Answer:
0.9g/L.
Step-by-step explanation:
To calculate the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K, we can use the ideal gas law:
PV = nRT
where P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles of gas, R is the universal gas constant (0.08206 L·atm/(mol·K)), and T is the temperature in Kelvin (K).
We can rearrange this equation to solve for the number of moles of gas:
n = PV / RT
Next, we can use the molar mass of H2S (34.08 g/mol) to convert the number of moles to mass:
mass = n × molar mass
Finally, we can divide the mass by the volume to obtain the density:
density = mass/volume
Let's assume a volume of 1 L (since the volume is not given in the question). Then we have:
P = 0.7 atm
T = 322 K
R = 0.08206 L·atm/(mol·K)
molar mass of H2S = 34.08 g/mol
First, we calculate the number of moles of H2S using the ideal gas law:
n = PV / RT
n = (0.7 atm) (1 L) / (0.08206 L·atm/(mol·K) × 322 K)
n = 0.0265 mol
Next, we calculate the mass of H2S using the number of moles and the molar mass:
mass = n × molar mass
mass = 0.0265 mol × 34.08 g/mol
mass = 0.9 g
Finally, we calculate the density of H2S:
density = mass/volume
density = 0.9g/1 L
density = 0.9 g/L
Therefore, the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K is approximately 0.9g/L.