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natulia [17]
3 years ago
13

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

the carbon dioxide if the temperature and pressure change to 75 oC and 15 kPa.
Chemistry
1 answer:
Dimas [21]3 years ago
5 0

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

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Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

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  • V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
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