What occurs during one half-life?
2 answers:
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The new concentrations of and
are 0.25M and 19M
Calculation of number of moles of each component,
Molarity of = number of moles/volume in lit = 0. 500 M
Number of moles = molarity of × volume in lit = 0. 500 M× 0.025 L
Number of moles of = 0.0125 mole
Molarity of = number of moles/volume in lit = 0. 38 M
Number of moles = molarity of × volume in lit = 0. 38 M× 0.025 L
Number of moles of = 0.95 mole
Calculation of new concentration at volume 50 ml ( 0.05L)
Molarity of = number of moles/volume in lit = 0.0125 mole/0.05L
Molarity of = 0.25M
Molarity of = number of moles/volume in lit = 0.95mole/0.05L
Molarity of = 19 M
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<u>Answer:</u> The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.
<u>Explanation:</u>
We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.
To calculate the volume of KBr, we use the formula:
Mass of the solution = 100 grams
Density of KBr solution = 1.03g/mL
Volume of the solution = ? mL
Putting values in above equation, we get:
Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.
In 97.08 mL of solution, mass of KBr present is 8.3 grams.
So, 41.2 mL of solution will contain = of KBr.
Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.