Science, asked by pradhanaayushree999, 5 months ago

8. How much quantity of silver chloride (AgCl) can be obtained from 1.0 g silver nitrate ? [Ag = 108 ;
Cl = 35.5).
[Ans. 0.844 g)

Answers

Answered by anveshadeshmukh68

3

Ag = 108 ; Cl = 35.5

Ans. 0.844 g

Answered by ziafirdous91

12

Answer:

AgNO3 + HCL __ AgCl + HNO3

170 143.5

quantity of the AgCl formed = 143.5 × 1 /170

= 0.844 g (Ans).

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