(Each and every questions are not answered, they will be added soon and in any chaos please comment)

1    Use of the Data Booklet is relevant to this question. 

     What could be the proton number of an element that has three unpaired electrons in each of its 

     atoms? 

     A    5                 B    13                C    15                D    21 

Solution:

To solve the question, you need to make the box diagrams of the above listed elements and hence fill it.

Since, the question asks you to give the element which have three unpaired electrons, we need to find out the half filled "p" orbital as it only can have three unpaired or half filled orbital i.e.

proton number 15=1s2 2s2 2p6 3s2 3p3 (box diagram will show you the three boxes of p orbital consist of only one electron in each

solution:

bond breaking process is endothermic (positive value) and bond formation is exothermic (negative value). So, lets calculate the enthalpy changes of reaction P, Q, R, S.

P- Br-Br bond gets broken, hence, enthalpy change is +193 KJ/mol

Q- Cl-Cl bong is formed, so enthalpy change is -244 KJ/mol

R- C-Cl bond is formed, so enthalpy change is -340KJ/mol

S- C-H bond is broken hence enthalpy change is +410 KJ/mol

Now, lets sort this out and we get the order R<Q<P<S

answer is c

solution:

first thing to do in this type of question is to draw the hess cycle so we too will do that here:

the equation will be 

2ΔHf=(+38) + (-214)

      =-176/2

      =-88

the answer is B

      

solution:

note that NH4 NO3 is the ionic compound hence it charge is NH4 + and NO3-  i.e. the oxidation number of nitrogen is -3 and +5 respectively. 

and on reactant side the oxidation number of nitrogen is +1.

hence the change in oxidation no. is 

+1-5=-4 and +1-(-3)=+4

so the answer is D

7 The Haber process for the manufacture of ammonia is represented by the following equation. 

solution:

Since, the reaction is exothermic it releases heat and and the temperature of system increases. According to Le Chatelier's principle, the equilibrium reaction strives to resist the change an hence to decrease the temperature the reaction shifts backward.

now note that the pressure is different, when the reaction shifts backward the pressure increases and this change is also resisted and the reaction shifts forward too.
so A is answer.

solution:

1 mole of the metal reacts with excess of chlorine to produce one mole of the compound.

since, the no., of moles is same, it can be written as

n(X)=n(XCl2

or, 2.920/mr=5.287/(mr+71)

or, mr=87.6....which is strontium

so the answer is D

solution:
lets calculate the no. of moles of each compound and then multiply the answer by 24 (because 1 mole of gas occupies volume of 24dm3 at NTP.
you will get answer C




10   The table gives the concentrations and pH values of the aqueous solutions of two compounds, X
     and Y. Either compound could be an acid or a base.

                                                                          X                         Y

                                 concentration            2 mol dm–3         2 mol dm–3

                                 pH                                    6                           9

     Student P concluded that X is a strong acid.

     Student Q concluded that the extent of dissociation is lower in X(aq) than in Y(aq).

     Which of the students are correct?

     A     both P and Q

      B    neither P nor Q

     C     P only

      D    Q only


solution:
the pH of strong acid is less so P cant be true.
and the pH of base is high and since the pH is 9 it can comparatively stronger base which ,means its extent of dissociation is high. so Q is right.
hence the answer is D.






15    The percentage of ammonia obtainable, if equilibrium were established during the Haber process,
     is plotted against the operating pressure for two temperatures, 400 °C and 500 °C.

     Which diagram correctly represents the two graphs?

solution:
the chemical equation of Haber process is:
N2 + 3H2 ↔2NH3
When pressure is increased, the reaction resists the change and tries to oppose the change which can be done by forward reaction. hence, when  pressure increases the reaction shifts forward and ammia yield increases. (c and d incorrect)
Haber process is exothermic so when temp decreases the reaction shifts forward to oppose the change and hence ammonia yield increases.
so answer is A.





24    What is formed when propanone is refluxed with a solution of NaBH4?

     A     propanal

      B    propan-1-ol

      C    propan-2-ol

      D    propane

solution:
ketones are formed when two degree alcohols are oxidised.
NaBH4 is the reducing agent. Hence, ketones when reacted with it gets reduced, the reaction is analogous to the backward oxidation reaction of ketones.
So, two degree alcohol is formed which is propan-2-ol.
answer is C




solution:
the given compound is ester. the -O- (of ester) bond gets broken and -OH gets added to the broken compound when ester reacts with H20.
so in this case the -O- bond is between C3H7 and CH3-C=O which are the decomposed compounds too.
and in C3H7, OH gets added and we get a product C3H7OH
B is the answer.

solution:
lets make a dispalyed formula of 1,1- Dichloroethene.

In polymerization, the C=C bond gets broken and the carbocation is the repeat unit for polymers.
the polymer with this repeat unit will look like:

write the structural formula of the compound which is A.
Note: A and C look similar but in C the third and fourth carbon possess only 1 H and Cl which is incorrect.