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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Water can't cool at a single temperature. It must start at a higher temperature, and drop to a lower temperature in order to cool. Unless we know the other temperature, there is no way to calculate the amount of thermal energy released.
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Effect of increasing surface area on the rate of a reaction. ... Increasing the surface area of a solid reactant exposes more of its particles to attack. This results in an increased chance of collisions between reactant particles, so there are more collisions in any given time and the rate of reaction increases.
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