Question

question 8
a sample of pure oxygen gas has a pressure of 795 torr. what is the pressure of the oxygen in units of atmospheres? (3 significant
figures)
add your answer
integer, decimal, or e notation allowed
question 9
calculate the standard enthalpy of formation for diamond given that: (the unit is kj/mol) and your answer should have 2 significant figures).
c (graphite) + o₂ (g) → co₂ (g) δh° = -393.5 kj/mol.
c (diamond) + o₂ (g) → co₂ (g) δh° = -395.4 kj/mol
add your answer
integer, decimal, or e notation allowed
question 10
sulfur hexafluoride (sf₆) is a colorless and odorless gas. calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43
l at 69.5°c.
add your answer
integer, decimal, or e notation allowed

Explanation:

Question 8

Step1: Unit conversion factor

We know that \(1\ atm = 760\ torr\).

Step2: Calculate pressure in atm

Let \(P\) be the pressure in atm. Using the formula \(P=\frac{795\ torr}{760\ torr/atm}\)

$$P=\frac{795}{760}=1.04605\approx1.05\ atm$$

Step1: Write the equations

Equation (1): \(C(\text{Graphite})+O_{2}(g)
ightarrow CO_{2}(g)\quad\Delta H_{1}=- 393.5\ kJ/mol\)
Equation (2): \(C(\text{diamond})+O_{2}(g)
ightarrow CO_{2}(g)\quad\Delta H_{2}=-395.4\ kJ/mol\)

Step2: Manipulate the equations

We want to find \(\Delta H\) for \(C(\text{Graphite})
ightarrow C(\text{diamond})\).
Reverse equation (2): \(CO_{2}(g)
ightarrow C(\text{diamond})+O_{2}(g)\quad\Delta H_{2}' = 395.4\ kJ/mol\)
Add equation (1) and the reversed equation (2):
\(C(\text{Graphite})+O_{2}(g)+CO_{2}(g)
ightarrow CO_{2}(g)+C(\text{diamond})+O_{2}(g)\)
Simplify to \(C(\text{Graphite})
ightarrow C(\text{diamond})\)
\(\Delta H=\Delta H_{1}+\Delta H_{2}'=-393.5 + 395.4=1.9\ kJ/mol\)

Step1: Convert temperature to Kelvin

\(T=(69.5 + 273.15)K=342.65\ K\)

Step2: Use the ideal gas law \(PV = nRT\)

We know \(n = 1.82\ mol\), \(V=5.43\ L\), \(R = 0.0821\ L\cdot atm/(mol\cdot K)\)
Rearrange for \(P\): \(P=\frac{nRT}{V}\)

$$P=\frac{1.82\times0.0821\times342.65}{5.43}$$

$$P=\frac{1.82\times0.0821\times342.65}{5.43}=\frac{1.82\times28.131565}{5.43}=\frac{51.2094483}{5.43}\approx9.43\ atm$$

Answer:

\(1.05\)

Question 9