Calculate the density for camphor gas (C10H16O) at 12 Torr and 80 C.

0.084g/L

Explanation:

First, let us obtain an expression for the density. This is illustrated below:

PV = nRT. (1)

V = nRT/P (2)

But n = Mass(m) /Molar Mass (M)

n = m/M

Substituting the value of n into equation 2

V = nRT/P

V = mRT/MP

Divide both side by m

V/m = RT/MP (3)

Invert equation 3

m/V = MP/RT (4)

But Density = Mass(m) /volume(V)

Density = m/V

Replace m/V with Density in equation 4

m/V = MP/RT

Density = MP/RT

The following data were obtained from the question:

P = 12 Torr

760torr = 1atm

12torr = 12/760 = 0.016atm

T = 80°C = 80 +273 = 353K

R = 0.082atm.L/K /mol

Molar Mass (M) of C10H16O = (12x10) + (16x1) + 16 = 120 + 16 + 16 = 152g/mol

Density =?

Density = MP/RT

Density = (152 x 0.016)/(0.082x353)

Density = 0.084g/L

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.

output?

explanation: