1).What is the pressure in a 1.5 L vessel containing 0.25 moles of O2 if the temperature is 37 degrees celcius?
2).The molar mass of MgCO3 is 84g/mole. How many moles are in 12.60g of MgCO3?
#1. Use PV = nRT. Don't forget to change C to Kelvin. Kelvin = C + 273.
#2. # mol = g/molar mass.
Would you please further explain the steps to sloving #2?
Thanks
There isn't anything to explain.
# mols = grams/molar mass. Plug in grams from the problem (12.6) and molar mass from the problem (84) and divide on your calculator. I get something like 0.15 or so
# mols = 12.6/84 = ??. If you prefer to do these by the dimensional method, then
# mols = 12.6 g x (1 mol/84g) = ?? mols. (Note how the unit we don't want (grams) cancel and the unit we want to keep (mols) doesn't cancel. We have converted, using a factor, grams to mols.
What is the pressure in a 1.5 L vessel containing 0.25 moles of O2 if the temperature is 37oC? (R, the universal gas constant, is 0.082 L atm / mole K.)
To find the pressure in a vessel, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in units of pressure, such as atm)
V = volume (in units of liters)
n = number of moles
R = ideal gas constant (0.0821 atm·L/mol·K)
T = temperature (in units of Kelvin)
1) To find the pressure in the vessel containing O2:
Given:
V = 1.5 L (volume)
n = 0.25 moles (number of moles)
T = 37 degrees Celsius (temperature)
We need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 37 + 273.15 = 310.15 K
Now let's plug the values into the Ideal Gas Law equation and solve for P:
PV = nRT
P * 1.5 L = 0.25 moles * 0.0821 atm·L/mol·K * 310.15 K
Simplifying the equation, we have:
P = (0.25 moles * 0.0821 atm·L/mol·K * 310.15 K) / 1.5 L
P ≈ 4.28 atm
Therefore, the pressure in the 1.5 L vessel containing 0.25 moles of O2 at 37 degrees Celsius is approximately 4.28 atm.
2) To find the number of moles in 12.60g of MgCO3:
Given:
Molar mass of MgCO3 = 84g/mol
Mass of MgCO3 = 12.60g
To find the moles, we can use the formula:
moles = mass / molar mass
Let's substitute the given values into the formula:
moles = 12.60g / 84g/mol
Calculating the result:
moles ≈ 0.15 mol
Therefore, there are approximately 0.15 moles in 12.60g of MgCO3.