An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution? Kb = 0.51°C/m for water. Enter your answer with 2 decimal places and no units.

Answers

Answer 1
Answer:

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:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_(solute) = Given mass of solute (CsCl) = 8.00 g

M_(solute) = Molar mass of solute (CsCl) = 168.4  g/mol

W_(solvent) = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2* 0.51^oC/m* (8.00* 1000)/(168.4g/mol* 92)\n\n\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53


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What reaction conditions most effectively conver a cabocxylic acid to a methly ester?

Answers

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.

The destruction of the ozone layer by chlorofluorocarbons (CFC’s) can be described by the following reactions:ClO(g) + O3(g) ? Cl(g) + 2 O2(g) ?H°rxn = –29.90 kJ2 O3(g) ? 3 O2(g) ?H°rxn = 24.18 kJDetermine the value of heat of reaction for the following:Cl(g) + O3(g) ? ClO(g) + O2(g) ?H°=_____________?

Answers

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

-------------------------------------------------------------------------------------------------

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

Final answer:

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.

Explanation:

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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How would you put this word equation into a balanced chemical equation: Aluminum nitrate reacts with potassium sulfate to produce aluminum sulfate and potassium nitrate.

Answers

Answer:

2Al(NO_{3} )_{3} + 3K_{2}SO_{4} ----> Al_{2}(SO_{4})_{3} + 6KNO_{3}

Explanation:

2Al(NO_(3) )_(3) + 3K_(2)SO_(4) ----> Al_(2)(SO_(4))_(3) + 6KNO_(3)

A carbon dioxide sample weighing 44.0 g occupies 32.68 l at 65°c and 645 torr. what is its volume at stp?

Answers

Combined gas law,

(P_(1)V_(1))/(T_(1)) =  (P_(2)V_(2))/(T_(2))

((645torr)(32.68 L))/(338 K) = ((760 torr)V_(2))/(273 K)

v_(2)=22.4 L

So, the gas will occupy 22.4 L at STP

At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.

Answers

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:

C_(H_2S)=K_H* p_(liquid)

where,

K_H = Henry's constant = 0.087M/atm

p_(H_2S) = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

C_(H_2S)=0.087M/atm* 2.42atm\n\nC_(H_2S)=0.2105M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}

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:

0.2105M=\frac{\text{Mass of hydrogen sulfide}* 1000}{34g/mol* 400.0mL}\n\n\text{Mass of }H_2S=(0.2105* 34* 400)/(1000)=2.86g

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

How many moles of atoms are in 9.00 g of 13c? express your answer numerically in moles?

Answers

Number of moles is defined as the ratio of given mass in grams to the molar mass of compound.

Number of moles =(Given mass in g)/(Molar mass)

Now, put the value of given mass of ^(13)C in grams and molar mass of ^(13)C in g/mol i.e. 13 g/mol.

Thus,

number of moles =(9.00 g)/(13 g/mol)

= 0.692 mol

Hence, number of moles of ^(13)C = 0.692 mol