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e-lub [12.9K]
3 years ago
7

Both hydrogen sulfide (H2S) and ammonia (NH3)

Chemistry
2 answers:
sveticcg [70]3 years ago
6 0

Answer:

The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster. Effusion rate is inversely proportional to molar mass. NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.

Explanation:

faust18 [17]3 years ago
4 0

Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.

Effusion rate is inversely proportional to molar mass.

NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.

Explanation: change up your response a bit

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8 0
2 years ago
The rate constant of a reaction is 4.7×10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C?
vodomira [7]

Answer:

k is 3,18*10⁻² s⁻¹ at 75°C

Explanation:

following Arrhenius equation:

k= k₀*e^(-Ea/RT)

where k= rate constant , k₀= frequency factor , Ea= activation energy , R= universal gas constant T=absolute temperature

then for T₁=25°C =298 K

k₁= k₀*e^(-Ea/RT₁)

and for T₁=75°C = 348 K

k₂= k₀*e^(-Ea/RT₂)

dividing both equations

k₂/k₁= e^(-Ea/RT₂+Ea/RT₁ )

k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )]

replacing values

k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )] = 4,7*10⁻³ s⁻¹ *e^[-33.6*1000 J/mol /8.314 J/molK*(1/ 348 K -1/298 K )] = 3,18*10⁻² s⁻¹

thus k is 3,18*10⁻² s⁻¹ at 75°C

6 0
4 years ago
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