What is the volume of 45.0g of nitrogen monoxide, NO, at 20 C and a pressure of 740 mm Hg?
Please someone can check my work is right or not?
NO = 14 + 16 = 30
54.0g / 30 = 33.60 L
273K/293K = 0.932 atm
740 mm Hg = 740 torr = 0.974 atm
33.60L x 0.932 x 0.974 = 30.50 L
Answer; 30.50L right or not? Than how to do? I am lost :(
PV=nRT
(0.974 atm)(V)=n(o.o821)(293K)
n= (45.0g)/(30.0g)=1.5 moles NO
(0.974atm)(v)=(1.5)(o.o821)(293K)
solve.
NO = 14 + 16 = 30
54.0g / 30 = 33.60 L The problem quotes 45.0 g and not 54.0 g
273K/293K = 0.932 atm The Kelvin is 273 + 20 = 293 K. I don't know how you divided temperature by temperature and came out with atm.
740 mm Hg = 740 torr = 0.974 atm
33.60L x 0.932 x 0.974 = 30.50 L This lat line is not correct at all.
The response by anonymous is correct.
Thank you anonymous and DrBob222!
no prob :)
To calculate the volume of a gas, you can use the ideal gas law equation, which is PV = nRT. In this equation, P stands for pressure, V stands for volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the given values to the appropriate units. The pressure is given as 740 mm Hg, which can be converted to atm by dividing by 760 (since 1 atm = 760 mm Hg). So, the pressure is 0.974 atm.
To convert the temperature from Celsius to Kelvin, you need to add 273. Therefore, 20 C + 273 = 293 K.
Now, let's calculate the number of moles of nitrogen monoxide (NO). The molar mass of NO is 14 + 16 = 30 g/mol. Given that the mass of NO is 45.0 g, we can divide this by the molar mass to find the number of moles: 45.0 g / 30 g/mol = 1.50 mol.
Now, plug these values into the ideal gas law equation: PV = nRT.
V = (nRT) / P
V = (1.50 mol * 0.0821 atm L/mol K * 293 K) / 0.974 atm
By performing these calculations, you will find the volume of nitrogen monoxide to be approximately 45.09 L (rounded to two decimal places).
Therefore, based on your calculations, your answer of 30.50 L is not correct.