## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 3 Chemical Reactions - Study Questions - Page 137g: 93

#### Answer

See answer below.

#### Work Step by Step

a) As: +3 in $As_2S_3$, +5 in $H_3AsO_4$ S: -2 in $As_2S_3$, 0 in $S$ N: +5 in $HNO_3$, +2 in $NO$ b) In 100 g of the compound: As: $16.199\ g/74.9216\ g/mol=0.216\ mol$ Ag: $69.946\ g/107.8682\ g/mol=0.649\ mol$ O: $13.855\ g/15.999\ g/mol=0.866\ mol$ Dividing by the smallest one: Ag: $0.649/0.216=3.00$ O: $0.866/0.216=4.00$ Hence the empirical formula is: $Ag_3AsO_4$

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