In 42 hrs, 4.32 g of argon-35 decay to leave only 0.54 g. What is the half-life of argon- 35?

o 21 hrs

O 7 hrs

O 84 hrs

14 hrs

14 hrs.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 42 hrs

Original amount (N₀) = 4.32 g

Amount remaining (N) = 0.54 g

Half-life (t½) =…?

Next, we shall determine number of half-lives required for 4.32 g of the isotope to decay to 0.54 g. This can be obtained as follow:

Original amount (N₀) = 4.32 g

Amount remaining (N) = 0.54 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

0.54 = 1/2ⁿ × 4.32

Cross multiply

0.54 × 2ⁿ = 4.32

Divide both side by 0.54

2ⁿ = 4.32 / 0.54

2ⁿ = 8

Express 8 in index form with 2 as the base

2ⁿ = 2³

n = 3

Thus, it took 3 half lives for 4.32 g of the isotope to decay to 0.54 g.

Finally, we shall determine the half-life of the isotope. This can be obtained as follow:

Time (t) = 42 h

Number of half-lives (n) = 3

Half-life (t½) =

n × t½ = t

3 × t½ = 42

Divide both side by 3

t½ = 42 / 3

t½ = 14 hrs

Thus, the half-life of the isotope is 14 hrs

gas

explanation:

smoke is the matter of a gas

colorless gas, exhibiting a blue glow when placed in an electric field

b) solid

its a type of rock