What volume will 73g of dry ammonia gas occupy at 11 degrees C and 720 mm HG?
How uch air is needed to dilute the ammonia to 10ppm?
To determine the volume of 73g of dry ammonia gas at 11 degrees C and 720 mm Hg, you can use the ideal gas law equation: PV = nRT.
First, convert the temperature from degrees Celsius to Kelvin by adding 273.15:
T = 11°C + 273.15 = 284.15 K
Next, convert the pressure from mm Hg to atm by dividing by 760 (1 atm = 760 mm Hg):
P = 720 mm Hg ÷ 760 mm Hg/atm = 0.947 atm
The ideal gas constant (R) is 0.0821 L·atm/(mol·K).
Now we can solve for the volume (V) in liters:
V = (nRT) / P
To find the value of n (number of moles), divide the mass (m) of ammonia gas by its molar mass (M):
n = m / M
The molar mass of ammonia gas (NH3) is 17.03 g/mol.
Substituting the values into the equation:
n = 73 g / 17.03 g/mol = 4.29 mol
Now we can calculate the volume (V):
V = (4.29 mol) * (0.0821 L·atm/(mol·K)) * (284.15 K) / 0.947 atm
V = 11.63 L
Therefore, 73g of dry ammonia gas will occupy approximately 11.63 liters at 11 degrees C and 720 mm Hg.
-----------------
To calculate the amount of air needed to dilute the ammonia to 10 ppm (parts per million), we need to know the initial concentration of ammonia and the desired volume in order to find the moles of ammonia.
Let's assume we have an initial volume of ammonia gas (V_initial) and we want to dilute it to a final volume (V_final) while maintaining the ammonia concentration at 10 ppm.
1 ppm is equal to 1 part per million, or 1 part in 1 million parts. This means that for every 1 million parts of air, there will be 10 parts of ammonia.
So, to calculate the moles of ammonia needed for the desired concentration:
Moles of Ammonia = (10 ppm) * (V_final) / 10^6
Next, we need to find the volume of air needed to dilute the ammonia to the desired concentration. To do this, we use the ratio of initial ammonia volume to the volume of air:
V_air = V_final - V_initial
Therefore, the amount of air needed to dilute the ammonia to 10 ppm is V_air.
Note that this calculation assumes ideal mixing and uniform distribution, so actual dispersion may vary.