not jxT greater..doesnt this seems to bother u
 
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