OCR Chemistry Salters B F331 27th May 2016 (Old specification)

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I did:

192/23 = 8.347826.......
Ans X 100 = 834.7826....g

Because you have to work out the mass of air and you already know how much oxygen you have in that air. (I don't know if that is right)

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I did that and then for the next part did you do 1/74 and times that by 834.7826? 74 because that was the Mr of C4H10O

Reply 121

username1084917

Anyone else put 0.23 for the first calculation, 0.059 for the second and put 0.74 for the Kno3 percentage moles one?

Please reply.

Reply 122

Average_Andrew

solid im pretty sure, as solubility of group 2 hydroxides decreases up group 2

Reply 123

username1084917

A Grade boundaries? 46?

Reply 124

Kwaks

Original post by Kira Yagami

Do you think, if they're nice I put get the mark for the correct equation too?

Oh man lol what is wrong with me... Like I do the 'hard' part right...

I would give you the mark tbh. The equation was correct and 'M' could stand for any group 2 metal. And Mg is a group 2 metal. (That's if I was an examiner)

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Reply 125

Kwaks

Original post by Kaz_96

I did that and then for the next part did you do 1/74 and times that by 834.7826? 74 because that was the Mr of C4H10O

Yep that's exactly what I did. And I got a mass of 11g or so???

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Reply 126

Kwaks

Original post by Average_Andrew

solid im pretty sure, as solubility of group 2 hydroxides decreases up group 2

That's carbonates?

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Reply 127

Daribig

Original post by Kwaks

Yep that's exactly what I did. And I got a mass of 11g or so???

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Yep i got around 11.8

Reply 128

Rustybl

Original post by Average_Andrew

solid im pretty sure, as solubility of group 2 hydroxides decreases up group 2

For this question it was unclear whether the water was in liquid or in gaseous form, Mg reacts with water to form Mg(OH)2 where as it reacts with steam to form MgO. So since nuclear reactors are very hot I made the assumption that the water would be in gaseous form and subsequently probably lost both of the marks for that question.

Reply 129

Kaz_96

Can anyone remember the working out for the energy density q? I think it was number of moles x the enthalpy of combustion value they gave you?

Reply 130

JgR1997

The Uranian decays to Thorium? Is this right

Reply 131

username1084917

I put 0.74 for the Kno3, is this correct?

Reply 132

Kwaks

Original post by Kira Yagami

Anyone else put 0.23 for the first calculation, 0.059 for the second and put 0.74 for the Kno3 percentage moles one?

Please reply.

I got:
1st = 835g
2nd = 11g
KNO3 = 34%

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Reply 133

Kwaks

Original post by JgR1997

The Uranian decays to Thorium? Is this right

I got the same as you

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Reply 134

Kaz_96

Original post by Kwaks

I got:
1st = 835g
2nd = 11g
KNO3 = 34%

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For the KNO3 it was just mass/Mr for the 3 involved, then do the number of moles of KNO3/ (Moles of KNO3 + the other 2) x 100?

Reply 135

JgR1997

Is there an unofficial mark scheme anywhere?

Reply 136

Kwaks

Original post by Kaz_96

Can anyone remember the working out for the energy density q? I think it was number of moles x the enthalpy of combustion value they gave you?

I did:
1000/Mr
And X enthalpy change of combustion

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Reply 137

Rustybl

Original post by JgR1997

The Uranian decays to Thorium? Is this right

Correct.

Reply 138

Kwaks

Original post by Kaz_96

For the KNO3 it was just mass/Mr for the 3 involved, then do the number of moles of KNO3/ (Moles of KNO3 + the other 2) x 100?

Yeah, I think so... What did you get as the final answer???

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Reply 139

KINGYusuf

Does anyone think I'll get a mark for the energy density Q? It was supposed to be near to 30000 or 36000 or something but then I divided this by 1000 so I got 36.something idk

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