6. The given mass of helium occupies a volume of 250 cm3 at 10 OC and
0.2 atm. What will be its volume at 12
OC and 1.4 atm?
A
29 76 cm3

V2 = 35.967cm^3

Explanation:

Given data:

P1 = 0.2atm

P2 = 1.4atm

V1 = 250cm^3

V2 = ?

T1 = 10°C + 273 = 283K

T2 = 12°C + 273 = 285K

Apply combined law:

P1xV1/T1 = P2xV2/T2 ...eq1

Substituting values:

0.2 x 250/283 = 1.4 x V2/285

Solve for V2:

V2 = 14250/396.2

V2 = 35.967cm^3

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.