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1 year ago

1. What is the molarity of a solution prepared by dissolving 3.11 grams of NaOH in

Chemistry

1 answer:

Elena L [17]1 year ago

3 0

The molarity of a solution prepared by dissolving 3.11 grams of NaOH in

enough water to make 300 milliliters of solution is 0. 256 mol/ L

Explanation:

<h3>What is Molarity?</h3>

Molarity (M) is the amount of a substance in a given volume of solution. It is otherwise known as the number of moles of solute in a certain amount of solution.

The unit of molarity is mol/L

<h3>Formula for calculating molarity :</h3>

Molarity = number of moles of solute divided by the volume of the solution\

<h3>Parameters </h3>

Molarity = ?

Number of moles is calculated using :

Mass of the solute = 3.11 grams of NaOH

Molar mass of the solute = Na + O + H

Let's input the atomic numbers of the elements: Na (23) , O (16) , H (1)

Molar mass of NaOH = 23 + 16 + 1 = 40 g/ mol

Number of moles = 3.11 ÷ 40 = 0.077 mol

Volume of the solution is measured in Liters

Let's convert 300 milliters to liters = 300 divided by 1000 = 0.3 Liters

Molarity = 0.077 ÷ 0.3 = 0. 256 mol/ L

Therefore, the Molarity of the solution is 0. 256 mol/ L

Read more on molarity here :

brainly.com/question/26873446

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

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Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

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Which of the following atoms has the greatest atomic radius? Oxygen (o) fluorine (F) Chlorine (CL) Carbon (C)

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Periodic Trend:

The Atomic radius of atoms generally decreases from left to right across a period

Group Trend:

The atomic radius of atoms generally increases from top to bottom within a group. As atomic number increases down a group, there is a increase in the positive nuclear charge, however the co-occurring increase in the number of orbitals wins out, increasing the atomic radius down a group in the periodic table

Answer :

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<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the molar mass of the unknown gas </h3>

The following data were obtained from the question:

Rate of unknown gas (R₁) = R
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The molar mass of the unknown gas can be obtained as follow:

R₁/R₂ = √(M₂/M₁)

R / 3.4R = √(16 / M₁)

1 / 3.4 = √(16 / M₁)

Square both side

(1 / 3.4)² = 16 / M₁

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Divide both side by (1 / 3.4)²

M₁ = 16 / (1 / 3.4)²

M₁ = 184.96 g/mol

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