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skad [1K]
3 years ago
14

Write the concentration equilibrium constant expression for this reaction. cr2o−27 (aq) + 6i− (aq) + 14h+ (aq) → 3i2 (s) + 2cr+3

(aq) + 7h2o (l)
Chemistry
1 answer:
dezoksy [38]3 years ago
6 0
The equilibrium constant of reaction, usually denoted as K, is a unit of ratio. The ratio involves concentrations of products to reactants. But you also have to incorporate their stoichiometric coefficients in the reaction as their respective exponents. Note that substances in their aqueous state are the ones that are included only in the expression. To properly show you how it's done, consider this equilibrium reaction:

aA (aq) + bB (l) ⇆ nN (aq)

Since only reactant A and product N are aqueous, the equilibrium constant for this reaction is:
K = [N]ⁿ/[A]ᵃ
where the [] brackets denotes concentration in molarity

Now, let's apply this to the given equation:
Cr₂O²⁻ (aq) + 6 I⁻ (aq) + 14 H⁺ (aq) → 3 I₂ (s) + 2 Cr³⁺ (aq) + 7 H₂O<span> (l)
</span>I think there is a typographical error because Cr₂O²⁻ has a negative 2 charge rather than -27. Remember that only substances in aqueous states are included in the K expression. Therefore, the expression for K is:

K = [Cr³⁺]² / [Cr₂O²⁻][I⁻]⁶[H⁺]¹⁴
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A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o
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Answer:

0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

  Mass of water = 355 g

                 ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

(a) <em>Moles of glucose </em>

n = 10.20 g × (1 mol/180.16 g)

  = 0.056 62 mol

(b) <em>Kilograms of water </em>

m = 355 g × (1 kg/1000 g)

   = 0.355 kg

(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

  = 0.056 62 mol/0.355 kg

  = 0.1595 mol·kg⁻¹

(d) <em>Freezing point depression </em>

ΔT_f = 1 × 1.86 × 0.1595

        = 0.297 °C

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You've made a hot drink by dissolving a teaspoon of instant coffee and a teaspoon of sugar in a cup of hot water. Which of the f
ollegr [7]

Answer: The statement, you've just prepared an aqueous solution is true.

Explanation:

When one or more number of substances are dissolved in a solvent like water then solution formed is called an aqueous solution.

For example, when a hot drink is made by dissolving a teaspoon of instant coffee and a teaspoon of sugar in a cup of hot water is an aqueous solution.

Here, both coffee and sugar are solute whereas hot water is the solvent.

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