Answer:
C4H8O4
Explanation:
To determine the molecular formula, first, let us obtain the empirical formula. This is illustrated below:
From the question given, we obtained the following information:
C = 45.45%
H = 6.12%
O = 48.44%
Divide the above by their molar mass
C = 45.45/12 = 3.7875
H = 6.12/1 = 6.12
O = 48.44/16 = 3.0275
Divide by the smallest
C = 3.7875/3.0275 = 1
H = 6.12/3.0275 = 2
O = 3.0275/3.0275 = 1
The empirical formula is CH2O
The molecular formula is given by [CH2O]n
[CH2O]n = 132.12
[12 + (2x1) + 16]n = 132.12
30n = 132.12
Divide both side by the coefficient of n i.e 30
n = 132.12/30 = 4
The molecular formula is [CH2O]n = [CH2O]4 = C4H8O4
The chemical reaction would be written as
2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
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We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.
150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4
Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:
1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2
For the first task, lighting a campfire can be used. Quantities of firestarter and kindling do not need to be precise. For the second task, the process of titration can be used. In a titration, the precise amount of a substance may be determined by adding a precise amount of a reactive species
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3
i.e 100 tonnes of CaCO3 .
mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2
mass % of CO2 in CaCO3 = (44/100)×100 =44%
i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.
therefore, 44% of CO2 is present in CaCO3.
Answer:
The answer to be filled in the respective blanks in question is
3 and 1
Explanation:
So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.
Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1
