A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

<u>Explanation:</u>

Given, the initial value of the sample, $A_0$ = 150mg

Final value of the sample or the quantity left, A = 18.75mg

Time = 11.4 days

The amount left after first half life will be ½.

The number of half-life is calculated by the formula

$\frac{A}{A_{0}}=\left(\frac{1}{2}\right)^{N}$

where N is the no. of half life

Substituting the values,

$\frac{18.75}{150}=\left(\frac{1}{2}\right)^{N}$

$\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{N}$

On equating, we get, N = 3

Therefore, 3 half-lives have passed.

7 0

3 years ago

What is Loschmidt’s number? How is it related to Avogadro’s number?

xxMikexx [17]

Loschmidt's constant is referred to as Avogadro's number (they are the same). Their values are 6.02 times 10 to the -34 power.

4 0

3 years ago

List three atoms or ions that have the same number of

Charra [1.4K]

Answer:

a b and c

Explanation:

S is in period 3 so it has 3 electrons

Al is in period 3 so it has 3 electrons

P is in period 3 so it has 3 electrons

Kr is in period 4 so it has 4 electrons

Cs is in period 6 so it has 6 electrons

8 0

1 year ago