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Rudiy27
2 years ago
12

The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r

→ or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
What is the rate law for this reaction?
a. Rate = [1-1
b. Rate = K[OCI-11-2
c. Rate = K[OCI,20-1
d. Rate = K[OCH
e. Rate = K[OC1-20-1
Chemistry
1 answer:
Vinil7 [7]2 years ago
5 0

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

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Write 3 to 5 sentences about predicting the properties of acids and bases​
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cause of the properties of their aqueous solutions. Those properties are outlined below:

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Acids react with bases to produce a salt compound and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The products of this reaction are an ionic compound, which is labeled as a salt, and water.

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Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can.

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8 0
3 years ago
What is the mass, in grams, of solute in 255mL of a 0.0525M solution of KMnO4 (MM =
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Mass = 2.12 g

Explanation:

Given data:

Volume of KMnO₄ = 255 mL (255/1000 = 0.255 L)

Molarity = 0.0525 M

Mass in gram = ?

Solution:

First of all we will calculate the number of moles.

<em>Molarity = number of moles of solute / volume in litter</em>

0.0525 M = number of moles of solute / 0.255 L

Number of moles of solute = 0.0525 M ×0.255 L

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Mass = 0.0134 mol × 158.04 g/mol

Mass = 2.12 g

4 0
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