To what temperature, in degree celsius, must 10.0 grams of ammonia gas (NH3) have to be heated in a 15.0 L container in order for it to exert a pressure of 3.50 atm?

Use PV = nRT.

n = grams/molar mass
T will be in kelvin.

To find the temperature to which the ammonia gas must be heated, we can use the Ideal Gas Law equation:

\(PV = nRT\)

where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin

First, we need to determine the number of moles of ammonia gas (NH3). We can use the formula:

\(n = \frac{m}{M}\)

where:
n is the number of moles
m is the mass of the gas (in grams)
M is the molar mass of the gas (NH3)

The molar mass of ammonia (NH3) is calculated as follows:
1 atom of Nitrogen (N) weighs 14.01 g/mol
3 atoms of Hydrogen (H) weigh 1.01 g/mol each, so a total of 3.03 g/mol.
Therefore, the molar mass of ammonia (NH3) is 14.01 + 3.03 = 17.04 g/mol.

Let's calculate the number of moles of ammonia gas:
\(n = \frac{10.0 \, \text{g}}{17.04 \, \text{g/mol}}\)

\( n \approx 0.59 \, \text{mol}\)

Now, we can rearrange the Ideal Gas Law equation to solve for temperature (T):

\(T = \frac{{PV}}{{nR}}\)

The ideal gas constant R is 0.0821 L·atm/(K·mol).

Let's substitute the known values into the equation:

\(T = \frac{{(3.50 \, \text{atm})(15.0 \, \text{L})}}{{(0.59 \, \text{mol})(0.0821 \, \text{L·atm/(K·mol)})}}\)

Solving the equation will give us the temperature in Kelvin. To convert it to Celsius, subtract 273.15:

\(T \approx 476 \, \text{K}\)

Converting to Celsius:

\(T \approx 476 \, \text{K} - 273.15 \, \text{K} \approx 202.85 \, \text{°C}\)

Therefore, the ammonia gas must be heated to approximately 202.85 °C in order to exert a pressure of 3.50 atm in a 15.0 L container.

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