I need help starting this problem:

A piece of dry ice (solid carbon dioxide) with a mass of 4.50 g is sealed inside a 2.0 L bottle, and allowed to sublime. At 22.0 degrees Celsius, what is the partial pressure of the CO2 in the bottle? (Assume that all of the CO2 in the sealed bottle came from the dry ice- the amount already present from the air will be negligible).

Thank you in advance! :)

Convert 4.50 g CO2 to moles. moles = grams/molar mass

Use PV = nRT to calculate P.

To solve this problem, we need to use the ideal gas law equation, which is:

PV = nRT

Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K for this problem)
T = temperature in Kelvin

We know the volume of the bottle is 2.0 L and the temperature is 22.0 degrees Celsius. However, we need to convert the temperature to Kelvin. To do this, we add 273.15 to the Celsius temperature:

T = 22.0 + 273.15 = 295.15 K

Next, we need to determine the number of moles of CO2 present in the bottle. We do this by dividing the mass of the dry ice by its molar mass, which is 44.01 g/mol for CO2. The molar mass of CO2 is the sum of the atomic masses of carbon (C) and oxygen (O) atoms in one molecule of CO2.

mass of dry ice = 4.50 g
molar mass of CO2 = 44.01 g/mol

n = mass of dry ice / molar mass of CO2
n = 4.50 g / 44.01 g/mol

Now, we can substitute the values into the ideal gas law equation:

PV = nRT
P (2.0 L) = (4.50 g / 44.01 g/mol) * (0.0821 L·atm / mol·K) * 295.15 K

We can solve this equation to find the partial pressure, P.

P = ((4.50 g / 44.01 g/mol) * (0.0821 L·atm/mol·K) * 295.15 K) / 2.0 L

By calculating this expression, we'll find the partial pressure of CO2 in the bottle at 22.0 degrees Celsius.