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lys-0071 [83]
2 years ago
9

A sample of crude sodium iodide was analysed by the following rection.

Chemistry
1 answer:
amm18122 years ago
6 0

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

Let's consider the following unbalanced redox reaction.

I⁻ + SO₄²⁻ → I₂ + H₂S

  • The oxidation number of I goes from -1 (I⁻) to 0 (I₂) so it is oxidized.
  • The oxidation number of S goes from +6 (SO₄²⁻) to -2 (H₂S) so it is reduced.

The corresponding half-reactions are:

I⁻ → I₂

SO₄²⁻ → H₂S

We will perform the mass balance adding OH⁻ and H₂O where appropriate.

2 I⁻ → I₂

6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Then, we will perform the charge balance adding electrons where appropriate.

2 I⁻ → I₂ + 2 e⁻

8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Finally, we will multiply the first half-reaction by 4 and the second by 1, and add them.

4 × (2 I⁻ → I₂ + 2 e⁻)

1 × (8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻)

------------------------------------------------------------

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

Learn more: brainly.com/question/2671074

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kati45 [8]

Explanation:

The given data is as follows.

T_{c} = 25^{o}C,  T_{h} = 350^{o}C

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Using this relation we will calculate the efficiency as follows.

                 Efficiency = [1 - (\frac{T_{c}}{T_{h}} )]

                        = 1 - (\frac{25}{350})

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7 0
3 years ago
Cyclopropene decomposes to propene when heated to 500 C, calculate rate constant for the first order reaction 0 min 1.48 mmol/L
Dmitry_Shevchenko [17]

Answer:

k = 0.0306 min-1

Explanation:

The table is given as;

Time, Concentration

0 1.48

5 1.27

10 0.98

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The integrated rate law for a first order reaction is given as;

ln [A] = -kt + ln [Ao]

where;

[A] = Final Concentration

[Ao] = Initial Concentration

k = rate constant

t = time

In the table, taking the first two sets of values;

t = 5

k = ?

[Ao]  = 1.48

[A] = 1.27

Inserting into the equation;

ln(1.27) = - k (5) + ln(1.48)

ln(1.27)  - ln(1.48) = -5k

-0.1530 = -5k

k = -0.1530 / -5

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