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Phoenix [80]
3 years ago
12

In a fission reaction of U-235, there was a release of 9.20 × 1011 kJ of energy. What amount of mass in kilograms would have bee

n lost? Recall that c = 3 × 108 m/s.
1.02 × 10-2
1.02 × 10-5
1.02 × 109
8.28 × 10-2
8.28 × 1022
Chemistry
2 answers:
Hatshy [7]3 years ago
3 0
<span>e = mc^2 
 
m = e / c^2 

Plug in your numbers in the appropriate units.
</span><span>m=1.02 × 10^-5

pulled from another response, hope it helps</span>
Aleonysh [2.5K]3 years ago
3 0
The answer is 1.02 x 10-5 B on plato
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11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

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4 0
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In a redox chemical reaction, one species gets reduced and another gets oxidized. Manganese element is reduced in this reaction.

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The oxidation number of Mn in permanganate ion was +8 on the left side and decreased to +4 on the right side of the equation. Potassium permanganate is an oxidizing agent that has reduced the manganese ion of the permanganate ion.

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