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3 years ago

Determine the volume of 3.30 mol of a gas at 25°C and 0.995 atm.

Chemistry

1 answer:

Alexxx [7]3 years ago

4 0

Answer:

V = 81.14 L

Explanation:

Given data:

Volume of gas = ?

Number of moles = 3.30 mol

Temperature of gas = 25°C

Pressure of gas = 0.995 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will convert the temperature.

25+273 = 298 K

now we will put the values in formula:

V = 3.30 mol 0.0821 atm.L/ mol.K 298 K / 0.995 atm

V = 80.74 L. atm / 0.995 atm

V = 81.14 L

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Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

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Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

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