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bonufazy [111]
2 years ago
10

Calculate the density of a sample of 1.00 mole of nh3 at 793mmhg and -9.00 c

Chemistry
1 answer:
xeze [42]2 years ago
8 0

0.164 g/L is the density of a sample of 1.00 mole of NH_3 at 793mmhg and -9.00 degrees celcius.

<h3>What is density?</h3>

Density is the mass of a unit volume of a material substance. The formula for density is d = \frac{M}{V}, where d is density, M is mass, and V is volume.

Given data:

n = 1.00 mole

P=793 mm hg =1.04342 atm

T=-9.00 degree celcius = -9.00 + 273= 264 K

V=?

Using Ideal Gas Law equation:  

PV = n R T      

R = gas constant = 0.082057 L-atm/(mol-K)

(1.04342 atm)(V) = 5 X 0.082057 L-atm/(mol-K)  X 264 K

V = 103.67 Liters

Now calculate density:

Mole weight of NH_3 = 1.00 mole

So, the mass of NH_3 = 17.031 g

Density =  \frac{mass}{volume}  

Density =  \frac{17.031 g}{ 103.67 Liters}  

= 0.164 g/L

Hence, 0.164 g/L is the density of a sample of 1.00 mole of NH_3 at 793mmhg and -9.00 degrees celcius.

Learn more about the density here:

brainly.com/question/15164682

#SPJ1

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The required mole ratio of  NH₃ to N₂ in the given chemical reaction is 2:4.

<h3>What is the stoichiometry?</h3>

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