KyleH123
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https://gyazo.com/d533a7ffaa65115ff16addc40811a9cd
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KyleH123
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Unlike any of my other questions, any tips? :-)

https://gyazo.com/11c01ab154f7a47e15cbfc4a5fc99289


*whistles* CheeseIsVeg
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Plagioclase
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Firstly, limestone is not a mineral and whoever wrote that question should feel thoroughly ashamed of themselves.

However to actually answer your question, percentage purity is what percentage of a substance is made of what it is supposed to be. So to calculate it, you divide the mass of the useful component (e.g. calcium carbonate) by the mass of the entire thing (e.g. the limestone block) and multiply it by 100. If the limestone was completely pure then the mass calcium carbonate would be equal to the mass of limestone. If the mass of calcium carbonate is anything less, then the limestone is not pure.
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Pigster
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Your Q makes no sense, it says 2.00 limestone. I'm guessing you missed off the 'g'.

Also your working for part a) is wrong: n = c x v (times not divide), although your answer is nearly correct. Since it is 0.034125, you shouldn't have shown it 0.03412, although to 3SF it is still 0.341.

In part b) you've rounded 0.01705 to 0.171 for some reason, rather than 0.0171

If I'm right, your 2.00 g of limestone contains 0.0171 g of CaCO3. What is that as a percentage?
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CheeseIsVeg
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(Original post by Plagioclase)
Firstly, limestone is not a mineral and whoever wrote that question should feel thoroughly ashamed of themselves.

However to actually answer your question, percentage purity is what percentage of a substance is made of what it is supposed to be. So to calculate it, you divide the mass of the useful component (e.g. calcium carbonate) by the mass of the entire thing (e.g. the limestone block) and multiply it by 100. If the limestone was completely pure then the mass calcium carbonate would be equal to the mass of limestone. If the mass of calcium carbonate is anything less, then the limestone is not pure.
ugh you spoil my fun
KyleH123 was doing violin practice, sorry
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KyleH123
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(Original post by Pigster)
Your Q makes no sense, it says 2.00 limestone. I'm guessing you missed off the 'g'.

Also your working for part a) is wrong: n = c x v (times not divide), although your answer is nearly correct. Since it is 0.034125, you shouldn't have shown it 0.03412, although to 3SF it is still 0.341.

In part b) you've rounded 0.01705 to 0.171 for some reason, rather than 0.0171

If I'm right, your 2.00 g of limestone contains 0.0171 g of CaCO3. What is that as a percentage?
0.00855%?
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Pigster
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(Original post by KyleH123)
0.00855%?
To quote Plagioclase, "and multiply it by 100".
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KyleH123
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(Original post by Pigster)
To quote Plagioclase, "and multiply it by 100".
oh no it contains 0.0171 moles

17.1 grams

so I think it comes to 85.5% purity

NO. What have I done actually
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Pigster
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1.71 g is 85.5 % of 2.00 g
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KyleH123
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(Original post by Pigster)
1.71 g is 85.5 % of 2.00 g
OH. 1 decimal point wrong in my working and everything falls apart.

Cheers.
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