a silver coin contains 50%of silver.the number of Ag atoms present in 21.6g of silver coin is...PLZZ ANSWER THIS QUESTION 100 POINTS.......
Answers
AgNO_3+NaCl\rightarrow AgCl+NaNO_3AgNO
3
+NaCl→AgCl+NaNO
3
Moles of AgCl = \frac{\text{mass of AgCl}}{\text{MOlar mass of AgCl}}=\frac{14.35 g}{143.32 g/mol}=0.1001 mol
MOlar mass of AgCl
mass of AgCl
=
143.32g/mol
14.35g
=0.1001mol
in one molecular formula of AgCl there are one Ag atom then in 0.1001 mol of AgCl will contain 0.1001 mole of Ag metal.
Mass of Ag metal = Moles of Ag × Molar mass of Ag
=0.1001 moles\times 107.87 g/mol=10.79 g0.1001moles×107.87g/mol=10.79g
\%\text{of Ag}=\frac{\text{mass of Ag}}{\text{total mass of silver coin}}\times 100=\frac{10.79 g}{21.6 g}\times 100=49.9\%%of Ag=
total mass of silver coin
mass of Ag
×100=
21.6g
10.79g
×100=49.9%
The weight of AgCl is 14.35g then percentage of silver in coin is 49.9 %.
Answer:
1g silver coin contains 1/2g of silver
21.6g silver coin contains 10.8 gm of silver
No.of atoms of Ag present are=(10.8/108)×6.022×10^23
=6.022×10^22 (or) N/10
Explanation: