Answer: It has to be 56.10564
Explanation:1 grams KOH to mol = 0.01782 mol
10 grams KOH to mol = 0.17824 mol
20 grams KOH to mol = 0.35647 mol
30 grams KOH to mol = 0.53471 mol
40 grams KOH to mol = 0.71294 mol
50 grams KOH to mol = 0.89118 mol
100 grams KOH to mol = 1.78235 mol
200 grams KOH to mol = 3.5647
Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to its temperature if<u> the temperature and the number of particles are constant.</u>
<h3>Further Explanation</h3><h3>Boyles’s law </h3>
- This gas law states that the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
- Therefore, when the volume of an ideal gas is increased at constant temperature then the pressure of the gas will also increase.
- Mathematically; Volume α 1/Pressure
Vα1/P
- Therefore, constant k, is = PV
<h3>Other gas Laws</h3><h3>Gay-Lussac’s law </h3>
- It states that at constant volume, the pressure of an ideal gas I directly proportional to its absolute temperature.
- Thus, an increase in pressure of an ideal gas at constant volume will result to an increase in the absolute temperature.
<h3>Charles’s law</h3>
- It states that the volume of a fixed mass of a gas is directly proportional to absolute temperature at constant pressure.
- Therefore, an increase in volume of an ideal gas causes a corresponding increase in its absolute temperature and vice versa while the pressure is held constant.
<h3>Dalton’s law </h3>
- It is also known as the Dalton’s law of partial pressure. It states that the total pressure of a mixture of gases is always equivalent to the total sum of the partial pressures of individual component gases.
- Partial pressure refers to the pressure of an individual gas if it occupies the same volume as the mixture of gases.
Keywords: Gas law, Boyles's law, pressure, volume, absolute temperature, ideal gas
<h3>Learn more about:</h3>
Level: High school
Subject: Chemistry
Topic: Gas laws
Sub-topic: Boyle's Law
Answer:
The calculated concentration of sodium thiosulphate solution will be less than the actual value.
Explanation:
When IO3^2- solution is added to KI solution, I2 gas is released ,then sulphuric acid is now added to facilitate reduction. In order to prevent the escape of iodine (I2) gas ,the solution must immediately be titrated with thiosulphate.
If the solution is not immediately titrated with thiosulphate, the concentration of iodine available in the system decreases. When this occurs, it will also cause a decrease in the amount of iodine available to react with thiosulphate thus decreasing the concentration of thiosulphate obtained from calculation
Answer: 100C of heat is needed.
Explanation: That is the heating point of water
