Correct Answer - A::B
a. 1 mol `S = 32 g`
b. `(12 xx 3 xx 10^(23))/(6 xx 10^(23)) = 6 g`
c. `2 g` atom of `N = 28 g`
d. `7.0 g Ag`
ii.b. mol of `A = (50)/(10) = 5`, mol of `B = (50)/(20) = 2.5`
`A:B = 5:2.5 = 2:1`
Formula `A_(2) B`.
iii.a. Weight of `H_(2) O = V xx d = 0.0018 mL xx 1 = 0.0081 g`
`n_(H_(2)O) = (0.0018)/(18) = 10^(-4) mol = 6.023 xx 10^(23) xx 10^(-4)`
`= 6.023 xx 10^(19)`