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3 years ago

100.0 mL of a 0.500 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution

Chemistry

1 answer:

andreyandreev [35.5K]3 years ago

3 0

Answer:

0.1 M

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 100 mL

Initial concentration (C1) = 0.5 M

Final volume (V2) = 500 mL

Final concentration (C2) =?

Using the dilution formula C1V1 = C2V2, the new concentration of the solution can be obtained as follow:

C1V1 = C2V2

0.5 × 100 = C2 × 500

50 = C2 × 500

Divide both side by 500

C2 = 50/500

C2 = 0.1 M

Therefore, the new concentration of the solution is 0.1 M

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Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation

nikdorinn [45]

Answer: The isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

Explanation:

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

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To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is $_{56}^{139}\textrm{Ba}$

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is $_{38}^{89}\textrm{Sr}$

Hence, the isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

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