What color is an orange?

Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Us

madam [21]

Answer:

$d=0.92\frac{kg}{m^{3}}$

Explanation:

Using the Ideal Gas Law we have $PV=nRT$ and the number of moles n could be expressed as $n=\frac{m}{M}$ , where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

$PV=\frac{m}{M}RT$

If we pass the V to divide:

$P=\frac{m}{V}\frac{RT}{M}$

As the density is expressed as $d=\frac{m}{V}$ , we have:

$P=d\frac{RT}{M}$

Solving for the density:

$d=\frac{PM}{RT}$

Then we need to convert the units to the S.I.:

$T=100^{o}C+273.15$

$T=373.15K$

$P=1bar*\frac{0.98atm}{1bar}$

$P=0.98atm$

$M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}$

$M=0.0289\frac{kg}{mol}$

Finally we replace the values:

$d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}$

$d=9.2*10^{-4}\frac{kg}{L}$

$d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}$

$d=0.92\frac{kg}{m^{3}}$

5 0

4 years ago

The pressure on 150 mL of a gas is increased from 500 mm Hg to 700 mm Hg at constant temperature. What is the new volume of the

Ivanshal [37]

Answer:

Option A. 107 mL

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 150 mL

Initial pressure (P₁) = 500 mmHg

Final pressure (P₂) = 700 mmHg

Temperature = constant

Final volume (V₂) =?

The final volume of the gas can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

500 × 150 = 700 × V₂

75000 = 700 × V₂

Divide both side by 700

V₂ = 75000 / 700

V₂ = 107 mL

Therefore, the final volume of the gas is 107 mL.

4 0

3 years ago

Write the chemical symbol for an element in Period 6 and Group 2A

Nataly_w [17]

The answer is Ba (56)

3 0

4 years ago

Which phase is heat energy being released?

scZoUnD [109]

The answer is B
Vaporization

6 0

3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

3 years ago