Question

Que volumen ocupa 0.015mg de NH3 a CNPT

Answer 1

Respuesta:

V = 0.00002 L de NH3

Explicación:

que volumen ocuparán 80 moles de CO2 en CNPT

aplicar la ley de los gases ideales:

V x P = n x R x T

DATOS:

n = ¿?

V (L)= ¿?

R = 0.0821 (L atm / mol K)

T =  273  K

P = 1 atm

1. calcular masa en g

masa = 0.015 / 1000 = 0.000015 g

2. calcular moles de NH3

n = masa/Mm

Mm NH3 = 17 g/mol

n = 0.000015 g / 17 g/mol

n = 8.823 x 10⁻⁷ moles

3. calcular volumen que ocupa

V = n x R x T

            P                        

V = (8.823 x 10⁻⁷ mol)  x  0.0821 (L atm / mol K) x 273 K

                              1 atm

V = 0.00002 L de NH3

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                      1 mol   -----   22.4 L

   8.823 x 10⁻⁷  mol  -----      x

                          x = 2 X 10⁻⁵ L de NH3