Question

If a substance has a concentration of 5.80 M when a reaction begins, what will the concentration be after seven half-lives?

Answer 1

The concentration of the substance after seven half-lives would be approximately 0.04534 M.

Step-by-step explanation:

Given that the substance has a concentration of 5.80 M when the reaction begins and we want to find the concentration after seven half-lives, we can use the formula for the amount of reactant remaining after n half-lives of a first-order reaction, which is (1/2)n times the initial concentration.

So, if we substitute n = 7, the concentration after seven half-lives would be (1/2)7 times the initial concentration.

Converting it to decimal form, (1/2)7 = 0.0078125. Multiplying this by the initial concentration of 5.80 M, we get a concentration of approximately 0.04534 M after seven half-lives.

Answer 2

The concentration of a 5.80 M substance after seven half-lives would be 0.0453 M, calculated by multiplying the initial concentration by (1/2)⁷.

Step-by-step explanation:

If a substance has a concentration of 5.80 M when a reaction begins, the concentration after seven half-lives can be calculated using the concept that after each half-life, the concentration reduces to half. This concept implies that after n half-lives, the concentration of a substance becomes (1/2)ⁿ times its original concentration.

To find the concentration after seven half-lives, we apply this formula with n equal to 7:

• Concentration after 7 half-lives = initial concentration × (1/2)⁷
• Concentration after 7 half-lives = 5.80 M × (1/128)
• Concentration after 7 half-lives = 0.0453 M

The concentration of the substance after seven half-lives would therefore be 0.0453 M.