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never [62]
3 years ago
7

Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kin

etics and the rate of degradation was measured to be 0.12 day−1 (at a particular sun- light intensity and temperature). Given this, how long does it take for the concentration of hypochlorous acid to reach nondetectable levels (0.05 mg · L−1) if the initial concentration were 3.65 mg · L−1?
Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
3 0

Answer:

35.75 days

Explanation:

From the given information:

For first-order kinetics, the rate law can be expressed as:

\mathsf{In \dfrac{C}{C_o} = -kt}

Given that:

the rate degradation constant = 0.12 / day

current concentration C = 0.05 mg/L

initial concentration C₀ = 3.65 mg/L

\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}

㏑(0.01369863014) = -(0.12) t

-4.29 = -(0.12)

t = -4.29/-0.12

t = 35.75 days

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Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
Aluminum can react with oxygen gas to produce aluminum oxide, what time of reaction is this?
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This is known as a synthesis reaction. A+B->C
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2 years ago
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Answer:

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5 0
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Rank the elements according to highest ionization energy:<br><br> Be, C, O, Ne, B, Li, F, N
aleksklad [387]
Given:
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C   - Carbon      - 11,2603
O   - Oxygen      - 13,6181
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5 0
3 years ago
Read 2 more answers
(Thermodynamics)
frutty [35]

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

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1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

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3 years ago
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