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Lerok [7]
4 years ago
15

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

barrier for the reaction?\
Chemistry
1 answer:
maria [59]4 years ago
3 0

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\

\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\

\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\

\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}

E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\

E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}

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Answer:

1.Reactivity

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3.Shape

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Explanation:

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You are asked to write observations about a 100g sample of Nitrogen-16 before it decays, but you're running late. In order to ma
san4es73 [151]

Answer:

You cannot make observations if you are 57 seconds late into the lab.

Explanation:

The atomic nucleus can split by decay into 2 or more particles as a result of the instability of its atomic nucleus due to the fact that radioactive elements possess an unstable atomic nucleus.

Now, the primary particles which are emitted by radioactive elements in order to make them decay are alpha, beta & gamma particles.

The half life equation is;

N_t = N₀(½)^(t/t_½)

Where:

t = duration of decay

t_½ = half-life

N₀ = number of radioactive atoms initially

N_t = number of radioactive atoms remaining after decay over time t

We are given;

t = 57 secs

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Now, half life of Nitrogen-16 from online sources is 7.2 seconds. t_½ = 7.2

Thus;

N_t = 100(1/2)^(57/7.2)

N_t = 0.4139g

We are told that In order to make observations, you require at least .5g of material.

The value of N_t you got is less than 0.5g, therefore you cannot make observations if you are 57 seconds late.

5 0
3 years ago
The emission spectrum of cesium contains two lines whose frequencies are (a)
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The lines are violet and blue respectively.

<h3>What is the energy?</h3>

We know that the energy of the photon could be obtained by the use of the equation;

E = hf

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f = frequency

For the first line;

E = 6.6 * 10^-34 Js * 3.45 x 10^14 Hz = 2.3 * 10^-19 J

Given that;

E = hc/λ

λ = hc/E

λ = 6.6 * 10^-34 * 3 * 10^8/2.3 * 10^-19

λ = 8.61 * 10^-7 m or 861 nm

The color is violet

For the second line;

E = 6.6 * 10^-34 Js * 6.53 xx 10^14 Hz

E = 4.3 * 10^-19 J

E = hc/λ

λ = hc/E

λ =  6.6 * 10^-34 * 3 * 10^8/4.3 * 10^-19

λ = 4.60 * 10^- 7 m or 460 nm

The color is blue

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Which statement describes the relationship between microscopic and macroscopic phenomena?
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Answer:

It's d

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