A rigid 3.80 L sealed vessel contains 0.650 mol Ne, 0.321 mol Kr, and 0.190 mol Xe. Find the density of the mixture in g/L.

17.09g/L

Explanation:

Density = total mass of elements/ volume

We need to find the mass of each mixture constituents using their molar mass:

mole = mass/molar mass

For Neon (Ne) which contains 0.650mol;

0.650 = mass/20.18

mass = 0.650 × 20.18

mass = 13.12g

For Krypton (Kr) which contains 0.321mol;

0.321 = mass/83.79

mass = 0.321 × 83.79

mass = 26.89g

For Xenon (Xe) which contains 0.190mol;

0.190 = mass/131.3

mass = 0.190 × 131.3

mass = 24.95g

Total mass = 13.12g + 26.89g + 24.95g = 64.96g

Density = total mass / volume

Density = 64.96g / 3.80L

Density of the mixture = 17.09g/L

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