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Chemistry- need help mcq questions plz!!!

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sunstar:
I'm stuck in june 2005 Q7,10,11,31,32
Nov.2002 Q8, 11, 39. ::) ??? ::)
so, if anyone can help me asap i shall be thankful.

ashish:

--- Quote from: sunstar on November 12, 2010, 10:40:41 pm ---I'm stuck in june 2005 Q7,10,11,31,32
Nov.2002 Q8, 11, 39. ::) ??? ::)
so, if anyone can help me asap i shall be thankful.

--- End quote ---

7) you must use HEss's law

you can manipulate these equations in order to do it
here is one way

ICl(s) + Cl2(g) ---88---> ICl3
                ^                                                       ^
                |                                                        |
                |(0.5*(+14))                                        |(x)
                |

                Cl2 + I2------------------

NOte: i multiplied by o.5 as i have to use 1 mole of ICl to get get 1 mole of ICl3 and in the question enthalpy change for 2moles of ICl is given

now solving for x
x= 7-88
x=-81
ans = c

11)A because the concentration must become a constant value,

31)B

Al has a configuration of 2.8.3(1s2 2s2 2p6 3s2 3p1)
in its outermost shell it has only 3 electrons in excited state .(1s2 2s2 2p6 3s1 3p2)

                                                Cl
                                                ..
                                            Cl:Al:Cl

trigonal planer since there is no lone pairs of electron, i have shown only bonding  electrons


C has e- configuration of 2.4(1s2, 2s2, 2p2)
C in the excited state (1s2, 2s1, 2p3)
C+ will have 2.3(1s2, 2s1, 2p2)
 
                                     H
                                     ..
                                  H:C:H

no lone pair of electrons here also, trigonal planer

P 2.8.5
1s22s2 2p6 3s2 3p3
in this case in the excited state the configuration will be the same

                                                            H
                                                            ..
                                                         H:P:
                                                            ..
                                                            H

there is a lone pair of electrons this is not a trigonal planar
Pyramidal shape, 107 degree

QU10 and 32 i have not reply :(

NOv02
NO.8

PV = nRT
P=nRT

n=moles
 
for hydrogen no. of moles = 2/1
                                   =2
P=2RT/V

A) moles of   Deuterium= 2/2
                               =1
P=RT/V

B)moles=4/2
          =2
P=2RT/(V/2)
P=4RT/V

C moles of H=1
   moles of deuterium= 1
P=(1+1)RT/V
P=2RT/V

hence answer is C

11,39 i have not reply :(
  


                                                            
 



      

Deadly_king:

--- Quote from: sunstar on November 12, 2010, 10:40:41 pm ---I'm stuck in june 2005 Q7,10,11,31,32
Nov.2002 Q8, 11, 39. ::) ??? ::)
so, if anyone can help me asap i shall be thankful.

--- End quote ---

I'll only answer those that ashish did not. ;)

Jun 05

10. The forward reaction is endothermic as indicated by the positive enthalpy change.

Hence according to Le Chatelier's principle, an increase in temperature will only favour the forward reaction in order to oppose the change. This will result in an increase in pressure since there will be two moles of NO2 gas formed at the expense of only one mole of N2O4 gas.

An increase in pressure only occurs when the volume increases or when the vessel's volume decreases. In this case the volume of the mixture will increase. From here we can already eliminate C and D.

A gas always expand when temperature is increased. Expansion also implies an increase in volume.

Hence answer is B

32. We'll proceed by elimination.

1. The mass of Y used in the experiment was ---> That's true since even as temperature continues to rise, the mass of vapour does not increase further showing that all the solid has turned into vapour.

2. The pressure of the vapour was constant for all the temperatures above . ----> That is not the case. Increase in temperature leads to expansion. This will cause the volume as well as the pressure to rise.

3. Liquid appeared at temperature -----> It may be possible but there is nothing on this graph which indicates it was the case. So we cannot take it for granted. ;)

Therefore answer is D

Nov 02

11. Kp = (H2).(I2) / (HI)2 at equilibrium.

It has been given that at equilibrium mole of HI has been dissociated. This means that ( - ) moles is left. As for the number of moles of products; H2 = /2 and I2 = /2

Hence Kp = (/2).(/2) / ( - )2

This ends up to Kp = 2 / 4( - )2

Hence answer is D

39. Ethanol dissociates to form the corresponding ions as shown :
CH3CH2OH <=> CH3CH2O- + H+

When H2SO4 is added, there is an increase in H+ ions. Hence according to Le Chatelier's principle, the equilibrium will shift to the left forming more ethanol at the expense of its corresponding ions.

Hence the ions present will be mainly HSO4- and CH3CH2O+H2 as H+ combines with ethanol.

The first one is rejected since it will have the tendency to form ethanol and will not be present.

Hence answer is C.

Hope it helps :)

sunstar:
Can u please explain me again june 2005 Q11 and Q31
nov 2002 Q39 {How is the first one having the tendency to form ethanol } sorry still didn't understand and plz i need help in june 04 Q2, 12, 36,
thank you very much.

sunstar:

--- Quote from: sunstar on November 14, 2010, 12:07:45 pm ---Can u please explain me again june 2005 Q11 and Q31
nov 2002 Q39 {How is the first one having the tendency to form ethanol } sorry still didn't understand and plz i need help in june 04 Q2, 12, 36,
thank you very much.

--- End quote ---

plzzzz need help in these questions urgently. thank you.

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