What is the volume of
2.8 × 10^6 molecules He?
Answer in units of L
At STP, 22.4L contains 6.02E23 molecules He.
is it 1.686×10^30
Not even close.
22.4L contains 6.02E23 molecules; therefore, ?L contains 2.8E6 molecules. It MUST be less than 22.4L. right?
To find the volume of 2.8 × 10^6 molecules of helium, we can use the ideal gas law equation, which is PV = nRT. Here's how you can solve it step by step:
Step 1: Convert the given number of molecules to moles.
The Avogadro's number states that 1 mole of any substance contains 6.022 × 10^23 molecules. So, to find the number of moles, divide the given number of molecules by Avogadro's number:
Number of moles = (2.8 × 10^6 molecules) / (6.022 × 10^23 molecules/mole)
Step 2: Calculate the volume using the ideal gas law equation.
P = pressure (which is not given)
V = volume
n = number of moles
R = ideal gas constant (which is 0.0821 L·atm/(mol·K))
T = temperature (which is not given)
Since the pressure and temperature are not provided in the question, we cannot calculate the volume without additional information.
Therefore, the volume of 2.8 × 10^6 molecules of helium in units of liters cannot be determined without knowing the temperature and pressure.