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#1 xClairex

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Posted 28 May 2007 - 06:11 PM

2002 - question 3 - i think this is standard grade stuff not sure how to do this question can do this now smile.gif

2002 - q11,12 can do them now smile.gif

2002 - 25 i understand that it will be either A or D - the answer is D but why is it not A?

2002 - q29 its a matter of ratios but im not sure how as i got the number of neutrons would be 52 :S

2003 - again i think it is standard grade work - q4 and 5

2003-11 how do we know that it is D?

2003-13 I understand that it has to be covalent but i dont know how you would be able to tell if it was polar or non polar :S

2003-14 not sure at all on this one

any help would be much appreciated

Edited by xClairex, 28 May 2007 - 06:38 PM.



#2 son_of_a_gun

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Posted 28 May 2007 - 06:23 PM

2002: Q3 - 0.1 mol of magnesium, so 2x that for Hydrogen ions. Then i just did 0.02/4 = 0.05, change this to cm3, to get 50cm3 = b
11 - 50 for 1, so for 5, 250. 500-250 = 250. 3CO2 = 3 x 50cm3 = 150........150+250 = 400 = C

12 - 1 atom = 6.02 x 10^23
3 atoms = 3 x 6.02 x 10^23 = C






#3 ginneswatson

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Posted 28 May 2007 - 06:31 PM

QUOTE(son_of_a_gun @ May 28 2007, 07:23 PM) View Post
2002: Q3 - 0.1 mol of magnesium, so 2x that for Hydrogen ions. Then i just did 0.02/4 = 0.05, change this to cm3, to get 50cm3 = b
11 - 50 for 1, so for 5, 250. 500-250 = 250. 3CO2 = 3 x 50cm3 = 150........150+250 = 400 = C

12 - 1 atom = 6.02 x 10^23
3 atoms = 3 x 6.02 x 10^23 = C


Doing great.

Q12 might stil leave a few puzzled so I'll add a little more.

1g = 1/12 mole of carbon. However, 12 protons and neutrons per atom, so 12 x 1/12 takes us back up to 1 mole of particles. 6 x 1023.
3 quarks per particle so, 3 x 6 x 1023 = 1.8 x 1024 as son of a gun says. wink.gif

#4 xClairex

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Posted 28 May 2007 - 06:32 PM

thanks i understand how to do they questions now smile.gif


#5 ginneswatson

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Posted 28 May 2007 - 06:36 PM

QUOTE(xClairex @ May 28 2007, 07:11 PM) View Post
2002 - question 3 - i think this is standard grade stuff not sure how to do this question

2002 - q11,12

2002 - 25 i understand that it will be either A or D - the answer is D but why is it not A?

2002 - q29 its a matter of ratios but im not sure how as i got the number of neutrons would be 52 :S

2003 - again i think it is standard grade work - q4 and 5

2003-11 how do we know that it is D?

2003-13 I understand that it has to be covalent but i dont know how you would be able to tell if it was polar or non polar :S

2002-14 not sure at all on this one

any help would be much appreciated


I'll start at the end and work forward until I bump into someone else's answers.

2002 Q14 Reforming. Look for a change in structure without much change in molecular formula. Usually chains to rings or branched molecules.

B is the answer. Original molecule is an octane chain, C8H18. Second molecule is a 6 carbon chain with two methyl branches, C8H18.


#6 xClairex

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Posted 28 May 2007 - 06:38 PM

oops sorry i meant 2003 question 14


#7 ginneswatson

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Posted 28 May 2007 - 06:40 PM

2003 Q13

To interact with water, especially to the extent of removing a hydrogen to leave a hydroxide ion, you really have to be polar covalent.

Covalent and alkali really has to mean ammonia or its organic equivalents, the amines. Not just polar, but capable of hydrogen bonding.


#8 ginneswatson

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Posted 28 May 2007 - 06:44 PM

Brat! tongue.gif

2003 Q14

Copper(II) phosphate is (Cu2+)3(PO43-)2.

There are, therefore, 5 ions; 3 Cu2+ ions and 2 PO43- ions.

Answer = D

#9 ginneswatson

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Posted 28 May 2007 - 06:51 PM

2003 Q11

Solubility relies on both molecules having similar strengths of intermolecular forces. If one is much stronger than the other, then those molecules will choose to remain with their own kind and won't mix (dissolve).

CCl4 is covalent. The CóCl bonds are reasonably polar, but the tetrahedral shape means there is no 'positive end' and no 'negative end' so no polar-polar interactions, just Van Der Waals Forces


A, B & C are all ionic compounds. Some ionic compounds containing metals closer to the zig-zag line can be a bit more covalent but none of these are a problem.

D is the only covalent compound. It is therefore the 'most likely' to be soluble in CCl4

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#10 ginneswatson

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Posted 28 May 2007 - 06:58 PM

2003 Q 4

Basically just checking that you know that nitrogen is diatomic, N2, and thereafter problem solving.

1414N = 28

1415N or 1514N = 29

1515N = 30

3 different possible molecules. Answer C.

2003 Q5

Need to know your formulae and be logical.

NaCl and Na2SO4

0.6 mol of Cl ions will be partnered by 0.6 mol of Na ions.
0.2 mol of SO4 ions will be partnered by 2 x 0.2 = 0.4 mol of Na ions.

0.6 + 0.4 = 1.0 mol of Na ions. Answer D.

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#11 ginneswatson

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Posted 28 May 2007 - 07:03 PM

2002 Q29

neutron to proton ratio means 'take the number of neutrons and divide by the number of protons'.

Strontium is element 38 so protons = 38

Mass = 90, so 90 - 38 = 52 neutrons.

Ratio = 52/38 = less than 1.5 so answer D

#12 ginneswatson

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Posted 28 May 2007 - 07:07 PM

2002 Q25

The only hydrogen that can ionise is the hydrogen in an acid group. COOH. Answer D

A is an aldehyde group. The carbonyl (C=O) by itself couldn't cope with the negative charge left behind.


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#13 xClairex

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Posted 28 May 2007 - 07:27 PM

thank you smile.gif


#14 celticfcz

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Posted 28 May 2007 - 09:23 PM

picked some good questions claire! this has helped quite a bit! thanks!





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