Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ + H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
O moles - 0.64 mmol
Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol
In menthol
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C H O
0.64 1.28 0.064
x1000 x1000 x1000 to get whole numbers
640 1280 64
10 20 1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O
Question:
Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:
equations
Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.
Answer:
NO
It is present but not consumed
NO Lowers the activation energy of the reaction
Explanation:
A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction
Therefore, whereby NO is not consumed, it is the catalyst
It functions by lowering the activation energy
The balanced reaction is
Na2O + H2O --> 2NaOH
If 2.24 moles of sodium oxide react, that means 4.48 moles of NaOH is formed as it is a 1 to 2 stoichiometric relationship.
Now we multiply by the molar mass to get grams.
4.48 moles NaOH * (39.997 grams/1 mole) = 179.2 grams
Your answer is 179. grams.
Answer:
It is basically a way of telling you how to solve for different variables in the equation d=m/v
Explanation: