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2004 - Multiple Choice - Numerous Qs - HSN forum

2004 - Multiple Choice - Numerous Qs

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#1mujb

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Posted 23 May 2006 - 12:37 PM

hi, it will be handy if you have past papers.

the Qs are 1,4,5,78,12,13,15,19,36,37,38,39.

i know, too many questions but any help will be appreciated.

thanks

#2Pammy

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Posted 23 May 2006 - 02:21 PM

Question 1

Silicon Dioxide and Potassium Fluoride are covalent, therefore do not conduct electricity in any state.
Iodine has a melting point of 114 which is higher than Potassium's (64).

Question 4

(power x volume x concentration)
(P x V x C) acid = (P x V x C) alkali
(2 x 0.05 x 0.1) = (1 x V x 0.4)
0.01 = 0.4 V
0.025 cm = V

Question 5

You just need to find the mass of each of the atoms.
For example the first one, H O tells us there are 2 hydrogen ions, whose mass each is one, therefore the total mass of the hydrogen ion = 2, there is only one oxygen, and the mass number is 16, therefor total mass of atom= 18

IF go go through this for all of them you should get
1. 18
2. 19
3. 20
4. 20
5. 21
6. 22

From this you can see 3 & 4 are the same, which makes the answer D

Question 7

If there is 1 mole of Hydrogen and 1 mole of Iodine the ratio is 1:1
If 0.8 moles of Hydrogen remain, 0.2 moles has been used, 0.2 moles of Iodine have also been used.
0.2 + 0.2 = 0.4

Question 8

This is kinda hard to explain coz it's a graph but I'll have a go

The energy of the products is greater than the energy of the reactants, the enthalpy change is positive, which is endothermic.

The activation energy is the 'hump' at the top of the graph (activated complex) minus the original energy
so it will be,
120 - 40 = 80

Qustion 12

Look at the symmetry of each of the molecules

H Cl
l l
H-C- H O=C=O cant draw NH3 on this Cl-C-Cl
l l
H Cl

All of these are symmetrical except NH3, which means the polarity is canceled out througout the molecule as a whole

Question 13

I've done this before, but I've forgotten now so could someone enlighten me ??

Question 15

Again could someone show me the path to enlightenment !!

Question 19

I worked this out by using the process of elimination
ethane is C 2 H6 and Ethene is C2H4
So there is a loss of two Hydrogens (which is dehydrogenation)

So, it is not addition as we are not gaining anything (we are losing)
It is not hydrogenation as we are not gaining hydrogen (we are losing it)
and it is not oxidation as we are not increasing the O:H ratio (we are decreasing it)

Question 36

You should know Sodium Hydroxide is a strong alkali and Ammonia is a weak alkali
So the pH of NaOH is greater than that of ammonia
The higher the pH (of the alkalis) the better the conductor, so it has a higher conductivity

Question 37

IO - ---------> I
2IO - ---------> I (balance I)
2IO - --------> I + 6H O (balance O)
2IO - + 12H+ -----------> I + 6H O (balance H)

12 H+ needed
6 H2O needed

Question 38

The Magnesium is oxidised in this as it loses electrons (has a positive charge)
The Ammonia is reduced in this as it gains electrons (has negative charge)

Oxidising agent= Something which oxidises other substances and is itself reduced

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#3Nathan

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Posted 23 May 2006 - 03:28 PM

question 13

1 litre of nitrogen dioxide + 2 litres of oxygen oxygen is in excess, so ignote oxygen

if 1 litre = 2moles, then 1 litre of NO2 will be produced thus answer A

#4Michael

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Posted 23 May 2006 - 03:29 PM

13. 2NO + 02 -----> 2NO2
2volume 1volume 2 volume
1l 0.5l 1l

Basically its an excess question so you work out ratios and only half a litre of oxygen will react with 1 litre of NO and 1l of NO2 will be formed. AnswerA

#5Michael

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Posted 23 May 2006 - 03:48 PM

ok ive got 15. Its a displacement reaction and first thing you have to do is write a balanced equation.

Cu + 2AgNO3 ---------> 2Ag + Cu(NO3)2
1 mole Cu :2 moles Ag
63.5 :2x107=214g
5g :214/63.5 x 5 = 16.8

39. You just have to create nuclear equations and see if they balance

B is the one where they dont balance therefore it must be the answer. Its difficult for me to write them out on here but you really just have to follow the information provided.

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Posted 26 May 2006 - 10:40 AM

Can somebody explain 20, 24, and 32 please?

Oh, and also 16. There's always a question like this.. and I never know what to do..

#7Nathan

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Posted 26 May 2006 - 10:52 AM

20. the acid part is CH3CH2COOH (propanoic acid)

*remember that acids have the functional group COOH

24. you need something that only has a hydroxyl (OH) at one end

32. dunno :S

16. can rule out C...reforming only occurs between hydrocarbons
also B as it is an addition reaction
not sure about how to choose between A and D tho

#8Michael

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Posted 26 May 2006 - 11:04 AM

16. B is the addition of hydrogen so its not the answer, C is hydrolysis, D is cracking. Therefore A is reforming

32. I'd go for hydoxide because of the addition of H+ and that none of the others really make much sense lol. Not sure of the true logic behind it.

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