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Masteriza [31]
3 years ago
5

Determine the value of k' from the slope. These graphs were plotted in terms of absorbance, but the rate constant should be in t

erms of concentration. Beer-Lambert's law and the literature value for the molar absorptivity constant (87,000 M^-1cm^-1 can be used to convert the slope from units of absorbance/s to units of concentration/s. This value is the pseudo rate constant, k'.

Chemistry
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

k' = (2.0345 × 10⁻⁸) M/s.

Explanation:

Beer-Lambert's law explains that Absorbance is directly proportional to the concentration of the substrate under consideration.

A ∝c

A = εlc

where ε = molar absorptivity constant = 87,000 M⁻¹cm⁻¹

l = path length of the solution that the light passes through in cm = 1 cm for most cases.

c = concentration of the substrate.

In terms of rate terms,

A' = εlc'

where A' has units of Absorbance/s

c' = rate of reaction, units of concentration/s

A' = εlc'

From the graph, the slope of the graph = 0.00177 Absorbance/s

0.00177 = 87000 × 1 × c'

c' = (0.00177/87000)

c' = (2.0345 × 10⁻⁸) M/s.

This value is the pseudo rate constant, k'.

Hope this Helps!!!

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<span>a pure substance that can be separated into different elements by chemical means</span>
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What type of radioactive decay will the isotopes 13B and 188Au most likely undergo?
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Answer:

b. Beta emission, beta emission

Explanation:

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Now let us look at the N/P ratio of each atom;

For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6

For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4

For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.

For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.

8 0
3 years ago
Consider the following reaction:
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Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

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Electron configuration of sodium atom: ₁₁Na 1s² 2s² 2p⁶ 3s¹.

Electron configuration of chlorine atom: ₁₇Cl 1s² 2s² 2p⁶ 3s² 3p⁵.

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