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Ronch [10]
3 years ago
11

What is the Molarity of a solution of HNO3 if it contains 12.6 moles in a 0.75 L solution? *

Chemistry
2 answers:
Korolek [52]3 years ago
7 0

Answer:

M = 16.8 M

Explanation:

<u>Data:</u> HNO3

moles = 12.6 moles

solution volume = 0.75 L

Molarity is represented by the letter M and is defined as the amount of solute expressed in moles per liter of solution.

M=\frac{moles}{solution volume}

The data is replaced in the given equation:

M=\frac{12.6 mol}{0.75L}=16.8\frac{mol}{L}

Zigmanuir [339]3 years ago
3 0

Answer:

The correct answer is 16.8 M

Explanation:

The molarity (M) of a solution is the number of moles of solute in 1 liter of solution. The HNO₃ solution contains 12.6 moles of solute (in this case the solute is HNO₃) in 0.75 liters. In order to find the molarity, we have to divide the number of moles into the volume in liters as follows:

M = 12.6 moles/0.75 L = 16.8 moles/L = 16.8 M

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What is the definition of a chemical bond?
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3 years ago
What is the percent yield for the reaction below when
kumpel [21]

Answer:

Percent yield of reaction is<em> 150%.</em>

Explanation:

Given data:

Percent yield = ?

Actual yield of SO₃ = 586.0 g

Mass of SO₂ = 705.0 g

Mass of O₂ = 80.0 g

Solution:

Chemical equation:

2SO₂ + O₂      →     2SO₃

Number of moles of SO₂:

Number of moles = mass/ molar mass

Number of moles = 586.0 g/ 64.1 g/mol

Number of moles = 9.1 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 80.0 g/ 32g/mol

Number of moles = 2.5 mol

Now we will compare the mole of SO₃ with O₂ and SO₂.

                      SO₂          :          SO₃

                        2             :            2

                     9.1              :           9.1

                       O₂            :          SO₃

                        1              :            2

                     2.5             :           2×2.5 = 5

The number of moles of SO₃ produced by oxygen are less it will limiting reactant.

Theoretical yield of SO₃:

Mass = number of moles × molar mass

Mass = 5 mol × 80.1 g/mol

Mass = 400.5 g

Percent yield of reaction:

Percent yield = actual yield / theoretical yield  × 100

Percent yield = 586.0 g/ 400.5 g× 100

Percent yield = 1.5× 100

Percent yield = 150%

8 0
3 years ago
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