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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Answer:
The answer to your question is SO₂ + 3H₂ ⇒ H₂S + 2H₂O
Explanation:
Reaction
SO₂ + H₂ ⇒ H₂S + H₂O
Reactants Elements Products
1 Sulfur 1
2 Hydrogen 4
2 Oxygen 1
This reaction is unbalanced so we need to balance it.
SO₂ + 3H₂ ⇒ H₂S + 2H₂O
Reactants Elements Products
1 Sulfur 1
6 Hydrogen 6
2 Oxygen 2
Now, the reaction is balanced
Answer: A. 1,1,1
Explanation:
The coefficients that will balance the equation; NH4OH(aq) →H2O(l) + NH3(g), is 1, 1, 1, because it proves the total number of atoms of each element on the LHS and RHS of the equation are equal, hence balanced.
LHS RHS
N = 1 1
H = 5 5
O = 1 1
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