CHEM 1411 Sample Final Exam Notes

CHEM 1411 Sample Final Exam Notes

Part I: Multiple Choice Questions

  • Question 1: Significant figures in multiplication.

    • Calculation: 0.3614×2.75=0.993850.3614 \times 2.75 = 0.99385. Round to three significant figures because 2.75 has the least number of significant figures.
    • Answer: B. 0.994

  • Question 2: Scientific notation.

    • Express 75,030.2m75,030.2 m in scientific notation.
    • Answer: A. 7.50302×104m7.50302 \times 10^4 m

  • Question 3: Atomic number, mass number, and number of electrons in 109Ag+^{109}Ag^+.

    • Atomic number (Ag) = 47 (number of protons).
    • Mass number = 109.
    • Number of electrons = 47 - 1 (because of +1 charge) = 46.
    • Answer: A. 47, 109, 46

  • Question 4: Correct symbols for copper and cobalt.

    • Copper: Cu
    • Cobalt: Co
    • Answer: D. Cu, Co

  • Question 5: Number of nitrogen atoms in 175 molecules of Cu(NO<em>3)</em>2Cu(NO<em>3)</em>2.

    • Each Cu(NO<em>3)</em>2Cu(NO<em>3)</em>2 molecule contains 2 nitrogen atoms.
    • 175×2=350175 \times 2 = 350 nitrogen atoms.
    • Answer: C. 350

  • Question 6: Equivalent of 130 K in degrees Fahrenheit.

    • K=°C+273.15K = °C + 273.15, so °C=K−273.15=130−273.15=−143.15°C°C = K - 273.15 = 130 - 273.15 = -143.15 °C
    • °F=(°C×9/5)+32=(−143.15×9/5)+32=−225.67°F°F = (°C \times 9/5) + 32 = (-143.15 \times 9/5) + 32 = -225.67 °F
    • Answer: A. -225

  • Question 7: Atom with mass number 17, proton number 8, neutron number 9, and electron number 10.

    • Proton number 8 implies Oxygen (O).
    • Mass number 17 implies 17O^{17}O.
    • Electron number 10 implies a -2 charge (8 protons, 10 electrons).
    • Answer: B. 8O17^8O^{17} (-2 charge)

  • Question 8: Moles in 112 g of ethyl chloride (C2H5Cl).

    • Molar mass of C2H5Cl = 2(12.01)+5(1.01)+35.45=64.57g/mol2(12.01) + 5(1.01) + 35.45 = 64.57 g/mol
    • Moles = 112g/(64.57g/mol)=1.73mol112 g / (64.57 g/mol) = 1.73 mol
    • Answer: B. 1.74

  • Question 9: Grams of copper produced from 160 g of CuO.

    • CuO molar mass = 63.55 + 16.00 = 79.55 g/mol
    • Moles of CuO = 160 g / 79.55 g/mol = 2.01 mol
    • Since 1 mol CuO -> 1 mol Cu, we have 2.01 mol Cu
    • Grams of Cu = 2.01 mol * 63.55 g/mol = 127.74 g
    • Answer: C. 128 g

  • Question 10: Molecular formula of a compound with empirical formula C2H4O and molar mass 88 g/mol.

    • Empirical formula mass = 2(12) + 4(1) + 16 = 44 g/mol
    • Ratio = 88/44 = 2
    • Molecular formula = (C2H4O)2 = C4H8O2
    • Answer: B. C4H8O2

  • Question 11: Molecules that undergo complete dissociation to form OH- in water.

    • These are strong bases.
    • Answer: B. Strong Base

  • Question 12: Concentration of a solution made by dissolving 890 mg of NaCl in 100 ml of water.

    • 890 mg = 0.89 g
    • Moles of NaCl = 0.89 g / 58.44 g/mol = 0.0152 mol
    • Concentration = 0.0152 mol / 0.1 L = 0.152 M
    • Answer: B. 0.15 M

  • Question 13: Grams of NaCl in 500.0 ml of a 5.000M solution.

    • Moles of NaCl = 5.000 mol/L * 0.500 L = 2.500 mol
    • Grams of NaCl = 2.500 mol * 58.44 g/mol = 146.1 g
    • Answer: C. 146.25 g

  • Question 14: Molarity of chloride ion in 100.0 mL of 1.00 M Aluminum chloride, AlCl3, solution.

    • 1 mol of AlCl3 produces 3 mol of Cl-
    • [Cl-] = 3 * 1.00 M = 3.00 M
    • Answer: A. 3.00 M

  • Question 15: Oxidation number of Cr in Cr2O72− ion.

    • 2(Cr) + 7(O) = -2
    • 2(Cr) + 7(-2) = -2
    • 2Cr - 14 = -2
    • 2Cr = 12
    • Cr = +6
    • Answer: C. +6

  • Question 16: Element reduced in the reaction: NaI + 3HOCl -> NaIO3 + 3HCl.

