All categories

2 years ago

Justify the statement nuclear energy is harmful as well as useful

Chemistry

1 answer:

dmitriy555 [2]2 years ago

7 0

The beneficial effects of nuclear energy is usually overlooked because of its potentially devastating side-effects

You might be interested in

At what temperature is a gas likely to have the lowest rate of diffusion?

Morgarella [4.7K]

Answer:

150

Explanation:

A gas will have the lowest rate of diffusion the lowest temperature from a given set of temperature values.

Since 150 is the lowest value, then this represents the lowest rate of diffusion.

The rate of diffusion is related to the prevalent temperature. When gases diffuse, they move from regions of high concentration to the at of low concentration. They must have enough kinetic energy to move in order to diffuse.

The higher the temperature, the more the kinetic energy and higher rate of diffusion.

4 0

3 years ago

Read 2 more answers

Chemical energy is a form of A. kinetic energy only. B. both potential and kinetic energy. C. neither potential nor kinetic ener

svetlana [45]

wrong. If its for the study island its potential energy only.

6 0

3 years ago

You have 16.0 g of some compound and you perform an experiment to remove all of the oxygen, 11.2 g of iron is left. What is the

oksian1 [2.3K]

The empirical formula of this compound is equal to $Fe_{2}O_3$ .

<h3>Empirical formula</h3>

To calculate the empirical formula of a compound, it is necessary to know the number of moles present.

Therefore, we will use the molar mass of iron and oxygen to find the amount of moles, so that:

$MM_O = 16g/mol\\MM_Fe = 55.8g/mol$

$MM = \frac{m}{mol}$

Oxygen

$16 = \frac{4.8}{mol}$

$mol = 0.3$

Iron

$55.8 = \frac{11.2}{mol}$

$mol = 0.2$

Finally, as the empirical formula is composed of integers numbers of moles, just multiply the values by the smallest common factor to transform into an integer, so that:

O => $0.3 \times 10 = 3moles$

Fe => $0.2 \times 10 = 2moles$

So, the empirical formula of this compound is equal to $Fe_{2}O_3$

Learn more about empirical formula in: brainly.com/question/1363167

6 0

3 years ago

What percent of the earth's surface does the ocean cover? 37% 75% 78% 82%

kiruha [24]

About 75% of earth’s surface is covered by the ocean

4 0

3 years ago

Read 2 more answers

During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S

Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

<u>Explanation:</u>

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. The equation given by this law follows the equation:

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

We are given:

Molar mass of $^{235}UF_6=349.034348g/mol$

Molar mass of $^{238}UF_6=352.041206g/mol$

By taking their ratio, we get:

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}$

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}$

From the above relation, it is clear that rate of effusion of $^{235}UF_6$ is faster than $^{238}UF_6$

Difference in the rate of both the gases, $Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816$

To calculate the percentage increase in the rate, we use the equation:

$\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100$

Putting values in above equation, we get:

$\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%$

The above calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

3 0

3 years ago