Answer: The mass percent of hydrogen in ascorbic acid is 4.5 %
Explanation:
In 
, there are 6 carbon atoms, 8 hydrogen atoms and 6 oxygen atoms.
To calculate the mass percent of element in a given compound, we use the formula:
Mass of hydrogen = 
Molar Mass of ascorbic acid =
Putting values in above equation, we get:
Hence, the mass percent of of hydrogen in ascorbic acid is 4.5 %.
Answer:
Oxygen in hydrogen peroxide oxidizes from -1 to 0.
Explanation:
Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.
The given reaction is shown below as:
Manganese in 
has oxidation state of +7
Manganese in 
has an oxidation state of +2
It reduces from +7 to +2
Oxygen in hydrogen peroxide has an oxidation state of -1.
Oxygen in molecular oxygen has an oxidation of 0.
Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.
T<span>he </span>Andes<span> range has many active volcanoes, which are distributed in four volcanic zones separated by areas of inactivity. The </span>Andean<span> volcanism is a result of subduction of the Nazca Plate and Antarctic Plate underneath the South American Plate.</span>
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
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I think it's RbCl and CaO
