Answer:
0.311 mmol/L
Explanation:
109 μmol = 109*10^(-6) mol
109*10^(-6) mol = 109*10^(-6) mol*(10^3 mmol/1mol) = 109*10^(-3) mmol =
=0.109 mmol
350 mL = 0.350 L
0.109 mmol/0.350 L = 0.311 mmol/L
Answer
there are to answers one is an answer to a problem or a mixture of chemicals
Explanation:
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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Atomic mass of boron = 10.81
<h3>What are Isotopes?</h3>
Isotopes are variants of a particular element in which they have the same number of protons but differ in the number of neutrons in the atom.
So, here as we said we have isotopes which weigh 10.01 and 11.01.
Given,
relative abundance of B-10 = 10.1 amu
relative abundance of B- 11 = 11.01 amu
percentage of B-10 = 20%
percentage of B-11 = 80%
Then the relative atomic mass depends upon the relative abundance of various isotopes of that particular element. Suppose an element consists of two isotopes and average atomic mass is equal to
(Relativeabundance(1)×Atomicmass(1)+Relativeabundance(2)×Atomicmass(2)) / (Relativeabundance(1)+Relativeabundance(2))
Atomic mass of boron = (20 × 10.01 + 80 × 11.01) / (80 + 20 )
= 1081/100
= 10.81
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