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Chemistry: Post your doubts here!

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Q 14 A 5.00 g sample of anhydrous Group II metal nitrate loses 3.29 g in mass on strong heating
which metal is present ?

A magnesium
B calcium
C strontium
D barium

can some one explain

answer is B
 
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group II metal nitrate is going to produce metal oxide and nitrogen dioxide gas,the ratio between metal nitrate and metal oxide would be 1:1
5/x+28+96 = (5-3.29)/x+16
equate it u will get the answer
 
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group II metal nitrate is going to produce metal oxide and nitrogen dioxide gas,the ratio between metal nitrate and metal oxide would be 1:1
5/x+28+96 = (5-3.29)/x+16
equate it u will get the answer

Please take a look at my questions as well
 
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf

8) I think that D will also be the answer. WHy is A the answer not D.
27) How is it 3:1 ratio?
33) WHy is C not correct. Cant hydrogen bonding occur in propanone anD propanal.
34) Where did Co2 come from?

PLease answer :)))
8) in both reactions HSO3¨ is accepting H+ and that is why its a base
33) hydrogen bond can only occur in atoms which have a polar hydrogen (hydrogen directly bonded to O,N or F)
34) atmospheric CO2,,, CO2 turns lime water milky
 
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Q39) B. Propane = C3H8 and the radical it forms is C3H7* (* being a radical). That means if you break a bond anywhere in options 1, 2 or 3 you should be able to get C3H7. It's possible in 1 and 2, but not 3 so B is right. If you still don't get it, let me know and I'll make a diagram.

Q38) A. 2 is wrong, therefore D is right. It's wrong because hydrocarbons are flammable, you wouldn't want to put them in fire extinguishers. :p Also, 1 is right because the more H atoms replaced by halogens in a carbon atom, the less flammable it is. 3 is wrong because it's an acid.

Q35) C. Halogen-halogen bond energy has nothing to do with the volality of the halogens. It's Van Der Waal forces + # of electrons.

Q33). Only activation energy does, so D. Enthalpy change has nothing to do with the rate of reaction, nor does Kc.

Q30) Use the mole concept here:

You have 0.67 moles of the alcohol and 0.50 moles of the acid. Since their ratios is 1:1, the alcohol is in excess. 0.50 moles of the acid/alcohol should therefore give you 0.5 moles of the ester. But 22g is only 0.25 moles, so the yield is 0.25/0.5 * 100 = 50%. C is right.

Q26) C. Split the ester into its consituent acid and alcohol, and the alcohol part is C2H5OH. That means A, B and D are wrong and C is right.

Q24) C. This is an SN1 reaction because it's a tertiary alcohol and the profile diagram for a tertiary alcohol will always be C.

Q17) This is sort of difficult for me to explain because there's lots of theory to it; you should learn about the oxidizing/reducing power of group 7 halogens, halides as well as H2SO4.

Q5) I think it's D, the more easier a halogen is vaporized, the better hydrogen bonding it has. I could be wrong, but that's what I think and the rest of the options don't seem right anyway.
 
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4 The equation below describes the equilibrium between
propan-2-ol and propanone in the gas phase.
CH3CH(OH)CH3(g) → CH3COCH3(g) + H2(g)
propan-2-ol propanone
Which one of the following statements about this equilibrium
is correct?
A Increasing the pressure decreases the yield of
propan-2-ol

B Increasing the pressure decreases the yield
of propanone

C Decreasing the pressure decreases the yield
of hydrogen

D Decreasing the pressure has no effect on the reaction

why is b correct.......increasing the pressure increases yeild of propan-2-ol ..........and will dissociate more to give propanone ... so how does the yeild of propanone decrease...?
 

Jaf

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Option 3 says that Ca ions react with acids ( hence the equilibrium will shift to the right, decaying the tooth)
That statement is, theoretically, false. That is because any acid (such as sulphuric acid) will 'react' with the ions and the Ca salt formed will be soluble, hence there will be no overall change.
That is okay, but what if you add carbonic acid: there will be carbonate ions in solution, which will react with the Ca ion forming insoluble CaCO3. Hence the equilibrium will shift.
Yes, you are missing something: the syllabus assumes you do not study Biology.
Oh you have no idea just how much AS chemistry, physics and biology overlap. Quite a few times I've gained marks in one subject because of one of the other two.
You should also know that carbonic acid is not an acid that frequently comes in contact with your teeth (unless you live on coke, that is...). I think we all realize that when there's a general substance named in the question (acids), they don't expect us to consider substances that would behave abnormally.
 

Jaf

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4 The equation below describes the equilibrium between
propan-2-ol and propanone in the gas phase.
CH3CH(OH)CH3(g) → CH3COCH3(g) + H2(g)
propan-2-ol propanone
Which one of the following statements about this equilibrium
is correct?
A Increasing the pressure decreases the yield of
propan-2-ol

B Increasing the pressure decreases the yield
of propanone

C Decreasing the pressure decreases the yield
of hydrogen

D Decreasing the pressure has no effect on the reaction

why is b correct.......increasing the pressure increases yeild of propan-2-ol ..........and will dissociate more to give propanone ... so how does the yeild of propanone decrease...?
Number of gas moles on the left hand side is 1.
Number of gas moles on the right hand side is 2.
Increasing the pressure favors backward reaction (so yield of propan-2-ol increases, yield of propanone and hydrogen decreases)
 
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Q39) B. Propane = C3H8 and the radical it forms is C3H7* (* being a radical). That means if you break a bond anywhere in options 1, 2 or 3 you should be able to get C3H7. It's possible in 1 and 2, but not 3 so B is right. If you still don't get it, let me know and I'll make a diagram.

.


kindly clarify q 39 its not clear to me by drawing diag ! plsss
 
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Oh you have no idea just how much AS chemistry, physics and biology overlap. Quite a few times I've gained marks in one subject because of one of the other two.
You should also know that carbonic acid is not an acid that frequently comes in contact with your teeth (unless you live on coke, that is...). I think we all realize that when there's a general substance named in the question (acids), they don't expect us to consider substances that would behave abnormally.
Tell me about it, I frequently use concepts of Bio in Chemistry (and vice versa), as well as using concepts of physics in maths (and vice versa).
But my argument still stands: when CIE sets questions they consider all the possible responses. Wouldn't they be aware of the fact that there is a probability that some candidates may think in terms of carbonic acid?
Which brings me to my next point: if what has been said above is true, then my theory of CaCO3 is false.
So what is the correct answer.
And again, the syllabus does not assume you know Bio, so stop explaining in those terms. I don't care if lactic acid is present in your buccal cavity; for all I know there could be traces of carbonic acid in the food we eat, that ever occur to you?
 
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