Ap chem 프린스턴 화학 summary part 4

 Answer Explanation:
Calculate moles of Carbon:

Moles of CO2 = 8.8 g / 44 g/mol = 0.2 mol
Moles of C = 0.2 mol
Calculate moles of Hydrogen:

Moles of H2O = 3.6 g / 18 g/mol = 0.2 mol
Moles of H = 0.2 mol × 2 = 0.4 mol
Calculate mass of C and H:

Mass of C = 0.2 mol × 12 g/mol = 2.4 g
Mass of H = 0.4 mol × 1 g/mol = 0.4 g
Calculate mass of Oxygen:

Mass of O = Total mass - Mass of C - Mass of H
Mass of O = 6.0 g - 2.4 g - 0.4 g = 3.2 g
Moles of O = 3.2 g / 16 g/mol = 0.2 mol
Determine mole ratio:

C : H : O = 0.2 : 0.4 : 0.2
Simplify by dividing by the smallest number (0.2):
C : H : O = 1 : 2 : 1
Empirical Formula: CH2O

Correct Answer: (B) 

Answer Explanation: The moles of nitrogen and hydrogen can be used to find the number of ammonia molecules formed. 
Using stoichiometry, the correct answer is 1.20 × 1024 molecules.

Correct Answer: (B)
            

 Answer Explanation:
Percent yield is calculated using:

Percent Yield
=
(Actual Yield/Theoretical Yield)×100

Correct Answer: (A) 

 Answer Explanation:
Calculate moles of CO2:

Moles of CO2 = 11.0 g / 44 g/mol = 0.25 mol
Calculate mass of Carbon:

Moles of C = Moles of CO2 = 0.25 mol
Mass of C = 0.25 mol × 12 g/mol = 3.0 g
Calculate mass percent of Carbon:

Mass percent = (Mass of C / Mass of hydrocarbon) × 100
Mass percent = (3.0 g / 9.0 g) × 100 = 33.3%
Closest Option:

The closest option is 36.4%
Correct Answer: (B) 

Answer Explanation:

Assume a 100 g sample for simplicity.

1. **Calculate moles of SiO2:**
   - Mass = 76 g
   - Molar mass of SiO2 = 28.1 (Si) + 2 × 16.0 (O) = 60.1 g/mol
   - Moles = 76 g / 60.1 g/mol ≈ 1.264 moles

2. **Calculate moles of Na2O:**
   - Mass = 12 g
   - Molar mass of Na2O = 2 × 23.0 (Na) + 16.0 (O) = 62.0 g/mol
   - Moles = 12 g / 62.0 g/mol ≈ 0.194 moles

3. **Calculate moles of CaO:**
   - Mass = 12 g
   - Molar mass of CaO = 40.1 (Ca) + 16.0 (O) = 56.1 g/mol
   - Moles = 12 g / 56.1 g/mol ≈ 0.214 moles

**Arrange in order of decreasing moles:**

1. **SiO2**: 1.264 moles (Most moles)
2. **CaO**: 0.214 moles
3. **Na2O**: 0.194 moles (Least moles)

Correct Answer: (A)
            

Answer Explanation: The loss in mass corresponds to the water content in the hydrate. 
Based on the mass loss, the formula is CuSO4·5H2O.

Correct Answer: (B)
            

Answer Explanation: The empirical formula is found by determining the simplest ratio of moles of sulfur to oxygen. 
The correct formula is SO2.

Correct Answer: (B)
            

Answer Explanation: Using stoichiometry and the molar volume of gases at STP, the mass of sodium is found to be 20 g.

Correct Answer: (C)
            

Answer Explanation: The masses suggest that the possible atomic mass of X could be 18.0, considering the different multiples. 
The correct answer is 18.0.

Correct Answer: (A)
            

Answer Explanation: When NaCl and Pb(NO3)2 are mixed, PbCl2 precipitates, leaving Na+ and NO3− in the solution. 
The correct answer is Cl− as it forms the precipitate.

Correct Answer: (B)
            

Answer Explanation: The correct net ionic equation involves the formation of a solid precipitate of copper carbonate. 
The correct answer is 2Cu+(aq) + CO32−(aq) → Cu2CO3(s).

