Increased irrigation demands: decrease the rate of transpiration increase the rate of surface water evaporation change the amoun
1 answer:
You might be interested in
Answer: The volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of (molar mass = 32.0 g/mol) is as follows.
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
Thus, we can conclude that the volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.