Question

One mole of agcl is dopped with 10^5 moles of cacl2

Answer 1

Answer:

1 mole of AgCl is doped with 105 moles of CaCl2.

One Ca2+ balances 2 Ag+. Thus, we have one cationic vacancy produced per CaCl2.

Number of Ag+ removed = Number of Ca2+ added/ 2 = 6.023×1023×1052=3.011×1028 .

Answer 2

Answer:

One mole of $Agcl$ is dopped with $10^{5}$ moles of $cacl_{2}$  $6.023*1023*1052=3.011*1028$

Explanation:

1 mole of $Agcl$ is doped with 105 moles of $cacl_{2}$.

One $ca_{2} ^{+}$ adjusts 2 $Ag^{+}$. Along these lines, we have one cationic opening created per CaCl2.

Number of $Ag^{+}$ eliminated = Number of $ca_{2} ^{+}$ added/2 =

                                                    $6.023*1023*1052=3.011*1028$ .

One mole of $Agcl$ is dopped with $10^{5}$ moles of $cacl_{2}$  $6.023*1023*1052=3.011*1028$

The project code is #SPJ3