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:
6 moles of electrons
Explanation:
Let us consider the species NO3− and ClO−. The NO3− is oxidized to NO the oxidation number of nitrogen is decreased from +5 to +2.
The oxidation number of chlorine is increased from +1 to +3. This implies that six electrons were transferred in the balanced reaction equation shown in the question. Hence the answer.
<span>d.2HNO3 (aq) + Sr(OH)2 (aq) → 2H2O (l) + Sr(NO3)2(aq)
4H </span>4H
8O 8O
2N 2N
1Sr 1Sr<span>
</span>
Answer:
102g
Explanation:
To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.
In this case, the given equation is already balanced.

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.
Mole= Mass ÷Mr
Mass= Mole ×Mr
<u>Method 1: using the </u><u>mass of glucose</u>
Mr of glucose
= 6(12) +12(1) +6(16)
= 180
Moles of glucose reacted
= 200 ÷180
=
mol
Amount of ethanol formed: moles of glucose reacted= 2: 1
Amount of ethanol
= 
=
mol
Mass of ethanol
= ![\frac{20}{9} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B9%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 
= 102 g (3 s.f.)
<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>
Mole of carbon dioxide produced
= 97.7 ÷[12 +2(16)]
= 97.7 ÷44
=
mol
Moles of ethanol: moles of carbon dioxide= 1: 1
Moles of ethanol formed=
mol
Mass of ethanol formed
= ![\frac{977}{440} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B977%7D%7B440%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 102 g (3 s.f.)
Thus, 102 g of ethanol are formed.
Additional:
For a similar question on mass and mole ratio, do check out the following!
Answer:
B.
Explanation:
electrons can be lost by one particle, and gained by another particle