4)
Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol
mass(HCl)= 3.65 g
use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(3.65 g)/(36.46 g/mol)
= 0.100 mol
Answer: B
5)
Molar mass of AgNO3,
MM = 1*MM(Ag) + 1*MM(N) + 3*MM(O)
= 1*107.9 + 1*14.01 + 3*16.0
= 169.91 g/mol
mass(AgNO3)= 32.46 g
use:
number of mol of AgNO3,
n = mass of AgNO3/molar mass of AgNO3
=(32.46 g)/(1.699*10^2 g/mol)
= 0.191 mol
Answer: A
Only 1 question at a time please. I have answered 1st two for you
locs are present in a sample of HCl with a mass of 3.65g? 4. How many...
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