Correct Answer: (A)
            

Answer Explanation: The oxidation number of hydrogen does not change in this reaction; it remains +1. 
Therefore, the statement that hydrogen changes from +1 to 0 is incorrect.

Correct Answer: (C)
            

Answer Explanation: Iron (III) carbonate, Fe2(CO3)3, precipitates out of solution. 
The correct answer is (B).

Correct Answer: (B)
            

Answer Explanation: In the reaction between AgNO3 and KBr, AgBr precipitates, leaving K+ and NO3− as the spectator ions. 
The correct answer is (C).

Correct Answer: (C)
            

Answer Explanation: Balancing the half-reaction shows that the coefficient for electrons is 1.

Correct Answer: (A)
            

 Answer Explanation:
Calculate moles of KClO3:

Molar mass of KClO3 = 39.1 (K) + 35.5 (Cl) + 48.0 (3×16 O) = 122.6 g/mol
Moles of KClO3 = 12.25 g / 122.6 g/mol ≈ 0.1 mol
Use stoichiometry to find moles of O2:

According to the balanced equation, 2 mol KClO3 produce 3 mol O2
Moles of O2 = 0.1 mol KClO3 × (3 mol O2 / 2 mol KClO3) = 0.15 mol
Calculate mass of O2:

Molar mass of O2 = 32.0 g/mol
Mass of O2 = 0.15 mol × 32.0 g/mol = 4.80 g
Correct Answer: (B) 

Answer Explanation: At STP, 1 mole of gas occupies 22.4 L. Using stoichiometry, 
the decomposition of 12.25 g of KClO3 produces approximately 3.36 L of O2 gas.

Correct Answer: (B)
            

Answer Explanation: According to the ideal gas law (PV = nRT), when the temperature is doubled while volume is constant, 
the pressure will also double due to the increased kinetic energy of the gas molecules.

Correct Answer: (A)
            

 Answer Explanation:
Calculate moles of NaCl:

Moles = Molarity × Volume (in liters)
Moles = 0.5 M × 0.5 L = 0.25 mol
Calculate mass of NaCl:

Molar mass of NaCl = 23.0 (Na) + 35.5 (Cl) = 58.5 g/mol
Mass = 0.25 mol × 58.5 g/mol ≈ 14.6 g
Correct Answer: (A) 

Answer Explanation:

Use the Ideal Gas Law:

PV = nRT

Where:

• P = Pressure = 2.0 atm
• V = Volume = ?
• n = Number of moles = 2.0 mol
• R = Ideal gas constant = 0.0821 L·atm/mol·K
• T = Temperature = 273 K

Solve for V:

V = (nRT) ÷ P

Plug in the values:

V = [ (2.0 mol) × (0.0821 L·atm/mol·K) × (273 K) ] ÷ 2.0 atm

Calculate the numerator:

Numerator = 2.0 × 0.0821 × 273 ≈ 44.8 L·atm

Now, compute V:

V = ^{44.8 L·atm}
2.0 atm  = 22.4 L

Correct Answer: (A)

Answer Explanation:

Use Boyle's Law (at constant temperature):

P₁V₁ = P₂V₂

Where:

• P₁ = 1.0 atm
• V₁ = 5.0 L
• P₂ = 2.5 atm
• V₂ = ?

Solve for V₂:

V₂ = (P₁V₁) ÷ P₂

Plug in the values:

V₂ = (1.0 atm × 5.0 L) ÷ 2.5 atm

Calculate:

V₂ = ^{5.0 atm·L}
2.5 atm  = 2.0 L

Correct Answer: (A)

Answer Explanation:

Use the Ideal Gas Law:

PV = nRT

Solve for n:

n = (PV) ÷ (RT)

Where:

• P = 2.0 atm
• V = 10.0 L
• R = 0.0821 L·atm/mol·K
• T = 300 K

Plug in the values:

n = (2.0 atm × 10.0 L) ÷ (0.0821 L·atm/mol·K × 300 K)

Calculate the denominator:

Denominator = 0.0821 × 300 ≈ 24.63 L·atm/mol

Compute n:

n = ^{20.0 atm·L}
24.63 L·atm/mol  ≈ 0.8116 mol

Rounded to two significant figures:

n ≈ 0.81 mol

Closest option:

0.80 mol

Correct Answer: (A)

Answer Explanation: When barium nitrate reacts with sodium sulfate, BaSO4 is formed as a precipitate 
since it is insoluble in water.

