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otez555 [7]
1 year ago
11

In a nuclear fission reaction a heavy nucleus divides to form smaller nuclei and one or more neutrons. Many nuclei can undergo f

ission, but the fission reactions of uranium-235 and plutonium-239 are the principal ones that generate energy in nuclear power plants. This problem deals with balancing the fission reaction of the uranium-235 isotope as it undergoes bombardment from a neutron.
A. When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation of two new nuclei and neutrons. The following equation is an example of one such fission process:
235 92U+10n→AZBa+9436Kr+310n
Enter the isotope symbol for the barium (Ba) nucleus in this reaction.
B. In another process in which 235 92U undergoes neutron bombardment, the following reaction occurs:
235 92U+10n→AZSr+143 54Xe+310n
Enter the isotope symbol for the strontium (Sr) nucleus in this reaction.
Chemistry
1 answer:
Effectus [21]1 year ago
7 0

Answer:

\frac{235}{92} U +\frac{1}{0} n -------> \frac{139}{36} Ba + \frac{94}{56} Kr + 3\frac{1}{0} n

\frac{235}{92} U + \frac{1}{0} n -------> \frac{90}{38} Sr  + \frac{143}{54} Xe + 3\frac{1}{0} n

Explanation:

In equation 1, equating the mass number (A) on both sides.

A = 235 + 1 = A + 94 + 3*1

236 = A + 94 + 3

A = 236 - 94 = 3

A = 139  

Equate the atomic numbers on both sides

92 + 0 = Z + 36 + 3*0

92 = Z + 36

Z = 92 - 36

Z = 56

In reaction 2, equating the mass number on both sides

235 + 1 = A + 143 + 3 *1

236 = A + 143 + 3

236 = Z + 146

Z = 90

Equatoing the atomic number of both sides

92 + 0 = Z + 54 + 3*0

92 = Z + 54

Z = 92 - 54

Z = 38.

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If 200mL of 0.60 M MgCl2 (aq) is added to 400mL of distilled water, what is the concentration of Mg2+(aq) in the resulting solut
lara [203]

Answer:

A.0.20M

Explanation:

c 1 V 1 = c 2 V 2

Initial Volume, V1 = 200 mL

Final Volume, V2 = 200 + 400 = 600 mL

Initial Concentration, c1 = 0.60 M

Final Concentration, c2= ?

Solving for c2;

c2 = c1v1 / v2

c2 = 0.60 * 200 / 600

c2 = 0.20M

3 0
1 year ago
Express each aqueous concentration in the unit indicated.
MAXImum [283]

Answer:

a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L =  9500 millileters (ml)

we consider 25 mg = 25 millileters

<em>ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)</em>

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

ppb of trichloroethylene = 3 × 10⁶ ppb

B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml =  10000  × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

<em>ppm = (mass of solute / mass of solvent) × 1000000 (1 million)</em>

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

<em>moles = mass / molar mass</em>

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

<em>Molarity = moles of solute / litres of solution</em>

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

7 0
2 years ago
Which of the following descriptions best describes a weak base?
Westkost [7]

Answer:

umm.. B. a base that generates a lot of hydroxide ions in water.

5 0
2 years ago
2KOH+H2SO4=k2SO4+2H2O Is a balanced equation, displaying the combination of potassium hydroxide with sulfuric acid increase pota
Lerok [7]

Answer:

The number of moles of potassium hydroxide, KOH required to make 4 moles of K₂SO₄ is 8 moles of KOH

Explanation:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄  and 2 moles of H₂O.

Therefore the number of moles of potassium hydroxide that will be needed to make 4 moles of K₂SO₄ is;

8KOH + 4H₂SO₄ → 4K₂SO₄ + 8H₂O

8 moles of KOH is required to make 4 moles of K₂SO₄.

6 0
1 year ago
kiran lit a candle. she placed a 100cm glass jar over the candle. the candle flame went out after 2 seconds. why did the flame g
Radda [10]

Answer:

The candle took up all the oxygen under the glass

Explanation:

Carbon dioxide molecules are heavier than air. Because of this, they push the oxygen and other molecules in the air out of the way as they sink down over the flame and candle. When oxygen is pushed away from the wick, it can't react with the wax anymore.

4 0
2 years ago
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