Answer: The boiling point of solution is 100.53
Explanation:
We are given:
8.00 wt % of CsCl
This means that 8.00 grams of CsCl is present in 100 grams of solution
Mass of solvent = (100 - 8) g = 92 grams
The equation used to calculate elevation in boiling point follows:
To calculate the elevation in boiling point, we use the equation:
Or,
where,
Boiling point of pure solution = 100°C
i = Vant hoff factor = 2 (For CsCl)
= molal boiling point elevation constant = 0.51°C/m
= Given mass of solute (CsCl) = 8.00 g
= Molar mass of solute (CsCl) = 168.4 g/mol
= Mass of solvent (water) = 92 g
Putting values in above equation, we get:
Hence, the boiling point of solution is 100.53
Answer:
Esterification reaction
Explanation:
When we have to go from an acid to an ester we can use the esterification reaction. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).
In this case, we need the methyl ester, therefore we have to choose the appropriate alcohol, so we have to use the methanol as reactive if we have to produce the methyl ester.
Answer:
ΔH°rxn = 54.08 kJ
Explanation:
Let's consider the following equations.
a) ClO(g) + O₃(g) ⇄ Cl(g) + 2 O₂(g) ΔH°rxn = –29.90 kJ
b) 2 O₃(g) ⇄ 3 O₂(g) ΔH°rxn = 24.18 kJ
We have to determine the value of heat of reaction for the following reaction: Cl(g) + O₃(g) ⇄ ClO(g) + O₂(g)
According to Hess's law, the enthalpy change in a chemical reaction is the same whether the reaction takes place in one or in several steps. That means that we can find the enthalpy of a reaction by adding the corresponding steps and adding their enthalpies. According to Lavoisier-Laplace's law, if we reverse a reaction, we also have to reverse the sign of its enthalpy.
Let's reverse equation a) and add it to equation b).
-a) Cl(g) + 2 O₂(g) ⇄ ClO(g) + O₃(g) ΔH°rxn = 29.90 kJ
b) 2 O₃(g) ⇄ 3 O₂(g) ΔH°rxn = 24.18 kJ
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Cl(g) + 2 O₂(g) + 2 O₃(g) ⇄ ClO(g) + O₃(g) + 3 O₂(g)
Cl(g) + O₃(g) ⇄ ClO(g) +O₂(g)
ΔH°rxn = 29.90 kJ + 24.18 kJ = 54.08 kJ
The heat of the reaction (ΔH°rxn) for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is calculated using Hess's Law. The sum of the heat of reversed first reaction and the second reaction provided is 54.08 kJ.
The chemistry question asks to determine the heat of the reaction for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). In Hess's Law, the heat of the reaction or ΔH for a reaction can be calculated from the sum of the heats of other reactions that sum to the desired reaction. In this case, we want to reverse the first reaction provided (which changes the sign of ΔH) and add it to the second reaction provided.
So, reversing the first reaction we get: Cl(g) + 2 O2(g) ? ClO(g) + O3(g) ?H°rxn = 29.90 kJ
Adding this to the second reaction: 2 O3(g) ? 3 O2(g), ?H°rxn = 24.18 kJ, gives the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). Adding the ΔH values gives the ΔH for this reaction: 29.90 kJ + 24.18 kJ = 54.08 kJ. So, ?H°rxn for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is 54.08 kJ.
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Answer:
2Al(NO_{3} )_{3} + 3K_{2}SO_{4} ----> Al_{2}(SO_{4})_{3} + 6KNO_{3}
Explanation:
Combined gas law,
So, the gas will occupy 22.4 L at STP
Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant =
= partial pressure of hydrogen sulfide gas = 2.42 atm
Putting values in above equation, we get:
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.2105 M
Molar mass of hydrogen sulfide = 34 g/mol
Volume of solution = 400.0 mL
Putting values in above equation, we get:
Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.
Number of moles is defined as the ratio of given mass in grams to the molar mass of compound.
Number of moles =
Now, put the value of given mass of in grams and molar mass of in g/mol i.e. 13 g/mol.
Thus,
number of moles =
= 0.692 mol
Hence, number of moles of = 0.692 mol