Correct Answer: (A)
            

Answer Explanation: When Na2SO4 and CaCl2 are mixed, CaSO4 precipitates, 
leaving Na+ and Cl− ions as the spectator ions.

Correct Answer: (A)
            

 Answer Explanation: When Na2CO3 and Mg(NO3)2 are mixed, they react to form magnesium carbonate (MgCO3), which is insoluble and precipitates out.
Reaction: Mg2+ + CO32− → MgCO3(s)

Both Mg2+ and CO32− ions are removed from the solution as they form the precipitate. However, due to the low solubility product of MgCO3, the concentration of CO32− ions remaining in the solution is significantly lower compared to Mg2+ ions.

Ions Remaining in Solution: Na+ and NO3−

Therefore, CO32− ions will NOT be present in significant amounts after equilibrium is established.

Correct Answer: (B) 

 Answer Explanation: The balanced equation is: N2 + 3H2 → 2NH3
Molar mass of N2 = 28 g/mol Moles of N2 = 14.0 g / 28 g/mol = 0.5 moles

Molar mass of H2 = 2 g/mol Moles of H2 = 6.0 g / 2 g/mol = 3.0 moles

From the equation, 1 mole of N2 requires 3 moles of H2. 0.5 moles of N2 would require 1.5 moles of H2.

Since we have 3.0 moles of H2, which is more than needed, N2 is the limiting reagent.

Moles of NH3 produced = 2 × 0.5 moles = 1.0 mole Molar mass of NH3 = 17 g/mol Mass of NH3 = 1.0 mole × 17 g/mol = 17.0 g

Correct Answer: (A) 

Answer Explanation:

1. **Calculate moles of Zn:**
   - Molar mass of Zn = 65.38 g/mol
   - Moles of Zn = 10.0 g / 65.38 g/mol ≈ 0.153 moles

2. **Use stoichiometry to find moles of H2:**
   - According to the balanced equation, 1 mole of Zn produces 1 mole of H2
   - Moles of H2 = 0.153 moles

3. **Calculate volume of H2 at STP:**
   - At STP, 1 mole of gas occupies 22.4 L
   - Volume = 0.153 moles × 22.4 L/mol ≈ 3.42 L

**Correct Answer:** **(A)**
        

Answer Explanation: Percent yield is calculated using the formula: 
(actual yield / theoretical yield) × 100. 
Here, percent yield = (20.0 g / 28.0 g) × 100 = 71.4%.

Correct Answer: (B)
            

Answer Explanation: Percent error is calculated using the formula:
| (theoretical - experimental) / theoretical | × 100.
Percent error = | (1.00 - 0.85) / 1.00 | × 100 = 15.0%.

Correct Answer: (C)
            

Answer Explanation: The balanced equation for the reaction is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Molar masses: H2SO4 = 98 g/mol, NaOH = 40 g/mol, Na2SO4 = 142 g/mol.
Moles of H2SO4 = 10.0 g / 98 g/mol = 0.102 moles.
Moles of NaOH = 10.0 g / 40 g/mol = 0.25 moles.
H2SO4 is the limiting reagent since it requires 2 moles of NaOH.
Moles of Na2SO4 produced = 0.102 moles.
Mass of Na2SO4 = 0.102 moles × 142 g/mol = 14.2 g.

Correct Answer: (B)
            

Answer Explanation: The balanced equation for the decomposition is:
CaCO3 → CaO + CO2
Molar mass of CaCO3 = 100 g/mol.
Moles of CaCO3 = 5.0 g / 100 g/mol = 0.05 moles.
At STP, 1 mole of gas occupies 22.4 L.
Volume of CO2 = 0.05 moles × 22.4 L/mol = 1.12 L.

