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Natalka [10]
2 years ago
12

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

at 25.0 °C and 1.50 atm.
Chemistry
1 answer:
Damm [24]2 years ago
4 0

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}
  • T= 25 C= 298 K (being 0 C=273 K)

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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7 0
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100.0 g of 4.0°C water is heated until its temperature is 37.0°C. If the specific heat of water is 4.184 J/g°C, calculate the am
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Answer:

13.8072 kj

Explanation:

Given data:

Mass of water = 100.0 g

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Final temperature = 37.0°C

Specific heat capacity = 4.184 j/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 37.0°C -  4.0 °C

ΔT = 33.0°C

Q = 100.0 g ×4.184 j/g.°C × 33.0°C

Q = 13807.2 j

Joule to KJ:

13807.2 j  × 1kj  /1000 j

13.8072 kj

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Explanation:

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