Chemistry, 18.11.2019 11:31 briarwilliams9668

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the mass of cobalt-60 in a sample decreased from 0.800g to 0.200g over a period of 10.5 years. from this information, calculate the half-life of cobalt-60.

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Answer from: amberunderhill

The half-life of cobalt-60 in this sample is 5.25 years.

Explanation:

Step One: How many half-live(s) in 10.5 years?

The mass of a radioactive element drop by one-half every half-life. This mass will drop by another one-half after another half-life, and so on so forth. In other words,

$\begin{array}{c|c}\text{Number of half-live(s) elapsed}& \dfrac{\text{Current Mass}}{\text{ Initial Mass}}\\0 & 1 \\ 1 & 0.5 \\ 2 & 0.25 \\ \dots & \dots \end{array}$ .

For this sample:

Initial Mass: 0.800 grams.Current Mass after 10.5 years: 0.200 grams.

$\dfrac{\text{Current Mass}}{\text{ Initial Mass}} = \rm \dfrac{0.200\; g}{0.800\; g} = 0.25$ .

Referring to the table, two half-lives have elapsed before the mass drop to 0.200 grams. There are two half-lives in 10.5 years.

Step Two: What's the length of each half-life?

There are two half-lives in 10.5 years. The length of each half-life will be

$\displaystyle \rm \frac{1}{2} \times 10.5 = 5.25\;years$ .

Alternative approach to step one: find the number of half-live(s) in 10.5 years with a scientific calculator.

In case the ratio between current mass and initial mass is between two entries, the tabular approach will no longer work. The number of half-lives in the 10.5 years can still be found using logarithm.

$\displaystyle \text{Number of half-lives} = \log_\frac{1}{2}{\left(\frac{\text{Current Mass}}{\text{Final Mass}}\right )} = \frac{\ln{\left(\dfrac{\text{Current Mass}}{\text{Final Mass}}\right )}}{\ln{\left(\dfrac{1}{2}}\right)}}$ .

For this question,

$\begin{aligned}\displaystyle \text{Number of half-lives} &= \frac{\ln{\left(\dfrac{\text{Current Mass}}{\text{Final Mass}}\right )}}{\ln{\left(\dfrac{1}{2}}\right)}}\\ &=\frac{\ln{\left(\dfrac{0.200}{0.800}\right)}}{\ln{\left(\dfrac{1}{2}}\right)}} \approx \frac{-1.38629}{-0.693147} \\&=2 \end{aligned}$ .

Answer from: Quest

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ 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 $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

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