CO2 and NaCl are both compounds
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
Electronegativity
Atomic radius
Ionization energy
The answer would be periods 6-7 :)
Answer:
37.25 grams/L.
Explanation:
- Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.
<em>M = (no. of moles of KCl)/(volume of the solution (L))</em>
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∵ no. of moles of KCl = (mass of KCl)/(molar mass of KCl)
∴ M = [(mass of KCl)/(molar mass of KCl)]/(volume of the solution (L))
∴ (mass of KCl)/(volume of the solution (L)) = (M)*(molar mass of KCl) = (0.5 M)*(74.5 g/mol) = 37.25 g/L.
<em>So, the grams/L of KCl = 37.25 grams/L.</em>
