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Romashka [77]
4 years ago
13

If three neutrons are produced when a neutron bombards a U- 235 atom, determine the second fission product isotope when the firs

t fission fragment is (a) xenon-133, (b) barium-144, (c) cesium-143, (d) tellurium-137, and (e) lanthanum-146.
Chemistry
1 answer:
alexira [117]4 years ago
8 0

Explanation:

235 U92 + 1n0 --> A + B + 3 1n0

when A = ;

(a) xenon-133

Mass number of B = 235 + 1 - 133 -3 = 100

Atomic number = 92 - 54 = 38

Element = Sr - Strontium

(b) barium-144

Mass number of B = 235 + 1 - 144 -3 = 89

Atomic number = 92 - 56 = 36

Element = Kr - Krypton

(c) cesium-143

Mass number of B = 235 + 1 - 143 -3 = 90

Atomic number = 92 - 55 = 37

Element = Rb - Rubidium

(d) tellurium-137

Mass number of B = 235 + 1 - 137 -3 = 96

Atomic number = 92 - 52 = 40

Element = Zr - Zirconium

(e) lanthanum-146

Mass number of B = 235 + 1 - 146 -3 = 87

Atomic number = 92 - 57 = 35

Element = Br - Bromine

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Why is the melting point of hydrogen flourine abnormally higher than other halogen acids.​
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3 years ago
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
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Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).
mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

             CuSO₄.5H₂O  →   CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

         Number of moles  = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

           Volume  = 250 x 10⁻³L

  Input the parameters and solve for number of moles;

        Number of moles  = 250 x 10⁻³  x  2.01 = 0.5mol

From the equation;

             1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O  

So  0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16)  = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

                                      = 0.5 x 249.6

                                     = 124.8g

The mass of CuSO₄.5H₂O is 124.8g

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4 years ago
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