# Can you help me with a chemistry question?

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#1
http://www.ocr.org.uk/Images/144862-...igher-tier.pdf
Page 12 I don't get it. I've looked at the mark scheme but still no help. Please help.Thanks.
0
6 years ago
#2
I don't do ocr but from what I've read for first question it's just the Mr of copper(2) divided by the Mr of the whole compound times by 100....so the Mr of copper is (63.5*2) divided by the Mr of the whole compound (63.5*2)+32 and times by 100 this gets you 79.9%..rounded. Is this correct?
0
6 years ago
#3
(Original post by JazzyT17)
http://www.ocr.org.uk/Images/144862-...igher-tier.pdf
Page 12 I don't get it. I've looked at the mark scheme but still no help. Please help.Thanks.
(Original post by GS98)
I don't do ocr but from what I've read for first question it's just the Mr of copper(2) divided by the Mr of the whole compound times by 100....so the Mr of copper is (63.5*2) divided by the Mr of the whole compound (63.5*2)+32 and times by 100 this gets you 79.9%..rounded. Is this correct?
Ar of coppper in the molecule divded by Mr of Cu2S and then multiply that by 100.
0
6 years ago
#4
(Original post by JazzyT17)
http://www.ocr.org.uk/Images/144862-...igher-tier.pdf
Page 12 I don't get it. I've looked at the mark scheme but still no help. Please help.Thanks.
do you have the link ot the mark scheme
because i hav an idea of how to do it, just want to check!thanks
0
6 years ago
#5
It's copper with a little 2 so it wouldn't be the Ar, it would be the Mr?
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6 years ago
#6
i guess fro part aii you just multitply the percentage by 1kg, to give 0.798kg of cu ?
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6 years ago
#7
(Original post by GS98)
It's copper with a little 2 so it wouldn't be the Ar, it would be the Mr?
ar is 63.5 mr is 127
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#8
(Original post by zhzstudent)
do you have the link ot the mark scheme
because i hav an idea of how to do it, just want to check!thanks
http://www.ocr.org.uk/Images/142598-...er-january.pdf
0
6 years ago
#9
Well I got the question right....and Ar , mr is also right
0
6 years ago
#10
yep i was right, you just times the percentage with the 1kg to get0.799 kg
0
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