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What is the volume, in mL, of 0.50 moles of ammonia (NH3) at STP?
A. 1120 L
B. 22.32 mL
C. 8500 mL
D. 11,200 mL

Respuesta :

zelinsky
For the purpose we will here use the ideal gas law:

p×V=n×R×T

V= ?
n = 0.5 moleT= 273.15 K (at STP)
p= 101.325 kPa (at STP)
R is  universal gas constant, and its value is 8.314 J/mol×K

Now when we have all necessary date we can calculate the number of moles:
V=nxRxT/p
V=0.5x8.314x273.15/101.325= 11.2 L = 11200 mL

Answer: D.

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