Chemistry- need help mcq questions plz!!!

[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]

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.

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]

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.

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. K_p = (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 K_p = (/2).(/2) / ( - )²

This ends up to K_p = 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]

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.

plzzzz need help in these questions urgently. thank you.

[Deadly_king]

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.

Nov 02 Q39

It will have a tendency to form ethanol since the equilibrium will be shifting to the left. Hence most H⁺ and CH₃CH₂O^- will combine together to form ethanol such that they'll be almost absent in the resulting mixture.

Jun 05 p1

11. I'll proceed by elimination.

B cannot be good since rate of formation of methanol is not gong to increase slowly. Rate of reaction given by the gradient should be highest when t = 0.

C also cannot be the answer since rate of reaction is not constant. As reactants decrease in quantity the rate will decrease as well.

D starts well but the concentration of methanol cannot come to a maximum and then decrease a bit until it becomes constant since this is not a reversible reaction.

This leaves us with A.

31. Trigonal planar is the structure of molecules having 3 bond pairs of electrons and zero lone pairs of electrons.

Aluminium has 3 electrons in its outermost shell and in AlCl₃, it is sharing all three of them, one with each Cl atom. In other words this molecule has got 3 bond pairs and 0 lone pairs. So it has a trigonal planar shape.

Carbon has 4 electrons in its outermost shell and CH₃ it is sharing 3 electrons and the + sign indicate that it has lost the remaining electron. Hence it is also trigonal planar in shape.

Phosphorus has got 5 electrons in its outermost shell. In PH₃, it is sharing 3 of them with each H atom such that a lone pair of electrons remains. Therefore, despite having 3 bond pairs it also has one lone pair of electrons which repel all the bond pairs to form a pyramidal shape. Hence it is not trigonal planar.

Answer is B since only 1 and 2 are good.

[Deadly_king]

Jun 04 p1

2. You just need to find the equation which balance appropriately.

The obvious answer is B since in the other 3 hydrogen is not balanced.

If you don't know how to balance an equation, you need to review your basics. Try here or here. They might help you.

12. Most rapid reaction is one with the lowest activation energy. So A, B and C are eliminated right away.

Hence answer is C.

36. Hydrogen iodide is not purple. It is the iodine vapour (at first iodine gas is emitted, it will then solidify since iodine exists as a solid at room temperature) which results from the decomposition of Hydrogen Iodide that is violet in colour.

So number 1 is rejected. The only answer left is C.

[$!$RatJumper$!$]

S05 Q1 Please... im SO lost in it.
Thank you

[Deadly_king]

S05 Q1 Please... im SO lost in it.
Thank you

Combustion of hydrocarbon : That was the first number in p2. So I doubt it'll come again.

Anyway, the equation is as follows.

Hydrocarbon + Oxygen ----> CO₂ + H₂O + O₂(excess)

Volume of hydrocarbon used : 10cm³
Volume of oxygen used : (70 - 20) = 50cm³
Volume of CO₂ formed : 30cm³
Volume of H₂O liberated : (80 - 50) = 30cm³

NOTE : The volume of H₂O liberated is calculated by subtracting the volume of CO2 and excess O2 from the total volumes of reactants.

Ratio of hydrocarbon to CO₂ is 10:30 ----> 1:3
So there need to be 3 carbon atoms in the hydrocarbon.

Since volume of CO₂ is equal to volume of H₂O, the hydrocarbon contains twice number of H atoms and hence is an alkene.

Answer is B

[$!$RatJumper$!$]

Many thankx

[Deadly_king]

Many thankx

Mention not dude

[$!$RatJumper$!$]

Answer is actually C.. :/

[elemis]

Answer is actually C.. :/

I think I remember this question. The answer is C3H8 right ?

I'll explain how.

[elemis]

10 C_xH_y + 50 O₂ ----------> 30 CO₂ + 40 H₂O

Since 10 cm3 of the Hydrocarbon is used along with 70 cm3 of Oxygen the hydrocarbon is the limiting agent. Hence, 10 moles of Hydrocarbon reacted.

Since 20 cm3 of Oxygen is left unreacted we insert 50 moles into our equation (above).

We are told there are 30cm3 of CO2 so we insert 30 moles into the equation.

Only H2O remains unbalanced. So we insert 40 moles of H2O to balance it.

We have a balanced equation.

We simplify the molar ratios by dividing by ten to give :

1 C_xH_y + 5 O₂ ----------> 3 CO₂ + 4 H₂O

Notice there are 3 moles of C on the products side as well as 8 (4*2) moles of H ?

Therefore, the hydrocarbon must be of formula C₃H₈

[$!$RatJumper$!$]

Thankx but why are there 40 moles of water?