[Atome]

Hello there,

I am having some trouble with the following challenging questions from the Avogadro exams and I would appreciate some help. My work is shown below.

Also, some of these questions are beyond what my chemistry classes have covered, so my plan for solving the problems may be naive.

Thank you very much.

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1. If 1.0 L of 0.10 M NaNO_{3 (aq)} and 1.0 L of 0.20 M Mg(NO_3)2(aq) are mixed, what are the final concentrations of Na⁺, Mg²⁺, and NO₃^-?

2. Which one of the following substances in aqueous solution is a nonelectrolyte?

a) NH₄Cl
b) CH₃OH
c) CO₂

3. Use bond dissociation enthalpies provided to calculate Change in Heat for the reaction:

3H₂ + N₂ -> 2NH₃

4. 10.0 g of water at 0°C is placed on top of 100 g of Cu at 100°C. What is the final temperature of the combined system (water and copper), if no heat is lost to the surroundings?

Note that heat capacity of Cu = 0.39 J/(gC)

Heat of fusion of ice = 334 J/g

Heat capacity of water = 4.18 J/(gC)

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My Work:

1. Since there are 0.30 mols of ions in total and 2.0 L of volume:

Concentration = 0.30 mols/2.0 L = 0.15 L

However, according to the answer, this is wrong.

2. Initially, I picked c) but the answer is b). I don't understand how b) can be the answer; wouldn't the hydrogen bonds in each methanol molecule help to conduct electricity?

3. Change in heat = heat of products - heat of reactants

                        = (389 kJ/mol x 6) - [(435 kJ/mol x 3) + 163 kJ/mol]

                        = 2334 kJ/mol - (1305 kJ/mol + 163 kJ/mol)

Since Change Heat is measured in kJ, I multiplied each enthalpy value by the molar coefficient to eliminate moles.

Change in heat = 4668 kJ - (3915 kJ + 163 kJ)

                     = 590 kJ

However, the answer is -83 kJ.

4. I am not sure what to do for this question. I tried to use mct = -(mct), but I do not know how the heat of fusion of ice would help.

[AWK]

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1. If 1.0 L of 0.10 M NaNO_{3 (aq)} and 1.0 L of 0.20 M Mg(NO_3)2(aq) are mixed, what are the final concentrations of Na⁺, Mg²⁺, and NO₃^-?

1. Since there are 0.30 mols of ions in total and 2.0 L of volume:

Concentration = 0.30 mols/2.0 L = 0.15 L

However, according to the answer, this is wrong.

Of course, wrong
Take into account 1. dilution, 2. complete dissociation of strong electrolytes

2. Which one of the following substances in aqueous solution is a nonelectrolyte?

a) NH₄Cl
b) CH₃OH
c) CO₂

2. Initially, I picked c) but the answer is b). I don't understand how b) can be the answer; wouldn't the hydrogen bonds in each methanol molecule help to conduct electricity?

Take into acount CO2 reacts with water to form carbonic acid which is a weak electrolyte

4. 10.0 g of water at 0°C is placed on top of 100 g of Cu at 100°C. What is the final temperature of the combined system (water and copper), if no heat is lost to the surroundings?

Note that heat capacity of Cu = 0.39 J/(gC)

Heat of fusion of ice = 334 J/g

Heat capacity of water = 4.18 J/(gC)

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My Work:




4. I am not sure what to do for this question. I tried to use mct = -(mct), but I do not know how the heat of fusion of ice would help.

10,0 g of water at 0 C or 10,0 g of ice at this temperature?
May be heat of fusion is not needed for calulations

[sjb]

3. Use bond dissociation enthalpies provided to calculate Change in Heat for the reaction:

3H₂ + N₂ -> 2NH₃

->8---snip

3. Change in heat = heat of products - heat of reactants

                        = (389 kJ/mol x 6) - [(435 kJ/mol x 3) + 163 kJ/mol]

                        = 2334 kJ/mol - (1305 kJ/mol + 163 kJ/mol)

Since Change Heat is measured in kJ, I multiplied each enthalpy value by the molar coefficient to eliminate moles.

Change in heat = 4668 kJ - (3915 kJ + 163 kJ)

                     = 590 kJ

However, the answer is -83 kJ.

I think you've already taken into account the stoichiometry in the first (multiplication) step, so no need to "remultiply the enthalpy values by molar coefficients"

Bonds formed 6 x N-H; bonds broken 1 x N#N (# being triple bond), and 3 x H-H

6 x BE(N-H) - [BE(N#N) - 3 x BE(H-H)]  = ?

S