Which type of perspective makes objects look similar to how they look in real life (select the best answer)

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

Now put all the given values in the above equation, we get:

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

Now put all the given values in the above equation, we get:

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

3 years ago

13. An aerosol spray can of deodorant with a volume of 0.410 L contains 3.0 g of propane gas (CH3) as propellant. What is the pr

qaws [65]

Answer: The pressure in the can is 4.0 atm

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = ?

V = Volume of gas = 0.410 L

n = number of moles = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{3.0g}{44.1g/mol}=0.068mol$

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $20^0C=(20+273)K=293K$

$P=\frac{nRT}{V}$

$P=\frac0.068mol\times 0.0820 L atm/K mol\times 293K}{0.410L}=4.0atm$

Thus the pressure in the can is 4.0 atm

8 0

3 years ago

Do elements have DNA?

jeyben [28]

Not most elements have DNA, only five "elements" have DNA. <span>The five "elements" are Carbon, Oxygen, Hydrogen, Nitrogen and Phosphorus. Good luck with your question I hope this is what you were looking for.</span>

6 0

3 years ago

How do sound waves travel? PLEASE HELP IF YOU WANT BRAINLEIST AND ME TO LIKE URE COMMENT!!

irina1246 [14]

The answer is A. The vibration caused by the waves through the air eventually weaken, which is why sound diminishes easily over distance.

8 0

3 years ago

Complete the charge balance equation for an aqueous solution of h2co3 that ionizes to hco−3 and co2−3.

Zielflug [23.3K]

The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The equation for aqueous solution of H₂CO₃ is

H₂CO₃ → H₂O + CO₂

The charge balance equation is

[HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]

Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]

Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
#SPJ4

8 0

2 years ago