Answer:
m AgCl = 28.395 g
Explanation:
- AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
∴ [AgNO3] = 0.156 mol/L
∴ V = 1.27L
⇒ mol AgNO3 = 0.156 mol/L * 1.27 L = 0.19812 mol AgNO3
mass AgCL:
⇒ m AgCl = 0.19812 mol AgNO3 * mol AgCl/molAgNO3 * 143.32gAgCl/molAgCl
⇒ m AgCl = 28.395 g
The structure of compound A would be solid that is dense enough for antimicrobial form