Answer :
There are approximately 8.98 grams of krypton (Kr) in the 4.39 L cylinder at 39.1 °C and 2.10 atm.
In a 4.39 L cylinder at 39.1 °C and 2.10 atm, the number of grams of Kr (krypton) can be calculated using the ideal gas law equation. The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.
First, let's convert the given temperature from Celsius to Kelvin by adding 273.15. So, 39.1 °C + 273.15 = 312.25 K.
Next, we need to rearrange the ideal gas law equation to solve for n (number of moles). The equation becomes n = PV / RT.
Substituting the given values into the equation:
P = 2.10 atm
V = 4.39 L
R = 0.0821 L·atm/mol·K (ideal gas constant)
T = 312.25 K
n = (2.10 atm) * (4.39 L) / (0.0821 L·atm/mol·K * 312.25 K)
Now, calculate the value of n.
n = 0.1073 mol
To convert moles to grams, we need to multiply the number of moles by the molar mass of krypton (Kr), which is 83.798 g/mol.
Grams of Kr = (0.1073 mol) * (83.798 g/mol)
Calculating the value:
Grams of Kr = 8.98 g
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