Write a balanced half-reaction describing the oxidation of solid chromium to aqueous chromium(IV) cations.

Round to 4 significant figures. 35.5450

IrinaVladis [17]

Answer:

no sé m i e r d a ..............

Explanation:

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8 0

3 years ago

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Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.

Firlakuza [10]

Answer:

[Ag⁺] = 1.3 × 10⁻⁵M

s = 3.3 × 10⁻¹⁰ M

Because the common ion effect.

Explanation:

1. Value of [Ag⁺] in a saturated solution of AgCl in distilled water.

The value of [Ag⁺] in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0

C -s +s +s

E X - s s s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × s

Ksp = s²

By substitution:

1.8 × 10⁻¹⁰ = s²

s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

[Ag⁺] = 1.3 × 10⁻⁵M ← answer

2. Molar solubility of AgCl(s) in seawater

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0.54

C -s +s +s

E X - s s s + 0.54

The new equation for the Ksp equation will be:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × ( s + 0.54)

Ksp = s² + 0.54s

By substitution:

1.8 × 10⁻¹⁰ = s² + 0.54s

s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

$s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}$

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

3. Why is AgCl(s) less soluble in seawater than in distilled water.

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0

3 years ago

The picture below shows one cell dividing into two cells.

vitfil [10]

all cells come from pre existing cells

4 0

3 years ago

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A solution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.15 ∘c. what is the molal concentration of glucose in this solution

Reil [10]

1.16 moles/liter The equation for freezing point depression in an ideal solution is Î”TF = KF * b * i where Î”TF = depression in freezing point, defined as TF (pure) â’ TF (solution). So in this case Î”TF = 2.15 KF = cryoscopic constant of the solvent (given as 1.86 âc/m) b = molality of solute i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1. Solving for b, we get Î”TF = KF * b * i Î”TF/KF = b * i Î”TF/(KF*i) = b And substuting known values. Î”TF/(KF*i) = b 2.15âc/(1.86âc/m * 1) = b 2.15/(1.86 1/m) = b 1.155913978 m = b So the molarity of the solution is 1.16 moles/liter to 3 significant figures.

7 0

3 years ago

What is true when an element is oxidized? It bonds with the hydroxide ion. It loses electrons to another element. It reacts with

Mama L [17]

Answer: It loses electrons to another element.

Explanation:- Oxidation is the process in which an element loses electrons and there is an increase in the oxidation state. On losing electrons it combines with a electronegative element such as oxygen, sulphur or nitrogen etc.

$Fe\rightarrow Fe^{2+}+2e^-$

Reduction is the process in which an element gains electrons and there is a decrease in the oxidation state.

$\frac{1}{2}O_2+2e^-\rightarrow O^{2-}$

8 0

3 years ago