    • Oxidation state of Cl in HOCl: +1
    • Oxidation state of Cl in HCl: -1
    • Chlorine is reduced.
    • Answer: A. Cl

  • Question 17: kJ of heat absorbed by 2 kg of water to raise the temperature from 18.0°C to 30.0°C.

    • q = m * c * ΔT
    • q = 2000 g * 4.184 J/g.°C * (30.0 - 18.0)°C = 100416 J
    • q = 100.416 kJ ≈ 100 kJ
    • Answer: B. 100 kJ

  • Question 18: Heat of formation of fructose, C6H12O6, given heat of combustion and heats of formation of CO2 and H2O.

    • C6H12O6 (s) + 6 O2(g) -> 6 CO2(g) + 6 H2O (l) ΔH° = -2812 kJ/mole
    • ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants)
    • -2812 = [6(-393.5) + 6(-285.83)] - [ΔH°f(C6H12O6) + 6(0)]
    • -2812 = -2361 - 1714.98 - ΔH°f(C6H12O6)
    • ΔH°f(C6H12O6) = -4075.98 + 2812 = -1263.98 kJ/mole
    • Answer: C. -1264 kJ

  • Question 19: Enthalpy change for the reaction with whole number coefficients.

    • 2CO2(g) + 3H2O(g) -> C2H6(g) + 7/2 O2 ΔH˚ = 1430 kj
    • Multiply by 2: 4CO2(g) + 6H2O(g) -> 2C2H6(g) + 7O2 ΔH˚ = 2 * 1430 kj
    • ΔH˚ = 2860 kj
    • Answer: D. 2860 kj

  • Question 20: ΔH˚ for equation 3 using Hess's Law.

    • 1. 2S (s) + 3O2 (g) -> 2SO3 (g) ΔH˚ = -790 Kj
    • 2. S (s) + O2 (g) -> SO2 (g) ΔH˚ = - 297 Kj
    • 3. 2SO2 (g) + O2 -> 2SO3 (g) ΔH˚ = ?
    • Reverse and multiply equation 2 by 2: 2SO2(g) -> 2S(s) + 2O2(g) ΔH˚ = +594 Kj
    • Add the modified equation 2 to equation 1:
      • 2S (s) + 3O2 (g) -> 2SO3 (g) ΔH˚ = -790 Kj
      • 2SO2(g) -> 2S(s) + 2O2(g) ΔH˚ = +594 Kj
      • 2SO2 (g) + O2 -> 2SO3 (g) ΔH˚ = -790 + 594 = -196 kj
    • Answer: A. −196 kj

  • Question 21: Heat exchanged when 12.5 g of CO reacts.

    • 2CO (g) + O2 (g) -> 2CO2 (g) ΔH˚ = -482 Kj
    • Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol
    • Moles of CO = 12.5 g / 28.01 g/mol = 0.446 mol
    • From the reaction, 2 moles of CO release 482 kJ of heat.
    • Heat exchanged by 0.446 mol of CO = (0.446 mol / 2 mol) * -482 kJ = -107.573 kJ
    • Answer: C. -107.6 kj

  • Question 22: Frequency of radiation with a wavelength of 0.589 pm.

    • c=λνc = \lambda \nu, where c is the speed of light, λ is the wavelength, and ν is the frequency.
    • ν=c/λν = c / \lambda
    • λ=0.589pm=0.589×10−12m\lambda = 0.589 pm = 0.589 \times 10^{-12} m
    • ν=(3×108m/s)/(0.589×10−12m)=5.09×1020s−1ν = (3 \times 10^8 m/s) / (0.589 \times 10^{-12} m) = 5.09 \times 10^{20} s^{-1}
    • Answer: D. 5.09×1020s−15.09 \times 10^{20} s^{-1}

  • Question 23: Maximum number of electrons that can occupy the subshell 4f.

    • f subshell has l = 3, which means it has 2l+1 = 2(3)+1 = 7 orbitals.
    • Each orbital can hold 2 electrons, so the f subshell can hold 7 * 2 = 14 electrons.
    • Answer: C. 14

  • Question 24: Most stable electron configuration for Cr.

    • Expected: [Ar] 4s2 3d4
    • Actual: [Ar] 4s1 3d5 (half-filled d-orbital is more stable)
    • Answer: C. [Ar]4s13d5

  • Question 25: Metallic character from right to left across a period.

    • Metallic character increases as you move from right to left.
    • Answer: A. Increases

  • Question 26: Element with the greatest electronegativity.

    • Electronegativity increases up and to the right on the periodic table.
    • Fluorine (F) is the most electronegative element.
    • Answer: A. F

  • Question 27: Total number of sigma and pi bonds.

    • (Note: the molecule structure is missing, thus I cannot provide the correct calculation)
    • Assuming structure is: CH3-CH=CH-CH2-C≡CH.
    • σ bonds: 3(C-H) + 1(C-C) + 1(C=C) + 2(C-H) + 1(C-C) + 2(C-H) + 1(C≡C) + 1(C-H) = 3+1+1+2+1+2+1+1 = 12
    • π bonds: 1 (from C=C) + 2 (From C≡C) = 3

  • Question 28: Which one violates the octet rule stability?

