# Help with Chemistry Equations Quiz Questions

• orlish
This makes nitrogen a poor fuel, as breaking these strong bonds requires a lot of energy.5. a. Using data from Thermo Props Table, ?Gr° for the reaction is -394 kJ/mol and K = 3.75 x 10^-26. b. The chance of finding CO2 molecules in a classroom is very low, as the equilibrium constant is extremely small. The atmospheric pressure and oxygen concentration are not enough to drive the reaction forward, so there is not enough oxygen present to produce a significant amount of CO2 molecules. c. We would not expect to get
orlish
I have a very difficult quiz that I need help with a few questions on. I'll list them below and if you can help with any of the bunch that would be extremely helpful! Thank you so much!

1. Consider two aqueous (HO)2SO2 solns, (i) & (ii). For (i), [H3O+] = 2/3Mi. For (ii), [H3O+] = 5/3Mi. WITHOUT calculation of [H3O+], explain why (i) is more acidic (has a higher [H3O+]) than (ii).

2. Can a saturated, buffered, aqueous solution of CaCO3 be prepared at a pH for which all forms of "carbonate" are present at appreciable concentration? Answer this by calculating the pH for this system at which ["H2CO3"]EQ = [CO32-]EQ. Show work and state your conclusion.

3. If the relative humidity is 100% at a T>0°C but <100°C, H20(l) --> H20(g) exists because, at these conditions, VP(H20(l)) = P(H2O(g)) since ?Gf(H20(l)) = ?Gf(H20(g)).
* * a. But, at these conditions, how does ?Hf(H20(l)) compare to?Hf(H20(g)), and how does ?Sf(H20(l)) compare to ?Sf(H20(g)) ? Why? (Write the rxn eqn corresponding to subscript "f" and then decide).

* * b. And, at these conditions, how does ?Gf°(H2O(l)) compare to ?Gf° (H2O(g))? Why?

4. For NO2(g) at 25C, ?Gf° =+51.0 kJ/mol. Tell why ?Gf° is such a large, positive number. To answer, consider ?Hf° and ?Sf° and their impacts on ?Gf°. Also consider Lewis structures and appropriate data. Finally is nitrogen useful as*fuel? Why or why not?

5. Consider: C(graphite) +O2(g) --> CO2 (g).
* * a. Use date from Thermo Props Table to calculate ?Gr°and K for this rxn at 25°C. Show all work.

* * b. Based on your K, what chance is had of finding CO2 molecules in a classroom (volume is ~107 L) producible by this reaction if we had 106g (~one ton) of graphie in an open vessel? Defend your answer (Remember that air is ~20% oxygen and atmospheric pressure is ~1.0 atm. So, is there enough O2 in the classroom to effect this conversion? And, should we expect the conversion to happen from a thermodynamic standpoint?) Show all pertinent work.

* * c. Would we really get any CO2 molecules from this experiment if we did it? Why or why not?

* * d. From a thermodynamic standpoint, if we did this experiment with diamond rather than graphite we'd have an ever better chance of producing CO2 molecules. Why?

6. For each of the following redox systems, write correct equations for the 1/2- rxns. Then combine the 1/2-rxns to get the net redox rxn.
* * a. Ga(s) does not react in 9 M Na2SO4 but reacts readily in 9 M (HO)2SO2 giving SO2(g) and a hydrated, octahedral complex ion among the products.
*
* * b. Sr(s) reacts vigorously in 1 M Na2SO4 giving a precipitate and a highly flammable gas among the products.

* *c. Ca(s) reacts vigorously in 1 M KHCO3 and, among the products, gives a precipitate and the same highly flammable gas as in b., above. {Note: For this rxn, the source of the gas is different from b.,above. Also, be aware that incompatible products cannot be produced in appreciable quantities by this (or any) rxn if the products can be in contact with each other. Keep this in mind on deciding on the equation for the net redox rxn for this system. Hint: Think about B/L A/B chem.}

7. Based solely on information from Q 6.,above, answer the following.
* * a. Would Sr(s) be expected to react violently in 9 M (HO)2SO4? Provide* meaningful comment.

* * b. How would the pH's compare for the resulting system of each of these redox rxns? Why?

8. a. Based on periodic trend considerations, how should the water-solubility of the precipitate formed in Q 6. b., above, compare to the water-solubility of gypsum? Why? (Hint: Are Epsom salts water soluble? How about barium sulfate? So?)

* *b. Again, with respect to Q 6.b., above, compared to Sr(s), would Mg(s) be expected to react more vigorously or less vigorously in 1 M Na2SO4? Why.

9. Imagine that an electochemical cell which operates spontaneously is constructed by coupling the*HO-(aq)|O2(g)|H2O(l) 1/2- cell at pH 14.00 with the Ag(s)|Ag+(aq) 1/2-cell.
* *a. Write the equation for the net cell rxn and calculate E°cell.

* *b. What would happen to E°cell if each 1/2-cell compartment is treated with excess 6 M* * HCl? Why? (Hint: Could this treatment make something happen to system polarity?)

