Answer:
There were originally 8 atoms of Potassium-40.
Explanation:
The half-life of a radioactive material is the time taken for half the original material to decay or the time required for a quantity of the radioactive substance to reduce to half of its initial value.
If the original material formed without any Argon-40, it means that the atoms originally present were Potassium-40 atoms.
Presently, there are 7 Argon-40 atoms for every 1 of Potassium-40, we can deduce the number of half-lifes the Potassium-40 has undergone as follows :
After one half-life, (1/2) there will be one Potassium-40 atom for every Argon-40 atom.
After a second half life, 1/2 × 1/2 = 1/4: there will be one Potassium-40 atom for every three atoms of Argon-40.
After a third half-life, 1/4 × 1/2 = 1/8: there will be one Potassium-40 atom for every 7 atoms of Argon-40.
Since there are 1/8 atoms of Potassium-40 presently, there were originally 8 atoms of Potassium-40.
First cell theory. All living things are composed of cells.
Second cell theory. Cells are the basic unit of structure and function in living things.
Third cell theory. All cells are produced from other cells
Answer:
92.9%
Explanation:
You have been given the actual yield of the reaction. First, you need to find the theoretical yield of the reaction. To do this, you need to (1) convert grams Fe₂O₃ to moles Fe₂O₃ (via molar mass from periodic table values), then (2) convert moles Fe₂O₃ to moles Fe (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe to grams Fe (via molar mass).
Once you have found the theoretical yield, you need to use the percent yield equation to calculate the final answer. This number should have 3 sig figs to match the given values.
<u>(Step 1)</u>
Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)
Molar Mass (Fe₂O₃): 159.684 g/mol
1 Fe₂O₃(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO₂(g)
Molar Mass (Fe): 55.845 g/mol
50.0 g Fe₂O₃ 1 mole 2 moles Fe 55.845 g
-------------------- x ------------------ x --------------------- x ---------------- = 35.0 g Fe
159.684 g 1 mole Fe₂O₃ 1 mole
<u>(Step 2)</u>
Actual Yield
Percent Yield = --------------------------- x 100%
Theoretical Yield
32.5 g Fe
Percent Yield = ---------------------- x 100% = 92.9%
35.0 g Fe
Answer:
With regards to the final composition of the mixture in the vessel at equilibrium, there will be very little amount of the reactants (BrOCl and BrCl), since some amounts of each have been used to form the products (Br2 and OCl).
