Answer:
Option C, (Actual yield ÷ percent yield) × 100
Explanation:
Theoretical yield is defined as the total amount of product formed for given reactants in a chemical reaction. It is an ideal case which assumes no exceptions or wastage.
The mathematical relation between the actual yield, percent yield and theoretical yield is as follows -
Where
P.Y. represents the percent yield a
M A.Y. represents the mass obtained from actual yield
M T.Y. represents the mass obtained from theoretical yield
Hence, if we rearrange the formula, we get -
Hence, option C is correct
2.258625 *10²³ oxygen atoms will be produced.
<h3><u>Explanation:</u></h3>
Decomposition reaction is defined as the type of reaction where one single reactant breaks to produce more than one product only by means of heat or other external factor.
Formula of magnesium oxide = MgO.
The molecular mass of magnesium oxide = 24 +16= 40.
So in 40 grams of magnesium oxide, number of molecules is 6.023 * 10²³.
So in 15 grams of magnesium oxide,, number of molecules is 6.023 *1023 * 15/40 = 2.258625 *10²³.
From one molecule of magnesium oxide, one oxide atom will be produced.
So number of oxide atoms with 100% yeild = 2.258625 *10²³
Answer:
18.84 g of silver.
Explanation:
We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole of sulphur contains 6.02×10²³ atoms.
1 mole of sulphur = 32 g
Thus,
32 g of sulphur contains 6.02×10²³ atoms.
Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.
From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.
Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.
This is illustrated below:
1 mole of silver = 6.02×10²³ atoms.
1 mole of silver = 108 g
108 g of silver contains 6.02×10²³ atoms.
Therefore, Xg of silver will contain 1.05×10²³ atoms i.e
Xg of silver = (108 × 1.05×10²³)/6.02×10²³
Xg of silver = 18.84 g
Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur
