What is the half‑life of an isotope that decays to 25% of its original activity in 85.6 h

42.9 h

Explanation:

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

Original amount (N₀) = 100%

Amount remaining (N) = 25%

Time (t) = 85.8 h

Half-life (t½) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Original amount (N₀) = 100%

Amount remaining (N) = 25%

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

25 = 1/2ⁿ × 100

Cross multiply

25 × 2ⁿ = 100

Divide both side by 25

2ⁿ = 100/25

2ⁿ = 4

Express 4 index form with 2 as the base

2ⁿ = 2²

n = 2

Thus, two half-lives has elapsed.

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

Time (t) = 85.8 h

Number of half-lives (n) = 2

Half-life (t½) =?

n = t / t½

2 = 85.8 / t½

Cross multiply

2 × t½ = 85.8

Divide both side by 2

t½ = 85.8 / 2

t½ = 42.9 h

Thus, the half-life of the of the isotope is 42.9 h

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.