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Chemistry difficult past papers questions!

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For the first image,
I1 = 870
I2 = 1800
I3 = 3000
I4 = 3600
I5 = 5800
I6 = 7000
I7 = 13200
there is a gradual increase in successive ionisation energies but there is a huge leap after the 6th ionisaton. That's because the 7th electron is being removed from either a complete shell or an octate.
This means the right element is in group VI
So right answer must be Te .
 
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For the first image,
I1 = 870
I2 = 1800
I3 = 3000
I4 = 3600
I5 = 5800
I6 = 7000
I7 = 13200
there is a gradual increase in successive ionisation energies but there is a huge leap after the 6th ionisaton. That's because the 7th electron is being removed from either a complete shell or an octate.
This means the right element is in group VI
So right answer must be Te .

OK GOT IT! THANKS midha.ch But what about remaining two o_O
 
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For the 3rd image, is the answer 5 - B?

If yes the reason is that for the each side chain there is 3 double bonds, so to saturate it all you would need 3 moles of Hydrogen. In the question it says ONE of its side chains is converted to bablabla (only has 2 double bonds - 1 double bond less than the original one), so to form that you would need ONE mole of H2.
For the second one they say TWO of the side chains is converted to blablalb (only has 1 double bond - 2 double bonds less than the original one), so to form that you would need TWO TIMES (because you are converting TWO side chains) times 2 (because you need 2 moles of H2 to get rid of those 2 bonds). Hence, if you add all up, to make glycerl trieleostearate you need 1 + 2*2 moles of H2 = 1 +4 = 5 moles of H2.
 
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The equilibrium one is C.
Coz at the start, there are 0.2 mol of H2 and 0.15 mol of I2.
Applying the algebraic method, the number of moles at equilibrium would be:
0.2-x mol of H2, 0.15-x mol of I2 and 2x mol of HI (As per the ratio, 2 moles of HI is produced)
But, the no of moles of HI at equilibrium is given in the question.
It is 0.26 mol. Thus, 2x=0.26 mol, and x=0.13 mol.
Substituting this value of x in the equilibrium moles equation, we get:
H2 = 0.07 mol, I2 = 0.02 mol and HI = 0.26 mol.
Substituting this in the Kc expression, we get the equation as stated in option C.

Tada!
 
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for the second one just count the number of double bonds that are broken and this is the number of hydrogen molecules that are used. c is the answer
 
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