The mole fraction of KBr in the solution is 0.0001
<h3>How to determine the mole of water</h3>
We'll begin by calculating the mass of the water. This can be obtained as follow:
- Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
- Density of water = 1 g/mL
- Mass of water =?
Density = mass / volume
1 = Mass of water / 400
Croiss multiply
Mass of water = 1 × 400
Mass of water = 400 g
Finally, we shall determine the mole of the water
- Mass of water = 400 g
- Molar mass of water = 18.02 g/mol
- Mole of water = ?
Mole = mass / molar mass
Mole of water = 400 / 18.02
Mole of water = 22.2 moles
<h3>How to de terminethe mole of KBr</h3>
- Mass of KBr = 0.3 g
- Molar mass of KBr = 119 g/mol
- Mole of KBr = ?
Mole = mass / molar mass
Mole of KBr = 0.3 / 119
Mole of KBr = 0.0025 mole
<h3>How to determine the mole fraction of KBr</h3>
- Mole of KBr = 0.0025 mole
- Mole of water = 22.2 moles
- Total mole = 0.0025 + 22.2 = 22.2025 moles
- Mole fraction of KBr =?
Mole fraction = mole / total mole
Mole fraction of KBr = 0.0025 / 22.2025
Mole fraction of KBr = 0.0001
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Besides producing hydrogen ions in water, all Arrhenius acids have a few things in common. They have pH values anywhere from 0 up to 7, they taste and smell sour and they will turn pH paper pink, red, or orange.
<h3>What Arrhenius acids?</h3>
A substance that raises the concentration of H+ ions in an aqueous solution is known as an Arrhenius acid. Traditional Arrhenius acids are highly polarized covalent substances that dissociate in water to form an anion (A-) and the cation H+.
Aqueous Arrhenius acids have distinguishing characteristics that serve as a useful definition of an acid. Acids can turn blue litmus red, produce aqueous solutions with a sour taste, and react with bases and some metals (like calcium) to generate salts. The Latin word acidus/acre, which means "sour," is where the word acid originates.
Although the precise definition solely refers to the solute, the term "acid" is sometimes used to refer to an aqueous solution of an acid that has a pH lower than 8.
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<u>Answer:</u>
<u>For 2:</u> The % yield of the product is 92.34 %
<u>For 3:</u> 12.208 L of carbon dioxide will be formed.
<u>Explanation:</u>
The percent yield of a reaction is calculated by using an equation:
......(1)
Given values:
Actual value of the product = 78.4 g
Theoretical value of the product = 84.9 g
Plugging values in equation 1:

Hence, the % yield of the product is 92.34 %
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(2)
Given mass of carbon dioxide = 24 g
Molar mass of carbon dioxide = 44 g/mol
Plugging values in equation 1:

<u>At STP conditions:</u>
1 mole of a gas occupies 22.4 L of volume
So, 0.545 moles of carbon dioxide will occupy =
of volume
Hence, 12.208 L of carbon dioxide will be formed.
1. 26
2. 115
THIS IS TO REACH 20 CHARACTERS