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 mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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stan bts

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol