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.
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 :(
5 answers
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.
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 :)
2 answers