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Maya Khoury
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Joined: Fri Apr 06, 2018 11:03 am


Postby Maya Khoury » Sun Apr 08, 2018 2:30 pm

Can someone explain why in Problem E. 23 part (a) after dividing 3.00 grams by 223.35 g, you multiply by (1 mol Cu²⁺/1 mol CuBr₂)? The question asks for the amount (in moles) of Cu²⁺ ions in 3.00 g of CuBr₂.

Thank you!

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Re: E.23

Postby Chem_Mod » Sun Apr 08, 2018 3:53 pm

Not all parts of the question seem to be posted, but I am assuming CuBr2 is in an aqueous solution. When CuBr2 dissolves, it dissolves into the respective ions by the following equation.

CuBr2 --> Cu2+ + 2Br-

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