* *c. Note that the anode material for the O2(g)-containing 1/2-cell has not been specified.* * * * * * * * * * *Suggest a material which could be used as this electrode and tell why the suggested material would be suitable.

1. The concentration of [H3O+] is directly related to the acidity of a solution. In solution (i), the [H3O+] is 2/3 of the initial concentration, while in solution (ii), the [H3O+] is 5/3 of the initial concentration. This means that solution (i) is more diluted than solution (ii), so the [H3O+] is lower and the solution is more acidic.

2. Yes, a saturated, buffered, aqueous solution of CaCO3 can be prepared at a pH for which all forms of "carbonate" are present at appreciable concentration. This occurs at a pH of 9.86, when the concentrations of H2CO3 and CO32- are equal. This can be calculated using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]). In this case, pKa = 9.86 and [A-] = [HA], so the pH = 9.86.

3. a. At 100% relative humidity, the vapor pressure of liquid water is equal to the atmospheric pressure, so VP(H20(l)) = P(H2O(g)). This means that there is no net driving force for the reaction H2O(l) --> H2O(g), so the Gibbs free energy for both phases is equal and the reaction is at equilibrium. Therefore, at these conditions, ?Gf(H20(l)) = ?Gf(H20(g)).

b. At these conditions, ?Gf°(H2O(l)) is equal to ?Gf°(H2O(g)) because the reaction is at equilibrium and there is no net driving force for the reaction to occur.

4. The large, positive value of ?Gf° for NO2(g) at 25C is due to the high energy required to break the strong N-O bonds in NO2. This is reflected in the high value of ?Hf° for NO2, indicating that a large amount of energy is required to form NO2 from its constituent elements. Additionally, the high entropy of NO2(g) compared to its constituent elements also contributes to the large positive value of ?Gf°.

Nitrogen is not useful as a fuel because it is a very stable molecule and requires a large amount of energy to break its bonds and release energy. This makes it a poor choice

1. The concentration of H3O+ in solution (i) is higher than in solution (ii) because the ratio of H3O+ to SO42- is higher in solution (i). This is because the concentration of SO42- in solution (i) is lower than in solution (ii), therefore, the H3O+ ions are not as attracted to the SO42- ions and are free to interact with the water molecules, leading to a higher concentration of H3O+.

2. No, a saturated, buffered solution of CaCO3 cannot be prepared at a pH where all forms of carbonate are present at appreciable concentration. This is because at this pH, the concentrations of H2CO3 and CO32- will be equal, but the concentration of HCO3- will be significantly lower. This means that the buffer system will not be able to maintain a stable pH, as the equilibrium between H2CO3 and CO32- will be shifted towards H2CO3, leading to a decrease in pH.

3. a. At these conditions, the enthalpy of formation of H2O(l) is lower than the enthalpy of formation of H2O(g), meaning that the reaction is exothermic and favors the formation of liquid water. Additionally, the entropy of formation of H2O(l) is higher than the entropy of formation of H2O(g), indicating that the liquid state has a higher degree of disorder and is more favorable at these conditions.

b. At these conditions, the free energy of formation of H2O(l) is lower than the free energy of formation of H2O(g). This is because the enthalpy of formation of H2O(l) is lower and the entropy of formation of H2O(l) is higher, making the formation of liquid water more thermodynamically favorable.

4. The large, positive value of ?Gf° for NO2(g) is due to the high enthalpy of formation (?Hf°) and low entropy of formation (?Sf°) of the gas. The high enthalpy of formation indicates that a large amount of energy is required to form NO2(g) from its constituent elements, while the low entropy of formation suggests that the gas is highly ordered and less thermodynamically stable. Therefore, nitrogen is not a useful fuel as it requires a large amount of energy to produce and is not easily converted into other

## 1. What is the purpose of a chemistry equations quiz?

A chemistry equations quiz is designed to test your understanding of chemical equations and your ability to balance and solve them. It also helps to reinforce your knowledge of important concepts and equations in chemistry.

## 2. How can I prepare for a chemistry equations quiz?

To prepare for a chemistry equations quiz, it is important to review your notes and textbook, practice solving different types of equations, and seek help from your teacher or tutor if needed. You can also use online resources, such as practice quizzes and videos, to supplement your studying.

## 3. What are some common types of chemistry equations that may be included in a quiz?

Some common types of chemistry equations that may be included in a quiz are: balancing chemical equations, solving stoichiometry problems, calculating molar masses, and determining percent composition of compounds.

## 4. How can I improve my performance on chemistry equations quizzes?

To improve your performance on chemistry equations quizzes, it is important to practice regularly, review your mistakes, and seek help when needed. You can also try different study techniques, such as creating flashcards or teaching the material to someone else.

## 5. Is it important to memorize chemistry equations for a quiz?

Memorization of chemistry equations is not the main focus of a quiz. Instead, it is more important to understand the concepts behind the equations and know how to apply them to different problems. However, some equations may need to be memorized, so it is important to review and practice them regularly.

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