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GrogVix [38]
3 years ago
5

Consider the cell described below at 259 K: Ni | Ni2+ (1.15 M) || Fe3+ (2.23 M) | Fe Given the standard reduction potentials fou

nd on the sheet attached to the exam, calculate the cell potential after the reaction has operated long enough for the [Fe3+] to have changed by 1.248 M
Chemistry
1 answer:
likoan [24]3 years ago
8 0

Answer:

i dont know sorry :(

Explanation:

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What are the products of the chemical reaction that occurs when zinc metal is added to an aqueous solution of hydrochloric acid?
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Zn(s) +2 HCl(aq) ---> ZnCl2(aq) +H2(g) (products are Zinc chloride and hydrogen gas)
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What does this weather map symbol represent
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B

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As the frequency of a wave increases, what happens to the energy the wave is carrying?
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A solution is prepared by adding 100 mL of 1.0 M HC₂H₃O₂(aq) to 100 mL of 1.0 M NaC₂H₃O₂(aq). The solution is stirred and its pH
anastassius [24]

Answer: Option (c) is the correct answer.

Explanation:

When an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This is because a buffer has the ability to not get affected by the addition of small amounts of an acid or a base. As a result, it helps in maintaining the pH of the solution.

In the give case, when we add the HCl then more number of protons will dissociate. This causes the acetate to react with the protons and leads to the formation of acetic acid.

We know that acetic acid is a weak acid and it dissociates partially or feebly. Therefore, no change in pH will take place.

Thus, we can conclude that equation H_{3}O^{+}(aq) + C_{2}H_{3}O^{-}_{2}(aq) \rightarrow HC_{2}H_{3}O_{2}(aq) + H_{2}O(l) represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH.

3 0
4 years ago
A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a
Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄  

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

                 BaC₂O₄ ⇌   Ba²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                   5.0 × 10⁻⁵     c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

                ZnC₂O₄ ⇌   Zn²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                 2.0 × 10⁻⁷      c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

                 Ag₂C₂O₄ ⇌   2Ag⁺   +   C₂O₄²⁻  

E/mol·L⁻¹:                      3.0 × 10⁻⁵       c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when   c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when   c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028       mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0
3 years ago
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