Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

Answer: The half life of the sample of phosphorus-32 is $14.28days^{-1}$

Step-by-step explanation:

The equation used to calculate the half life of the sample is given as:

$m(t)=m_o(1/2)^{t/h}$

where,

m(t) = amount of sample after time 't' = 356 g

$m_o$ = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

$356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}$

Hence, the half life of the sample of phosphorus-32 is $14.28days^{-1}$

7 0

2 years ago

Equation <img src="https://tex.z-dn.net/?f=%28z%20%2B%20%20%5Cfrac%7B6%7D%7B7%7D%20%29%20%2B%20%20%5Cfrac%7B2%7D%7B14%7D%20%2

ddd [48]

Given:

The equation is

$\left(z+\dfrac{6}{7}\right)+\dfrac{2}{14}=-1$

To find:

The solution of the given equation.

Solution:

We have,

$\left(z+\dfrac{6}{7}\right)+\dfrac{2}{14}=-1$

It can be written as

$\left(z+\dfrac{6}{7}\right)+\dfrac{1}{7}=-1$

$\left(z+\dfrac{6}{7}\right)=-1-\dfrac{1}{7}$

Multiply both sides by 7.

$7z+6=-7-1$

$7z+6=-8$

$7z=-8-6$

$7z=-14$

Divide both sides by 7.

$z=\dfrac{-14}{7}$

$z=-2$

Therefore, the value of z is -2.

5 0

2 years ago

The Latin Club charted a bus to travel to and participate in a state competition. The cost for each adult to ride the bus was $2

Bumek [7]

a+s =17 25a + 15s = 305

7 0

2 years ago