Question 1
acetone (ch3)2co is a volatile liquid that can be readily converted to vapour.
calculate the density of acetone vapour at 100,000 pa and 398 k in kg/m3. assume
ideal behaviour and use r=8.314 j/mol. k.
your
answer

1.76 kg/m³

Explanation:

1) Data:

a) Chemical formula: (CH₃)₂CO

b) P = 100,000 Pa

c) T = 398 K

d) D = ?

e) R = 8.314 J / mol.K

2) Principles and formulae:

a) Ideal behaviour of a gas means that it follows the ideal gas equation:

PV = nRT

b) Density, D = mass / volume

c) mass in grams = n × molar mass

3) Solution:

a) Transform the ideal gas equation to solve for the density:

PV = nRT                      ( n = mass in grass / molar mass) PV = (mass in grams / molar mass) RT(mass in grams / V) = P × molar mass / (RT)mass in kg × 1,000 (g/ kg) / V = P × molar mass / (RT)(mass in kg / V) =  P × molar mass / (1,000RT)D = P × molar mass / (1,000RT)

b) Molar mass of (CH₃)₂CO: 58.08 g/mol

c) Substitute and compute:

D = 100,000 Pa × 58.08 kg/mol / ( 1,000 × 8.314 J/mol.K × 398 K)

* (note that by dividing by 1,000 the molar mass is transformed from g/mol to kg/mol)

D = 1.76 kg/m³← answer

Ithink it’s one of those things that you put the object on

4- c, because all matter stays the same no matter what. it is still there. 5- b, because matter is known as mass. 6-c because, flammability is a chemical property. malleability is changing the shape of an object.