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
1 answer:
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×
= 11.36 moles, being 44
the molar mass of CO₂
- R= 0.082
- 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 × 298 K
Solving:
V= (11.36 moles×0.082 × 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:
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Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both and
will increase if the pressure is increased through compression. However, because
and
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient
.
As a result, the increase in pressure will have no impact on the value of . If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.