The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.
<h3>What is Combined gas law?</h3>
Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.
It is expressed as;
P₁V₁/T₁ = P₂V₂/T₂
Given the data in the question;
- Initial volume V₁ = 14.5L
- Initial pressure P₁ = 0.980atm
- Initial temperature T₁ = 20.0°C = 293.15K
- Final pressure P₂ = 740.mmHg = 0.973684atm
We substitute our given values into the expression above.
P₁V₁/T₁ = P₂V₂/T₂
( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂
14.21Latm / 293.15K = 13.92368Latm / T₂
14.21Latm × T₂ = 13.92368Latm × 293.15K
14.21Latm × T₂ = 4081.72679LatmK
T₂ = 4081.72679LatmK / 14.21Latm
T₂ = 287.24K
T₂ = 14.09°C
Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.
Learn more about the combined gas law here: brainly.com/question/25944795
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Answer:
52.9 KJmol-1
Explanation:
From;
log(k2/k1) = Ea/2.303 * R (1/T1 - 1/T2)
The temperatures must be converted to Kelvin;
T1 = 25° C + 273 = 298 K
T2= 35°C + 273 = 308 K
R= gas constant = 8.314 JK-1mol-1
Substituting values;
log 2 = Ea/2.303 * 8.314 (1/298 - 1/308)
Ea = 52.9 KJmol-1
Answer:
reaction B is the best one I would choose
