Higher chemistry 2014-2015

OP

Original post by jonathan14

Could anyone help me with these two questions please from the 2011 paper?

34, So basically you want forward reaction.

Increase in pressure shifts in direction which has less number of molecules in gas phase(which in this case, is to the right)
So the pressure must be increased.

Cooling refers to decrease in temperature
Warming refers to increase in temperature
Since the delta H is negative, this shows the reaction is an exothermic reaction
Decrease the temperature shifts the equilibrium in the direction of an exothermic reaction, so it must be cooling.

so the answer is A

Here is a little summary that sums up what you need to know about equilibrium:

Concentration:
Adding reactant/Removing product- equilibrium shifts to the right
Adding product/Removing reactant- equilibrium shifts to the left

Temperature:
Increase- shifts in the direction of endothermic reaction
Decrease- shifts in the direction of exothermic reaction

Pressure:
Increase- shifts in the direction which has less number of molecules in GAS PHASE
Decrease- shifts in the direction which has more molecules in GAS PHASE

Catalyst:
No effect on equilibrium position, equilibrium more rapidly achieved.

Reply 123

OP

Original post by jonathan14

Could anyone help me with these two questions please from the 2011 paper?

Oh and 38, it's either A or C
Look at the number of electrons of NO3-(3e-) and Zn(2e-)..
I'm not 100% sure about this one... :<

Reply 124

penguined

Original post by jonathan14

Sorry I'm still confused

The second condition requires the oxide of that element to be a gas. When states are involved in a question but no temperature is explicitly given, then it is safe to assume room temperature. From the data book, we know that silicon oxide has a boiling point well above room temp. Hence, it does not form a gasseous oxide - it will actually be a solid. In contrast, sulfur oxide has a melting point below room temp, hence it will form a gasseous oxide, therefore satisfying the second condition.

Reply 125

How would someone know as they are all very soluble?

Reply 126

Original post by HiddenNinja01

How would someone know as they are all very soluble?

These questions i do not get either, q2 they all work and can form a full outer shell by donating
and i dont know how to figure out 3?
Any help is appreciated :smile:

Reply 127

Original post by Junioh

Yeah~ Do you think you would lose marks if you used a slightly different(and strange/more complex way?) but got the same answer?
for instance, recurrence relationship, you do Un+1=aUn+b for several times instead of the limit way(L=b/1-a)
sorry if this is off-topic~ just curious :tongue:

If you are referring to the toad question in the maths exam I used limits in that one lol. I was done in literally 30 seconds and didnt understand why it was a five mark question so you were probably meant to do several RR's. I've seen in several maths marking schemes that you get marks if you do it a different but mathematically sound way. Same applies for chemistry calculations too.

Reply 128

Original post by Junioh

That's good~ never really liked the PPAs... arghhh

I swear the two PPA's we didn't do in class always come up in past papers and I have no notes on them! So annoying. Our school has the worst equipment..we had to do the quantitative electrolysis experiment out of old cracked margarine tubs and we had to stick our hands into the acid solution to make sure no solution fell out of the measuring cylinder lol

Reply 129

Original post by HiddenNinja01

How would someone know as they are all very soluble?

You need to swap the last parts. So the filtrate would be Silver Chloride I assume? For instance for A) Barium Chloride becomes Barium Chloride again when reacted with the filtrate (which is vs) and Silver Chloride (which is i) so a precipitate IS formed so it can't be A.

B) When reacted with the filtrate (silver chloride) it makes Potassium Chloride (vs) and Silver Nitrate (which is also vs) so in this instance NO precipitate is formed therefore the answer must be B.

I am assuming this is the right answer? Could you check the Marking scheme pls?

Reply 130

Original post by Zain-A

You need to swap the last parts. So the filtrate would be Silver Chloride I assume? For instance for A) Barium Chloride becomes Barium Chloride again when reacted with the filtrate (which is vs) and Silver Chloride (which is i) so a precipitate IS formed so it can't be A.

B) When reacted with the filtrate (silver chloride) it makes Potassium Chloride (vs) and Silver Nitrate (which is also vs) so in this instance NO precipitate is formed therefore the answer must be B.

I am assuming this is the right answer? Could you check the Marking scheme pls?

Yes you are correct! Thanks :smile:

Reply 131

Original post by HiddenNinja01

Yes you are correct! Thanks :smile:

If you need me to explain any preciptate questions I can lol one of the only things im good at

Reply 132

Original post by Zain-A

If you need me to explain any preciptate questions I can lol one of the only things im good at

I sure will do!
And if you could, can you help me out on the other 2 questions ive asked,
Would be very appreciated...i really suck at common knowledge questions, so if you or anyone else has a site for practice and explanation..then i really need them lol
Thanks again :smile:

Reply 133

AzarFan

1

Original post by HiddenNinja01

These questions i do not get either, q2 they all work and can form a full outer shell by donating
and i dont know how to figure out 3?
Any help is appreciated :smile:

Q2) Na2O is the correct answer. Na has a electron arrangement of 2,8,1 while Oxygen has an arrangement of 2,6. There are 2 Na ions and both will lose 1 electron to the oxygen resulting in all of the ions having an arrangement of 2,8,8. Making them isoelectronic. The other compounds when together wont have identical arrangements.

I'm sure that's correct anyway. Not too sure on Q3

(edited 8 years ago)

Reply 134

Original post by HiddenNinja01

I sure will do!
And if you could, can you help me out on the other 2 questions ive asked,
Would be very appreciated...i really suck at common knowledge questions, so if you or anyone else has a site for practice and explanation..then i really need them lol
Thanks again :smile:

Iso electro thingy is so confusing.. i never really understood it

Reply 135

Original post by AzarFan

Q2) Na2O is the correct answer. Na has a electron arrangement of 2,8,1 while Oxygen has an arrangement of 2,6. There are 2 Na ions and both will lose 1 electron to the oxygen resulting in all of the ions having an arrangement of 2,8,8. Making them isoelectronic. The other compounds when together wont have identical arrangements.

I'm sure that's correct anyway. Not too sure on Q3

Thanks alot!! :smile:

Reply 136

Original post by Zain-A

Iso electro thingy is so confusing.. i never really understood it

Azar fan summed it up perfectly, it will sure help you out too :smile:

Reply 137

Ed2015

I suck at Hess's Law so bad I find it so time consuming and never know what to write out.. Could anyone explain this? it is 15(b) from 2009. Don't understand the marking scheme that well either! ImageUploadedByStudent Room1432371276.812670.jpg

ImageUploadedByStudent Room1432371276.812670.jpg

Posted from TSR Mobile

Reply 138

OP

Original post by Ed2015

I suck at Hess's Law so bad I find it so time consuming and never know what to write out.. Could anyone explain this? it is 15(b) from 2009. Don't understand the marking scheme that well either! ImageUploadedByStudent Room1432371276.812670.jpg

Posted from TSR Mobile

Hess's Law takes practice, you'll eventually get there~ trust me :wink:

Formation of silane
Si(s) + 2H2(g) → SiH4(g) delta H1=?

Si(s) + O2(g) → SiO2(s) delta H2=-911 kJ

2H2(g) + O2(g) → 2H2O(g) delta H3=(-286)x2=-572 kJ

SiO2(s) + 2H2O(g) → SiH4(g) delta H4=1517 kJ

delta H1
=delta H2 + delta H3 + delta H4
= -911 + (-572) + 1517
= 34 kJ mol^-1

Reply 139

OP