contestada

A flashbulb of volume 2.70 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 30.0 °C. How many grams of O2(g) does the flashbulb contain?

Respuesta :

Willchemistry
P = 2.30 atm

Volume in liter = 2.70 mL / 1000 => 0.0027 L

Temperature in K = 30.0 + 273 => 303 K

R = 0.082 atm

molar mass O2 = 31.9988 g/mol

number of moles O2 :

P * V = n * R* T

2.30 * 0.0027 = n * 0.082 * 303

0.00621 = n * 24.846

n = 0.00621 / 24.846

n = 0.0002499 moles of O2

Mass of O2:

n = m / mm

0.0002499 = m / 31.9988

m = 0.0002499 * 31.9988

m = 0.008 g
Lanuel

Using the ideal gas equation, the flashbulb contains 8 grams of oxygen gas ([tex]0_2[/tex]).

Given the following data:

  • Volume = 2.70 mL
  • Temperature = 30.0 °C
  • Pressure = 2.30 atm
  • Ideal gas constant, R = 0.0821L⋅atm/mol⋅K

First of all, we would convert the temperature to Kelvin:

[tex]Temperature \;in \;Kelvins = 273 \;+ C\\\\Temperature \;in \;Kelvins = 273 \;+ 30[/tex]

Temperature = 303 K

To find how many grams of [tex]0_2 \; (g)[/tex] the flashbulb contain, we would use the ideal gas law equation;

[tex]PV = nRT[/tex]

Where;

  • P is the pressure.
  • V is the volume.
  • n is the number of moles of substance.
  • R is the ideal gas constant.
  • T is the temperature.

Making n the subject of formula, we have;

[tex]n = \frac{PV}{RT}[/tex]

Substituting the given parameters into the formula, we have;

[tex]n = \frac{2.30(2.70)}{0.0821(303)}\\\\n = \frac{6.21}{24.88}[/tex]

Number of moles, n = 0.25 moles.

Now, we would determine the mass:

  • Molecular mass of oxygen gas ([tex]0_2[/tex]) = 32

[tex]Mass = n[/tex] × [tex]Molecular \; mass[/tex]

[tex]Mass = 0.25[/tex] × [tex]32[/tex]

Mass = 8 grams

Therefore, the flashbulb contains 8 grams of oxygen gas ([tex]0_2[/tex]).

Read more: https://brainly.com/question/24290885

Otras preguntas