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2 years ago

How many moles are equal to 13.25 grams of barium permanganate

Chemistry

2 answers:

Mazyrski [523]2 years ago

4 0

Answer: The number of moles of given mass of barium permanganate is 0.035 moles.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Given mass of barium permanganate = 13.25 g

Molar mass of barium permanganate = 375.2 g/mol

Putting values in above equation, we get:

$\text{Moles of barium permanganate}=\frac{13.25g}{375.2g/mol}=0.035mol$

Hence, the number of moles of given mass of barium permanganate is 0.035 moles.

Dmitry_Shevchenko [17]2 years ago

3 0

The molas mass of Ba(MnO4)2 is 375.21 g/m
So what’s next it’s a simple stoichiometry conversion:)
We start with what’s given: 13.25 g and because we want to convert to moles moles will be on top and grams on the bottom. We will end up dividing 13.25 by 375.21; this gives us 0.0353 mol Ba(MnO4)2

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Answer:

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Question:

Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.

A. 0.272 mol

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E. 0.0681 mol

Answer:

The correct option is;

E. 0.0681 mol

Explanation:

The equation for the reaction is

Zn + HCl = H₂ + ZnCl₂

Vapor pressure of the liquid = 18 torr = 2399.803 Pa

Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm

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Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation

Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa

Volume of H₂ = 1.7 L = 0.0017 m³

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Answer:

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Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

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