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Oksi-84 [34.3K]
2 years ago
9

If a hazardous bottle is labeled 20.2 % by mass Hydrochloric Acid, HCl, with a density of 1.096 g/mL. Calculate the molarity of

the HCl solution.
a. 6.07 M
b. 0.22 M
c. 14.7 M
Chemistry
1 answer:
Brut [27]2 years ago
8 0

<u>Answer:</u> The molarity of HCl solution is 6.07 M

<u>Explanation:</u>

We are given:

20.2 % hydrochloric acid solution

This means that 20.2 grams of hydrochloric acid is present in 100 grams of solution

To calculate the volume for given density of substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Mass of solution = 100 g

Density of solution = 1.096 mL

Putting values in above equation, we get:

1.096g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.096g/mL}=91.24mL

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of hydrochloric acid = 20.2 g

Molar mass of hydrochlroic acid = 36.5 g/mol

Volume of solution = 91.24 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{20.2\times 1000}{36.5g/mol\times 91.24}\\\\\text{Molarity of solution}=6.07M

Hence, the molarity of HCl solution is 6.07 M

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Answer:

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

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Mass of methane burnt = 530 g

Moles of methane burnt = \frac{530 g}{16 g/mol}=33.125 mol

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ

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H₂Lv

Explanation:

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If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products
puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

  • <u>For a:</u>

The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

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As, K_{eq}>1, the reaction will be favored on the product side.

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Urea is a common fertilizer with the formula CO(NH2)2 that is sometimes used as a chemical de-icer for icy road surfaces in the
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Answer:

The answer to your question is  molality = 0.61

Explanation:

Freezing point is the temperature at which a liquid turns into a solid if a solute is added to a solution, the freezing point changes.

Data

Kf = 1.86 °C/m

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ΔTc = 1.13°C

Formula

ΔTc = kcm

Solve for m

m = ΔTc/kc

Substitution

m = 1.13 / 1.86

Simplification and result

m = 0.61

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