E. 23 b

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asalest 2K
Posts: 41
Joined: Wed Sep 30, 2020 9:39 pm

E. 23 b

Postby asalest 2K » Tue Oct 27, 2020 7:31 pm

why don't we divide by Avogadro's to get moles for b in:

E.23 Calculate the amount (in moles) of (a) Cu21 ions in 3.00 g of CuBr2; (b) SO3 molecules in 7.00 3 102 mg of SO3; (c) F2 ions in 25.2 kg of UF6; (d) H2O in 2.00 g of Na2CO3?10H2O.

Kimiya Aframian IB
Posts: 56
Joined: Wed Sep 30, 2020 9:34 pm

Re: E. 23 b

Postby Kimiya Aframian IB » Tue Oct 27, 2020 8:03 pm

asalest 2K wrote:why don't we divide by Avogadro's to get moles for b in:

E.23 Calculate the amount (in moles) of (a) Cu21 ions in 3.00 g of CuBr2; (b) SO3 molecules in 7.00 3 102 mg of SO3; (c) F2 ions in 25.2 kg of UF6; (d) H2O in 2.00 g of Na2CO3?10H2O.

Hi! so solve for moles for b, the first step is to convert mg to g. Now use the molar mass of SO3 to convert g to mol.
If you continue to solve for molecules, then you need to multiply by Avogadro's number (it's multiplication not division so that you can cancel out your units and get molecules. Hope this helps!

marisagaitan1G
Posts: 46
Joined: Wed Sep 30, 2020 9:47 pm
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Re: E. 23 b

Postby marisagaitan1G » Tue Oct 27, 2020 8:07 pm

You only use avogadro's number when converting to/from number of atoms/molecules. Since this question asks you to go from mg to moles, you simple go from mg to g to moles of SO3.

marisagaitan1G
Posts: 46
Joined: Wed Sep 30, 2020 9:47 pm
Been upvoted: 2 times

Re: E. 23 b

Postby marisagaitan1G » Tue Oct 27, 2020 8:09 pm

Disregard my answer, I misread the question. The person above me has the correct answer. So sorry!!


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