Physics, asked by ayush11aug, 10 months ago

How many silver atoms are present in a piece of jewellery weighing 10.78g ? Ag = 107.8 a.m.u

Answers

Answered by Thinkab13

3

Answer:

107.8g of Ag →6*10^23 atoms

10.78g of Ag→6*10^23*10.78/107.8

. =6*10^22

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