Question
A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding 30.00 mL of 5.912 x 10-1 M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004 x 10-1 M sodium Iodate, NaIO3
whats the percent yield?
whats the percent yield?
Answers
See you other post.
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