If 0.213 moles of argon occupies a volume of 652 mL at a particular temp and pressure, what volume would it occupy if 0.162 moles of argon were added at the same conditions?

V₂  = 495.89 mL

Explanation:

Given data:

Initial number of moles = 0.213 mol

Initial volume = 652 mL

Final number of moles = 0.162 mol

Final volume = ?

Solution:

V₁/n₁    =    V₂/n₂

By putting values,

652 mL/0.213 mol   =  V₂ /0.162 mol

V₂  = 652 mL 0.162 mol /0.213 mol

V₂  = 105.62 mL.mol /0.213 mol

V₂  = 495.89 mL

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o