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Multiple Choice - 2001 (Q3,Q21), 2002 (Q18), 2003 (Q3,Q14,Q37) - HSN forum

# Multiple Choice - 2001 (Q3,Q21), 2002 (Q18), 2003 (Q3,Q14,Q37)

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### #1Gillz

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Posted 27 May 2006 - 02:00 PM

Any help with how to tackle these questions would be much appreciated! Thanks in advance

2001: 3 and 21

2002: 18

2003: 3, 14 and 37

### #2d00pyd

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Posted 27 May 2006 - 02:31 PM

QUOTE(Gillz @ May 27 2006, 03:00 PM)

Any help with how to tackle these questions would be much appreciated! Thanks in advance

2001: 3 and 21

2002: 18

2003: 3, 14 and 37

Ok,

2001: 3
YOu need to write out the equation. You will see that there are 2 moles of NAOH for HsSo4. You can use the figures in the question to work out that there are 0.006 moles of NAOH. This is done by the conc x vol = moles formula. 20/1000 x 0.3. Use this to go through each of the choices, and you will need to get half 0.006 as per the ratio in the equation. So 10/1000 x 0.3 is half the moles and the answer is D. Hope that makes sense.

2001. 21. The best way to look at this (although draw them out if you have time) is that they are all odd numbers of carbon atoms, with the OH on the middle carbon. In this case the alkene bond can form either side but in terms of naming the molecule there is no difference as you always give the bond the lowest number. Therefore the answer is C beacause hexanol is an even number of C atoms and it will give a different name with the C=C bond going either way.

I'm going to post now to see if anyone else has done the others or I'll post them ///

### #3Pammy

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Posted 27 May 2006 - 02:35 PM

2001 Question 3

I just done
(V x C x P)acid = (V x C x P)alkali
2 (V x C) = (0.02 x 0.3 x 1)
2 (V x C) = 0.006
(V x C) = 0.003

so you go through each answer to find which V x C is 0.003

You don't have to make it 2(V x C), you could just have (V x C x P) acid = 0.006

Question 21

This is a drawing one really

If you draw each of these out you will see that dehydrating
A. will give you prop-1-ene, no matter what side the double bond is added to
B. will give you pent-2-ene, no matter what side the double bond is added to.
C. will give you hex-2-ene or hex-3-ene
D. will give you hept-3-ene, no matter which side the double bond is added to.

2002 Question 18

Work out the O:H ratios for each reaction for the reactants and the products

A. 1:6 ----------> 2:5
B. 1:8 -----------> 1:6
C. 1:8 -----------> 1:10
D 1:6 -----------> 2:6

These are easier to work out if you make the H term a common term eg. for A make it 5:30 -----> 12:30

2003 Question 3

I'm not sure how to explain this, but I always just remember that answer
I dont have a data book here, but I'm assuming Copper is below Iron in the electrochemical series, so it's not C.
Sodium Sulphate is something to do with the ions flowing.
Sorry, I can't really explain it !!!

Question 14

Sorry, I need a data book for tht !!

Question 37

An oxidising agent oxidises something
Oxidation is loss of electrons
So you are looking for something which has lost electrons
In C. Na ------> Na+
This means it has lost electrons

Hope that has helped, sorry I couldn't do a few of them !! xxx
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### #4d00pyd

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Posted 27 May 2006 - 02:38 PM

2002 18.

Reduction will be a decrease in the oxygen to hydrogen ratio. One way to do this is to go through and count the O and H atoms and see which one follows this rule. It's C because the O to H ratio decreases (because it gains Hydrogen). Of course, this fits with "reduction is gain, oxidation is loss".

2003:
4. Nitrogen is diatomic N2. So it can combine 3 ways in the air with the isotopes: 14N and 15N would be one way, 14 N 14 N another way, and 15N, 15N the final way.

14. The formula for copper (ii) phosphate will be CU3(PO4)2, beacuse CU2 + and PO43- are the ions. So there is 1 mole of molecule and 5 ions (3 copper and 2 phosphate). Had the question said how many in 2 moles of copper(II) phosphate, you would need to times each ions by 2.

Hopefully these make sense. Opps I answered 4 sorry - 3 is above!!

For the 2003 ques 3, I did this by elimination. If you add ions to the water (as in A) they carry electrical charge and will therefore speed it up. Carbon dioxide will be acidified in water and therefore H+ ions can carry the charge. The copper nail I'm not sur eof the logic - hopefully someone else will - and the glucose, it does nothing. A tricky one for sure!!

### #5d00pyd

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Posted 27 May 2006 - 02:57 PM

In 2003 37 simply looking for something which has been oxidised will not let you rule out D. Double check your answer by making sure that the oxidising agent has itself been reduced - ie. gain of electrons. Then D can be ruled out too. The H2 in C have been reduced as it is now the H - hydride ion.

### #6Gillz

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Posted 27 May 2006 - 03:28 PM

Help much appreciated! Thanks very much you both explain things very clearly

### #7Michael

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Posted 27 May 2006 - 03:29 PM

In 2003 the explanation for question 3 is:

Adding sodium sulphate(i.e. a salt) will increase corrosion(this is knowledge from standard grade).
Copper is below iron in the electrochemical series therefore it would increase the rate of corrosion.
Passing CO2 through water creates a weak acid which increases the corrosion rate.

### #8Gillz

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Posted 27 May 2006 - 03:53 PM

QUOTE(Dhesi @ May 27 2006, 04:29 PM)

In 2003 the explanation for question 3 is:

Adding sodium sulphate(i.e. a salt) will increase corrosion(this is knowledge from standard grade).
Copper is below iron in the electrochemical series therefore it would increase the rate of corrosion.
Passing CO2 through water creates a weak acid which increases the corrosion rate.