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
1)Delta H=(Delta H of reactants)-(Delta H of products)
2)And we know CO have 3 bond CO and CO2 have 2 bond that each of them are 2 bond, please see the picture!
so lets answer it:
Answer:
Hello! Some household items that are electroplated include kitchenware, such as metal pots and pans, door handles, mobile phones, and coins. An electroplated object is coated by electrolytic deposition with chromium, silver, or another metal.
Answer:
The ones with 8 protons
Explanation:
Since there are two of them with 8 protons, we can assume they are the same element. The first 8 proton element has 10 neutrons while the second has 11. This makes them isotopes of one another
Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y = 
= 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>
