The empirical formula is a formula of a compound showing the proportion of each element involved in the compounds but it does not represent the total number of atoms in the compound. It is the lowest number of ratio between the elements in the compound. In order, to determine the actual number of the atoms or the molecular formula of the compounds, we make use of the molar mass of the compound.
<span>To
determine the molecular formula, we multiply a value to the empirical formula.
Then, calculate the molar mass and see whether it is equal to the one
given (104.1 g/ mol). From the choices, the only valid options are b, d and e.
</span> molar mass
1 CH 13.02
8 C8H8 104.16
6 C6H6 78.12
Therefore the correct answer is option B.
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
<span>0.0292 moles of sucrose are available.
First, lookup the atomic weights of all involved elements
Atomic weight Carbon = 12.0107
Atomic weight Hydrogen = 1.00794
Atomic weight Oxygen = 15.999
Now calculate the molar mass of sucrose
12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol
Divide the mass of sucrose by its molar mass
10.0 g / 342.29208 g/mol = 0.029214816 mol
Finally, round the result to 3 significant figures, giving
0.0292 moles</span>
Answer:
Nitrogen and Oxygen make up 99% of the Earths atmosphere. Then Argon makes up the rest of the atmosphere.
The balanced chemical reaction:
C3H8 + 5O2 = 3CO2 + 4H2O
We are given the amount of the carbon dioxide to be produced. This will be the starting point of our calculations.
<span>43.62 L CO2 ( 1 mol CO2 / 22.4 L CO2 ) (5 mol O2 / 3 mol CO2 ) (
22.4 L O2 / 1 mol O2) = 72.7 L O2</span>
