SHOW WORK

How many moles of helium occupy 26.4 L at a pressure of 103.0 kPa, and 10 degrees Celsius?
A.) 0.03 mole
B.) 0.86 mole
C.) 1.16 mole
D.) 32.72 mole

PV = nRT

P = pressue
V = volume
n = moles
R = 8.31
T = Temperature (0c = 273K)

(103.0)(26.4) = n(8.31)(283)
2719.2 = n(2351.73)
2719.2 / 2351.73 = n
m = 1.16

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (in kPa)
V = Volume (in liters)
n = Number of moles
R = Gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin:

T(K) = T(C) + 273.15
T(K) = 10 + 273.15 = 283.15 K

Now, let's rearrange the ideal gas law equation to solve for the number of moles (n):

n = PV / RT

Substituting the given values into the equation:

P = 103.0 kPa
V = 26.4 L
R = 8.314 J/(mol·K)
T = 283.15 K

n = (103.0 kPa * 26.4 L) / (8.314 J/(mol·K) * 283.15 K)

Simplifying the units:

n = (103.0 * 26.4 * 10^3) / (8.314 * 283.15)

n = 3.62336 * 10^3 / 2350.6581

n ≈ 1.54 moles

Rounding to two decimal places, we find that approximately 1.54 moles of helium occupy 26.4 L at a pressure of 103.0 kPa and 10 degrees Celsius.

Therefore, the correct answer is not given in the options provided.