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3 years ago

How many milliliters of 5.50 m hcl(aq) are required to react with 9.55 g of zn(s)?

1 answer:

6 0

Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g) From the balanced chemical equation, 2 moles of HCl is required to react with 1 mole of Zn Mass of Zn = 9.55 g Molar mass of Zinc = 65.41 g/mol Number of moles = mass / molar mass Moles of Zn = 9.55 / 65.41 = 0.146 mol Moles of HCl required = 2 x moles of Zn = 2 x 0.146 = 0.292 mol

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Which metal used in hot water systems

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3 years ago

HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i

OleMash [197]

Answer: S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

$HNO_3+S\rightarrow H_2SO_4+NO$

On the reactant side:

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

On the product side:

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

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3 years ago

Use the Nernst equation to calculate the concentration of the unknown solution. Base this on your experimental voltage of 1.074

Hoochie [10]

Answer:

0.3793 M

Explanation:

The unknown metal is zinc. So the equation of the reaction is;

Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)

From Nernst equation;

E = E° - 0.0592/n log Q

[Cu2+] = 0.050179 M

n = 2

[Zn^2+] = ?

E = 1.074 V

E° = 0.34 - (-0.76) = 1.1 V

Substituting values;

1.074 = 1.1 - 0.0592/2 log [Zn^2+]/0.050179

1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179

-0.026 = -0.0296 log [Zn^2+]/0.050179

-0.026/-0.0296 = log [Zn^2+]/0.050179

0.8784 =log [Zn^2+]/0.050179

Antilog(0.8784) = [Zn^2+]/0.050179

7.558 = [Zn^2+]/0.050179

[Zn^2+] = 7.558 * 0.050179

[Zn^2+] = 0.3793 M

4 0

3 years ago