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).