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KonstantinChe [14]
3 years ago
10

a 100.0 g sample of a metal m is burned in air to produce 103.7 g of its oxide with the formula m2o. what is the atomic mass of

the metal?
Chemistry
1 answer:
Ahat [919]3 years ago
6 0
The molar mass should be 14.898g/mol. I used the equation 100gx(1/xg/mol)x(1mol/2mol)x(16g/mol+x)/1)=103.7 and solve for x. I found that equation using stoichiometry and the equation 2m+1/2O2-->m2O The molar mass of the metal I set to x and the molar mass of the metal oxide is 16+x.
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A 500.0 g block of dry ice (solid CO2, molar mass = 44.0 g) vaporizes at room temperature. Calculate the volume of gas produced
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Considering the ideal gas law, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

<h3>Definition of ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Ideal gas law</h3>

An ideal gas is characterized by absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Volume of gas</h3>

In this case, you know:

  • P= 1.50 atm
  • V= ?
  • n= 500 g×\frac{1 mole}{44 g}= 11.36 moles, being 44 \frac{g}{mole} the molar mass of CO₂
  • R= 0.082 \frac{atmL}{molK}
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Replacing in the ideal gas law:

1.50 atm×V = 11.36 moles×0.082\frac{atmL}{molK} × 298 K

Solving:

V= (11.36 moles×0.082\frac{atmL}{molK} × 298 K) ÷ 1.50 atm

<u><em>V= 184.899 L</em></u>

Finally, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

Learn more about the ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

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