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2 years ago

140 g of KCl is dissolved in 600 mL of water. What is the molarity?

Chemistry

1 answer:

FromTheMoon [43]2 years ago

7 0

Answer:

$\boxed {\boxed {\sf molarity \approx 3 \ M \ KCl}}$

Explanation:

Molarity is a measure of concentration in moles per liter. The formula is:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

We are given grams of solute and liters of solution, so we must convert both before calculating molarity.

<h3>1. Convert Grams to Moles</h3>

We convert grams to moles using the molar mass. This value is found on the Periodic Table. It is the same as the atomic mass, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We have the compound KCl, so we look up the molar masses of the individual elements.

Potassium (K): 39.098 g/mol
Chlorine (Cl): 35.45 g/mol

The compound does not contain subscripts, so we can add the molar masses together to find the molar mass of the compound.

Potassium chloride (KCl): 39.098+ 35.45= 74.548 g/mol

Use the molar mass as a ratio.

$\frac {74.548 \ g\ KCl}{1 \ mol \ KCl}$

Multiply by 140 grams of KCl.

$140 \ g\ KCl*\frac {74.548 \ g\ KCl}{1 \ mol \ KCl}$

Flip the ratio so the units of grams of KCl cancel.

$140 \ g\ KCl*\frac{1 \ mol \ KCl} {74.548 \ g\ KCl}$

$140 *\frac{1 \ mol \ KCl} {74.548 }$

$1.877984654 \ mol \ KCl$

<h3>2. Convert Milliliters to Liters</h3>

1 liter contains 1000 milliliters. Create another ratio.

$\frac{ 1 \ L}{1000 \ mL}$

Multiply by 600 milliliters (the value we are converting).

$600 \ mL *\frac{ 1 \ L}{1000 \ mL}$

The units of milliliters cancel.

$600 \ *\frac{ 1 \ L}{1000 }$

$0.6 \ L$

<h3>3. Calculate Molarity</h3>

Now we know the moles of solute and the liters of solution.

1.877984654 mol KCl and 0.6 L

Substitute the values into the molarity formula.

$molarity= \frac {1.877984654 \ mol \ KCl}{0.6 \ L}$

$molarity= 3.129974424 \ mol \ KCl/L$

The original measurements of grams and milliliters have 2 and 1 significant figures. We must round our answer to the least number of sig figs: 1.

For the number we found, that is the ones place. The 1 in the tenths place tells us to leave the 3 in the ones place.

$molarity \approx 3 \ mol \ KCl/L$

1 mole per liter is equal to 1 molar or M. Convert the units.

$molarity \approx 3 \ M \ KCl$

The molarity is approximately <u>3 M KCl.</u>

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Taking logarithm on both the sides, we get:

$\ln k=-\frac{Ea}{RT}+\ln A$ ............(1)

where,

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Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

$\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)$

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2 years ago

3. In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. Write a balanced

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Answer:

SO₂ + 0.5 O₂ + H₂O → H₂SO₄

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Explanation:

In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. The balanced chemical equation is:

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The molar mass of SO₂ is 64.07 g/mol. The moles of SO₂ corresponding to 2.50 g are:

2.50 g × (1 mol/64.07 g) = 0.0390 mol

The molar ratio of SO₂ to H₂SO₄ is 1:1. The moles of H₂SO₄ formed are 0.0390 moles.

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Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you t

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Answer:

Sample A is a mixture

Sample B is a mixture

Explanation:

For sample A, we are told that the originally yellow solid was dissolved and we obtained an orange powder at the bottom of the beaker. Subsequently, only about 30.0 g of solid was recovered out of the 50.0g of solid dissolved. This implies that the solid is not pure and must be a mixture. The other components of the mixture must have remained in solution accounting for the loss in mass of solid obtained.

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Explanation:

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Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

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Part A

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By the Nersnt equation:

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Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

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3 years ago