The heat which we supply to water during boiling is used to overcome these forces of attraction between the particles so that they become totally free and change into a gas. This latent heat does not increase the kinetic energy of water particles and hence no rise in temperature. takes place during the boiling of water.
Answer:
68.1% is percent yield of the reaction
Explanation:
The reaction of methane with oxygen is:
CH₄ + 2O₂ → CO₂ + 2H₂O
<em>Where 2 moles of oxygen react per mole of CH₄</em>
<em />
Percent yield is:
Actual yield (28.2g CO₂) / Theoretical yield * 100
To solve this question we need to find theoretical yield finding limiting reactant :
<em>Moles CH₄:</em>
15.1g CH₄ * (1mol / 16.04g) = 0.9414 moles
<em>Moles O₂:</em>
81.2g * (1mol / 32g) = 2.54 moles
For a complete reaction of 0.9414 moles of CH₄ are needed:
0.9414 moles CH₄ * (2 mol O₂ / 1mol CH₄) = 1.88 moles of O₂. As there are 2.54 moles, O₂ is in excess and <em>CH₄ is limiting reactant</em>
In theoretical yield, the moles of methane added = Moles of CO₂ produced. That is 0.9414 moles CO₂. In grams = Theoretical yield:
0.9414 moles CO₂ * (44.01g / mol) = 41.43g CO₂
Percent yield: 28.2g CO₂ / 41.43g CO₂ * 100=
<h3>68.1% is percent yield of the reaction</h3>
Well one mole of stuff, any stuff, including carbon dioxide, specifies
6.022
×
10
23
individual items of that stuff.
Explanation:
And thus we work out the quotient:
7.2
×
10
25
⋅
carbon dioxide molecules
6.022
×
10
23
⋅
carbon dioxide molecules
⋅
m
o
l
−
1
≅
120
⋅
m
o
l
carbon dioxide
.
This is dimensionally consistent, because we get an answer with units
1
m
o
l
−
1
=
1
1
mol
=
m
o
l
as required.
Answer:
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Explanation: