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jek_recluse [69]

3 years ago

5

What poisonous gas is produced by uranium in rocks underneath a house?

1 answer:

andrey2020 [161]3 years ago

6 0

Radon is produced by uranium rocks underneath a house.

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I NEED HELP ASAP!!!!!!!!!!!!!!!!!!!

Hoochie [10]

Answer:

a) 0

Explanation:

Each of the small dots surrounding the C1 represents one electron. These are where electrivity comes from. Since there is the same number of electrons in both atoms, the difference is 0 (because 6 electrons-6electrons= 0).

4 0

2 years ago

Read 2 more answers

A particle has 5 protons, 6 neutrons, and 5 electrons. What is its net charge.

Trava [24]

It would have a zero charge because it is a neutral atom. The number of electrons which is a negative charge is equal to the number of protons which is positive, they will cancel each other out hence meaning it will become neutral

4 0

3 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

Water that fills the cracks and spaces in underground soil and rock layers is calleda. salt water.

algol [13]

The answer is groundwater

8 0

3 years ago

In a certain variety of plants, red flowers appear in majority of plants and blue flowers appear only in a few plants. Which sta

Jlenok [28]

Answer:

C

Explanation:

3 0

3 years ago