The mole ratio of acetic acid to water in 100 g of vinegar is 0.015 : 0.985.
<h3>What is the mole ratio of acetic acid to water in 100 g of vinegar?</h3>
The mole ratio of acetic acid to water in 100 g of vinegar is determined from their percentage composition.
The percentage composition of acetic acid and water in vinegar is 5% acetic acid and 95% water.
In 100 g of vinegar, there are 5 g of acetic acid and 5 g of water.
Moles = mass/molar mass
molar mass of acetic acid = 62 g/mol
molar mass of water = 18 g/mol
moles of vinegar = 5/62 = 0.08
moles of water = 95/18 = 5.28
total moles = 5.36
Mole ratio of vinegar to water = 0.08/5.36 : 5.28/5.36
Mole ratio of vinegar to water = 0.015 : 0.985
In conclusion, the mole ratio is determined from the percentage composition of acetic acid and water in vinegar.
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9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen
The mass of nitrogen in the second sample is 23.33 g.
Answer:
Tincture of iodine, iodine tincture, or weak iodine solution is an antiseptic. It is usually 2–7% elemental iodine, along with potassium iodide or sodium iodide, dissolved in a mixture of ethanol and water. Tincture solutions are characterized by the presence of alcohol.
Explanation:
Which is a correct description of the organization of subatomic particles in atoms?
Protons and neutrons are tightly packed into a small nucleus. Electrons occupy the space
outside the nucleus
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