lodine-131 has a half life of about 8 days. If you had 60 grams of , how much would remain after 24 days?

7.5 grams

Explanation:

From the question,

(R/R')ᵇ = 2ᵃ Equation 1

Where b = half life, R = Original mass of iodine, R' = Remaining mass of Iodine after disintegration, a = Total time taken.

⇒ R/R' = 2ᵃ/ᵇ Equation 2

Given: R = 60 grams, a = 24 days, b = 4 days

Substitute these values into equation 2

60/R' = 2²⁴/⁸

60/R' = 2³

R' = 60/2³

R' = 60/8

R' = 7.5 grams

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west