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Irina-Kira [14]
3 years ago
11

hich of the following changes always leads to an increase in the rate constant for a reaction: Decreasing the temperature Decrea

sing the activation energy Making the value of ΔEΔE more negative Only one—(i), or (ii), or (iii)—increases the reaction rate. (i) and (ii) (i) and (iii) (ii) and (iii) All three—(i), (ii), and (iii)—increase the reaction rate.
Chemistry
1 answer:
GrogVix [38]3 years ago
4 0

Answer:

Only one—(i), or (ii), or (iii)—increases the reaction rate.

Explanation:

<em>Which of the following changes always leads to an increase in the rate constant for a reaction?</em>

  • <em>Decreasing the temperature. </em>NO. A lower temperature leads to a slower reaction because the molecules have less energy to react.
  • <em>Decreasing the activation energy</em>. YES. According to the Arrhenius equation, the lower the activation energy, the higher the rate constant.
  • <em>Making the value of ΔE more negative</em>. NO. A more negative ΔE means a reaction is more spontaneous but not faster.
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If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

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1. Discuss briefly system and surrounding using suitable examples, highlighting heat transfers. 14 marks​
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A system is a part of the <em>physical</em> universe defined <em>arbitrarily</em> for observation purposes.

Boundaries are a part of the <em>physical</em> universe that are around the system.

In a scientific sense, a system is a part of the <em>physical</em> universe whose boundaries, that is, the limit between the system and its surroundings, are defined <em>arbitrarily</em> for observation purposes.

A system contains at least a model, represented in a phenomenological way, and it can be isolated (no mass nor energy interactions), closed (no mass interactions) or open.

The surroundings are a part of the <em>physical</em> universe that are around the system.

An example is a coffee-maker, where coffee-maker the system and air represents the surroundings, the coffee-maker receives energy from a heat source to warm up itself and releases part of such energy to the air.

We kindly invite to check this question on systems and surroundings: brainly.com/question/6044762

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