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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This statement is false due to the fact that the ammonia gas has the lower molar mass.
Answer:
sugar and oxygen
Explanation:
sugar (glucose) and oxygen
Answer:
The final temperature will be "12.37°".
Explanation:
The given values are:
mass,
m = 0.125 kg
Initial temperature,
c = 22.0°C
Time,
Δt = 4.5 min
As we know,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 