the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment
theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations
we need to first find the theoretical yield
2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂
molar ratio of Na₂O₂ to Na₂CO₃ is 2:2
number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed
number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol
therefore number of Na₂CO₃ moles formed is - 0.10 mol
mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g
therefore theoretical yield is 10.6 g
percent yield = actual yield / theoretical yield x 100%
81.0 % = actual yield / 10.6 g x 100 %
actual yield = 10.6 x 0.81
actual yield = 8.59 g
therefore actual yield is 8.59 g
Answer:
Light is used by plants in a complex system that uses the light and several other resources to produce food. The more food= the more the plant grows, but too much food can be unhealthy for the plant. Some plants require more energy and others require less, this is because of the environment that they are in.
Explanation:
Answer:
The correct answer is B.2.52 mol
Explanation:
The molarity of a solution (M) expresses the number of moles of solute there is in 1 liter of solution. If we have the molarity (M) and the volume in liters (L), we simply multiply M x V to obtain the moles, as follows:
M = 2.21 = 2.21 moles/L
V = 1.14 L
M x V = 2.21 moles/L x 1.14 L = 2.5194 moles ≅ 2.52 moles
Answer:
One quart.
Explanation:
Let us consider the relation between all of the given units of volume and one litre.
a) one litre equals to 33.84 ounces
b) one litre equals to 2.11 pint
c) one litre equals to 0.95 quart
d) one litre equals to 4.54 gallons
Thus from above conversions it is clear that one litre is almost equal to or about 1 quart (0.95 approxes 1).
Answer:
Molecular formula is: X₆Y₃
Explanation:
First of all we think in centesimal composition:
25 % of X and 75 % of Y
This means, that in 100 g of compound we have 25 g of X and 75 g of Y.
Therefore, our compound is: XY
To determine the moles, we divide the % by the molar mass of each
25 g / 25 g/mol = 1 X
75 g / 150 g/mol = 0.5 Y
As we can not have decimal numbers we mutiply x2.
X₂Y
This is the empirical formula, which is not the same as molecular one. We calculated from 100 g of compound so now we finish the excersise by rules of three:
In 100 g of compound we have 1 moles of X
In 600 g we must have (600 . 1)/100 = 6 X
In 100 g of compound we have 0.5 mol of Y
In 600 g we must have (600 . 0.5)/100 = 3 Y
Molecular formula is: X₆Y₃
We can confirm it by the molar mass:
25 g/mol . 6 + 150 g/mol . 3 = 600 g/mol