Five more than two times the square root of a number is fifteen. What’s the number
2 answers:
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<u>Answer:</u> The mass of decane produced is
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
......(1)
Mass of hydrogen gas = 2.45 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1:, we get:
The chemical equation for the hydrogenation of decene follows:
As, decene is present in excess. So, it is considered as an excess reagent.
Thus, hydrogen gas is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of hydrogen gas produces 1 mole of decane.
So, 1.225 moles of hydrogen gas will produce = of decane
Now, calculating the mass of decane by using equation 1, we get:
Moles of decane = 1.225 mol
Molar mass of decane = 142.30 g/mol
Putting values in equation 1, we get:
Hence, the mass of decane produced is
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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