Correct Answer: (A)
            

Answer Explanation: The balanced equation for the reaction is:
2H2 + O2 → 2H2O
Molar mass of H2 = 2 g/mol, molar mass of H2O = 18 g/mol.
Moles of H2 = 10.0 g / 2 g/mol = 5 moles.
From the balanced equation, 2 moles of H2 produce 2 moles of H2O.
Therefore, 5 moles of H2 produce 5 moles of H2O.
Mass of H2O = 5 moles × 18 g/mol = 90.0 g.

Correct Answer: (A)
            

Answer Explanation: The balanced equation for the reaction is:
Mg + 2HCl → MgCl2 + H2
Molar mass of Mg = 24 g/mol, molar mass of HCl = 36.5 g/mol.
Moles of Mg = 20.0 g / 24 g/mol = 0.833 moles.
Moles of HCl = 10.0 g / 36.5 g/mol = 0.274 moles.
From the equation, 1 mole of Mg reacts with 2 moles of HCl.
0.833 moles of Mg would require 1.666 moles of HCl, but only 0.274 moles of HCl are present.
HCl is the limiting reagent.
Moles of Mg that react = 0.274 moles / 2 = 0.137 moles.
Mass of Mg that reacts = 0.137 moles × 24 g/mol = 3.3 g.
Mass of excess Mg = 20.0 g - 3.3 g = 16.7 g.

Correct Answer: (B)
            

Answer Explanation: 
1. Molar masses: Al = 27 g/mol, Fe2O3 = 159.7 g/mol, Fe = 55.85 g/mol.
2. Moles of Al = 15.0 g / 27 g/mol = 0.556 moles.
   Moles of Fe2O3 = 50.0 g / 159.7 g/mol = 0.313 moles.
3. From the balanced equation, 2 moles of Al react with 1 mole of Fe2O3.
   Since we have fewer moles of Fe2O3, it is the limiting reagent.
4. Moles of Fe produced = 2 × 0.313 moles = 0.626 moles.
5. Mass of Fe = 0.626 moles × 55.85 g/mol = 34.97 g.
6. Percent yield = 75%, so actual yield = 34.97 g × 0.75 = 26.23 g.

Correct Answer: (A)
            

Answer Explanation: 
1. Moles of CO2 = 33.0 g / 44 g/mol = 0.75 moles. Each mole of CO2 contains 1 mole of C, so moles of C = 0.75 moles.
2. Moles of H2O = 13.5 g / 18 g/mol = 0.75 moles. Each mole of H2O contains 2 moles of H, so moles of H = 0.75 × 2 = 1.5 moles.
3. Mole ratio of C:H = 0.75:1.5 = 1:2.
4. The empirical formula is CH2.

Correct Answer: (B)
            

Answer Explanation: 
1. Molar mass of Zn = 65.38 g/mol. Moles of Zn = 2.00 g / 65.38 g/mol = 0.0306 moles.
2. Balanced equation: Zn + 2HCl → ZnCl2 + H2. Thus, 1 mole of Zn produces 1 mole of H2.
3. Moles of H2 = 0.0306 moles.
4. Dry gas pressure = Total pressure - Vapor pressure of water = 780 mmHg - 24 mmHg = 756 mmHg.
5. Convert pressure to atm: 756 mmHg × (1 atm / 760 mmHg) = 0.995 atm.
6. Using the ideal gas law: PV = nRT. V = (nRT) / P.
   V = (0.0306 moles × 0.0821 L·atm/mol·K × 298 K) / 0.995 atm = 0.745 L.

Correct Answer: (A)
            

Answer Explanation: 
1. Moles of HCl initially added = 50.0 mL × 1.0 M = 0.0500 moles.
2. Moles of NaOH used in back titration = 30.0 mL × 0.50 M = 0.0150 moles.
3. Moles of excess HCl = 0.0150 moles.
4. Moles of HCl reacted with CaCO3 = 0.0500 moles - 0.0150 moles = 0.0350 moles.
5. Balanced equation: CaCO3 + 2HCl → CaCl2 + CO2 + H2O.
   Thus, 2 moles of HCl react with 1 mole of CaCO3.
6. Moles of CaCO3 = 0.0350 moles / 2 = 0.0175 moles.
7. Mass of CaCO3 = 0.0175 moles × 100 g/mol = 1.75 g.
8. Percentage purity = (1.75 g / 2.50 g) × 100 = 70.0%.

Correct Answer: (B)