Question

22.4 dm³ of dry ammonia gas contains 6 x10²² molecules at STP. Calculate the number of molecules in 56 cm³ of dry nitrogen at STP​

Answer 1

I'VE ROUNDED OFF SOME VALUES HERE AND USED SIMPLE FORMULA

HOPE IT HELPS..❤️❤️

Answer 2

The number of molecules in 56 cm³ of dry nitrogen at STP​ is $0.015\times 10^{22}$ molecules

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number  of particles.

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

a) According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = $22.4dm^3$ = 22.4 L

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of dry ammonia gas= ?

$n=\frac{PV}RT}=\frac{1\times 22.4}{0.0821\times 273}=1$

1 mole of dry ammonia gas at STP contain = $6.023\times 10^{22}$ molecules

b ) $PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = $56cm^3$ = 56 ml = 0.056 L    (1L=1000ml)

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of dry nitrogen gas= ?

$n=\frac{PV}RT}=\frac{1\times 0.056}{0.0821\times 273}=0.0025$

1 mole of dry nitrogen at STP​ contains = $6.023\times 10^{22}$ molecules

0.0025 moles of dry nitrogen at STP​ contains = $\frac{6.023\times 10^{22}}{1}\times 0.0025=0.015\times 10^{22}$ molecules

Learn more about ideal gas equation

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