Consider the thermochemical equation for the combustion of acetone (C3H6O), the main ingredient in nail polish remover: C3H6O(l)
1 answer:
Answer:
the energy produced or released = 2.4 X (-1790) = -4296 kJ
Explanation:
In the given question the heat of combustion is given as H°rxn = −1790 kJ
Thus when one mole of acetone undergoes complete combustion, it gives this much of energy.
The volume of acetone present in nail polish remover = 177 mL
The density of acetone = 0.788 g/mL
The mass of acetone present in nail polish remover = density X volume
the mass of acetone = 0.788 X 177 = 139.48 grams
the moles of acetone present =
The energy produced = moles present X enthalpy of reaction
the energy produced or released = 2.4 X (-1790) = -4296 kJ
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To make 1000. L of bleach solution 1.69×10⁴ L of Cl₂(g) at STP will be needed .
<h3>What is Balanced chemical equation?</h3>A balance chemical equation is an equation that contain same number of atoms as well as of each element on each side of reaction.
<h3>What is Density?</h3>It is ratio of mass to the volume.
Density=
Now,
The balanced chemical equation of preparing bleach from chlorine is as follow:
Cl₂(g) + 2NaOH(aq) → NaClO(aq) + NaCl(aq) + H₂O(l)
Now,
To calculate the Moles of NaClO in 1000L solution, first calculate it's mass.
Mass of NaClO = Density × Volume
= 1.07 × 1000L ×
×
= 56175g
Now, we have to calculate moles of NaClO
Hence,
Moles of NaClO =
=
= 754.63 mol
Now,
According to the reaction number of moles of Cl₂ is equal to the number of moles of NaClO.
Hence,
moles of Cl₂ = 754.63 mol
Now , in order to calculate litres of Cl₂ at STP,
We know that
1 mol Cl₂ = 22.4 L of Cl₂
754.63 mol of Cl₂ = 754.63 mol × 22.4 L
= 1.69×10⁴ L of Cl₂
Thus from the above conclusion we can say that , to make 1000. L of bleach solution 1.69×10⁴ L of Cl₂(g) at STP will be needed .
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<u>Answer:</u> The order of increasing boiling points follows:
<u>Explanation:</u>
The expression of elevation in boiling point is given as:
where,
= Elevation in boiling point
i = Van't Hoff factor
= change in boiling point
= boiling point constant
m = molality
For the given options:
- <u>Option 1:</u> 0.12 m
Value of i = 1 (for non-electrolytes)
So, molal concentration will be =
- <u>Option 2:</u> 0.02 m LiBr
Value of i = 2
So, molal concentration will be =
- <u>Option 3:</u> 0.05 m
Value of i = 3
So, molal concentration will be =
As, the molal concentration of is the highest, so its boiling point will be the highest.
Hence, the order of increasing boiling points follows: