What are equation for the chemical change that produces water from two hydrogen molecules and one oxygen molecule. And the react
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Answer:
The volume of the air is 0.662 L
Explanation:
Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:
If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:
In this case:
- V1= 0.730 L
- T1= 28 °C= 301 °K (0°C= 273°K)
- V2= ?
- T2= 0°C= 273 °K
Replacing:
Solving:
V2=0.662 L
<u><em>The volume of the air is 0.662 L</em></u>
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.
Learn more about molarity here:
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pH=4.625
The classification of this sample of saliva : acid
<h3>Further explanation</h3>The water equilibrium constant (Kw) is the product of concentration
the ions:
Kw = [H₃O⁺] [OH⁻]
Kw value at 25° C = 10⁻¹⁴
It is known [OH-] = 4.22 x 10⁻¹⁰ M
then the concentration of H₃O⁺:
pH=-log[H₃O⁺]
Saliva⇒acid(pH<7)