Tuesday
July 7, 2015

Homework Help: Chemistry

Posted by Chelsea on Friday, October 7, 2011 at 9:47am.

A student synthesized 6.895 g of barium iodate monohydrate by adding 30.00 mL of 5.912 x 10-1 M sodium iodate, NAIO3. 125 mL of distilled water as used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (at 4* C). The solubility of barium iodate monohydrate in 25* C water is .028 g per 100 mL of water; in 4* C water, it is .010 g per 100 mL of water.

1) What mass of product would you expect to isolate?

2) Calculate the percent error as a result of using 125 mL of 25* C, compared with the correct yield using 20 mL of 4* C water.

Please help, thanks!

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