There are
A. 6 significant figures
B.5 sig figs
C.2 sig figs
D.5 sig figs
- <u>Physical Properties :</u>
- It is a crystalline colorless solid structure it is soluble in solution such as ether, alcohol, and chloroform but insoluble in water.
- It is dangerous when it is dry and is very volatile and decomposes when it reaches 103 c.
2. <u>Chemical Properties :</u>
- It causes swelling of the skin and undergoes free radical products that are highly reactive.
- It is irritant when it contacts the skin it breaks down to yield oxygen and benzoic acid.
<u>Explanation:</u>
- Benzoyl Peroxide is widely used in industry at large scale.
- It is formed when there is a reaction between hydrogen peroxide and benzoyl chloride.
- Benzoyl peroxide is used in bread and cheese as the bleaching agent and also used in cosmetics as hair coloring.
- Because of its antibacterial property and a peroxide group it is applied to the human skin to treat acne.
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
= (1)
and we also know that:
n = M x
If we replace this expression in (1) we have:
x = x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
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As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane
Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ
X = 0.970 moles of Ocatne
So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol
Mass = 110.83 g of Octane