The new temperature (in °C) of the gas, given the data is –148.20 °C
<h3>Data obtained from the question </h3>
- Initial temperature (T₁) = 149.05 °C = 149.05 + 273 = 422.05 K
- Initial pressure (P₁) = 349.84 KPa
- Volume = constant
- New pressure (P₂) = 103.45 KPa
- New temperature (T₂) =?
<h3>How to determine the new temperature </h3>
The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:
P₁V₁ / T₁ = P₂V₂ / T₂
Since the volume is constant, we have:
P₁ / T₁ = P₂ / T₂
349.84 / 422.05 = 103.45 / T₂
Cross multiply
349.84 × T₂ = 103.45 × 422.05
Divide both side by 349.84
T₂ = (103.45 × 422.05) / 349.84
T₂ = 124.80 K
Subtract 273 from 124.80 K to express in degree celsius
T₂ = 124.80 – 273
T₂ = –148.20 °C
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The different is that the galvanic cell converts chemical energy into the electrical energy and the electrolytic cell coverts electrical energy into chemical energy
Answer:
Increase the pressure of the gas
Explanation:
According to the Pressure law, for a fixed mass of gas, at a constant volume (V), the pressure (P) is directly proportional to the absolute temperature (T).
From the kinetic molecular theory, gases are composed of particles which are in constant motion, colliding with themselves as well as with the walls of their container.
When the temperature of these gas molecules is increased, the molecules acquire more kinetic energy and the rate of collisions increases. Since the container cannot expand, the increase in pressure is due to the increase in collisions between the molecules of the gas as well as with the walls of their container.
Answer: The results agree with the law of conservation of mass
Explanation:
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. On the reactant side, the total mass of reactants is 14.3g and the total product masses is also 14.3g. That implies that no mass was !most in the reaction. The sum of masses on the left hand side corresponds with sum of masses on the right hand side of the reaction equation.
The compound is sodium chloride
