I need help please.
"what voluma [in mL] at STp will be occupied by 0.0035 mol of methane [CH4]"
THANK YOU SOO MUCH!
WEll, one mole is 22400 ml.
So do it in proportion:
.0035/1 =V/22400ml
is that for every mole in problems like these?
because i thought it was 6.022x10^23
To find the volume of a gas at STP (Standard Temperature and Pressure) using the given moles of gas, you can use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure in atmospheres (atm)
V = Volume in liters (L)
n = Number of moles of gas
R = Ideal Gas Constant (0.0821 L * atm / mol * K)
T = Temperature in Kelvin (K)
At STP, the conditions are defined as a temperature of 273.15 Kelvin (0 degrees Celsius) and a pressure of 1 atmosphere (1 atm).
Now let's solve the equation for volume (V):
V = (n * R * T) / P
Substituting the values:
n = 0.0035 mol (given)
R = 0.0821 L * atm / mol * K (ideal gas constant)
T = 273.15 K (STP temperature)
P = 1 atm (STP pressure)
V = (0.0035 mol * 0.0821 L * atm / mol * K * 273.15 K) / 1 atm
V = 0.0035 mol * 0.0821 L * 273.15 K
V ≈ 0.00360 L or 3.60 mL (approximately)
Therefore, approximately 3.60 mL of volume at STP will be occupied by 0.0035 mol of methane (CH4).