Answer: Ethyl Ethanoate can be used as a developing solvent. It’s safer.
Explanation:Di ethyl ether should be carefully used because it’s highly flammable and intoxicating when inhaled and can cause explosions because of its high reactivity to air and light.
To convert the mass of a compound to formula units, the conversion factor is Avogadro's number, 6.022 x10^23 formula units/ mol and its molar mass. In this case, we are given with 19.0 grams of magnesium chloride which has a mass of 95.21 g/mol. Hence the answer is 1.20 x 10^23 formula units.
Hello!
To find the mass of helium, we need to multiply the total moles by the mass of helium. We are given 2.714 moles of helium, and the mass of helium is about 4.00 grams. Now, we multiply the two values together to get the grams.
2.714 moles x 4.00 grams = 10.856 grams
According to the number of significant figures, 2.714 moles of helium has a mass of 10.9 grams (exact value: 10.856 grams).
The correct answer is alcohol. It is the common component in beer, wine and any liquor. Usually, alcohol is produced by fermentation of organic products containing glucose to produce alcohol, specifically ethanol, as the important product and the by-products water and carbon dioxide.
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
