A chemist places 10.0 grams of dry ice (CO 2 ) in a 2.5 L vacuum at 25˚C.
After it all sublimates, what would be the pressure inside the container?

P = 2.25 atm

Explanation:

Given data:

Mass of dry ice = 10 g

Volume of container = 2.5 L

Temperature = 25°C (25+273=298 K)

Pressure inside container = ?

Solution:

First of all we will calculate the number of moles of dry ice.

Number of moles = mass / molar mass

Number of moles = 10 g/ 44 g/mol

Number of moles = 0.23 mol

Now we will determine the pressure.

PV = nRT

P = nRT/V

P = 0.23 mol × 0.0821 atm.L/mol.K  × 298 K / 2.5 L

P = 5.63 atm  / 2.5

P = 2.25 atm

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