The activity of the sample when it was shipped from the manufacturer is 4.54 mCi
<h3>How to determine the number of half-lives that has elapsed </h3>
From the question given above, the following data were obtained:
- Time (t) = 48 hours
- Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
- Number of half-lives (n) =?
n = t / t½
n = 48 / 342.72
n = 0.14
<h3>How to determine the activity of the sample during shipping </h3>
- Number of half-lives (n) = 0.14
- Original activity (N₀) = 5.0 mCi
- Activity remaining (N) =?
N = N₀ / 2ⁿ
N = 5 / 2^0.14
N = 4.54 mCi
Thus, the activity of the sample during shipping is 4.54 mCi
Learn more about half life:
brainly.com/question/2674699
D. due to the the water it will bring sand with the water there for us is D.
Answer: The mass of 
and 
produced are 336.6 g and 183.6 g respectively.
Explanation:
The combustion reaction between propane and oxygen leads to formation of carbon dioxide and water.
Law of Conservation of mass states that the mass will remain constant for a balanced equation. This is carried out when the total number of atoms on reactant side is same as the total number of atoms on the product side. Thus the equation must be balanced.
a) 1 mol of propane produces = 3 moles of
Thus 2.55 mol of propane produces = ![\frac{3}{1}\times 2.55=7.65 moles of [tex]CO_2](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B1%7D%5Ctimes%202.55%3D7.65%20moles%20of%20%5Btex%5DCO_2)
mass of 
b) 1 mol of propane produces = 4 moles of
Thus 2.55 mol of propane produces = ![\frac{4}{1}\times 2.55=10.2 moles of [tex]H_2O](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B1%7D%5Ctimes%202.55%3D10.2%20moles%20of%20%5Btex%5DH_2O)
mass of 
The mass of and produced are 336.6 g and 183.6 g respectively.