    • Octet rule states atoms prefer to have 8 electrons in their valence shell
    • O- : has 7 valence electrons. Can form stable octet
    • Answer: C. O-

  • Question 29: Formal charge on sulfur.

    • Formal Charge = Valence Electrons - Non-bonding Electrons - (1/2 Bonding Electrons)
    • (Note: the molecule structure is missing, thus I cannot provide the correct calculation)

  • Question 30: Electron-pair geometry and molecular geometry of ammonia.

    • Ammonia (NH3) has 4 electron pairs around the central nitrogen atom (3 bonding, 1 lone pair).
    • Electron-pair geometry: Tetrahedral
    • Molecular geometry: Trigonal pyramidal
    • Answer: D. tetrahedral, trigonal pyramidal

  • Question 31: Moles of N2 gas occupying 33.6 liters at STP.

    • At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.
    • Moles of N2 = 33.6 L / 22.4 L/mol = 1.5 mol
    • Answer: C. 1.5 mol

  • Question 32: Unknown gas effusing four times faster than O2(g).

    • Graham's Law of Effusion: Rate1/Rate2 = √(M2/M1)
    • Rate(X) / Rate(O2) = 4
    • 4 = √(32 / M1), where M1 is the molar mass of the unknown gas
    • 16 = 32 / M1
    • M1 = 32 / 16 = 2 g/mol. Thus Hydrogen (H2)
    • Answer: A. H2

  • Question 33: Volume of air at 760 torr to put into a 15 L tire to reach 2.38 atm.

    • P1V1 = P2V2
    • P1 = 760 torr = 1 atm
    • P2 = 2.38 atm
    • V2 = 15 L
    • V1 = (P2V2) / P1 = (2.38 atm * 15 L) / 1 atm = 35.7 L
    • Answer: B. 35.7 L

  • Question 34: Molecule that cannot form a hydrogen bond.

    • Hydrogen bonds form between H and O, N, or F.
    • SiH4 cannot form hydrogen bonds.
    • Answer: A. SiH4

  • Question 35: Ice floats on water because of.

    • Ice floats because it is less dense than liquid water.
    • Hydrogen bonding in ice creates an open structure, which decreases the density.
    • Answer: C. Low density due to Hydrogen bonding

Part II: Show All Your Work For Complete Credit

Question 1

Convert 2435 ft/s to kilometers per hour.

  • 2435fts×1m3.2ft×1km1000m×3600s1h=2739.375kmh2435 \frac{ft}{s} \times \frac{1 m}{3.2 ft} \times \frac{1 km}{1000 m} \times \frac{3600 s}{1 h} = 2739.375 \frac{km}{h}
Question 2

Calculate the average atomic mass of copper.

  • Cu-63: 62.93 amu, 69.09%
  • Cu-65: 64.93 amu, 30.91%
  • Average atomic mass = (0.6909 * 62.93) + (0.3091 * 64.93) = 43.48 + 20.08 = 63.56 amu
Question 3

Determine the empirical formula of Phenobarbital (62.06% C, 5.21% H, 12.06% N, 20.67% O).

  • Assume 100 g sample: 62.06 g C, 5.21 g H, 12.06 g N, 20.67 g O
  • Convert to moles:
    • C: 62.06 g / 12.01 g/mol = 5.17 mol
    • H: 5.21 g / 1.01 g/mol = 5.16 mol
    • N: 12.06 g / 14.01 g/mol = 0.86 mol
    • O: 20.67 g / 16.00 g/mol = 1.29 mol
  • Divide by the smallest number of moles (0.86):
    • C: 5.17 / 0.86 = 6.01 ≈ 6
    • H: 5.16 / 0.86 = 6.00 ≈ 6
    • N: 0.86 / 0.86 = 1
    • O: 1.29 / 0.86 = 1.50 ≈ 1.5
  • Multiply by 2:
    • C: 6 * 2 = 12
    • H: 6 * 2 = 12
    • N: 1 * 2 = 2
    • O: 1.5 * 2 = 3
  • Empirical formula: C12H12N2O3
Question 4

Phosphoric acid preparation: PCl5 + 4H2O -> H3PO4 + 5HCl.

  • 0.360 mol PCl5, 2.88 mol H2O
  • (a) Limiting reactant:
    • Mole ratio PCl5: H2O = 1:4
    • Required H2O = 0.360 mol PCl5 * (4 mol H2O / 1 mol PCl5) = 1.44 mol
    • Since we have 2.88 mol H2O (more than 1.44), PCl5 is the limiting reactant.
  • (b) Moles of HCl formed:
    • Mole ratio PCl5: HCl = 1:5
    • Moles HCl = 0.360 mol PCl5 * (5 mol HCl / 1 mol PCl5) = 1.8 mol
Question 5

Calculate the heat of reaction for N2(g) + O2(g) -> 2NO(g) using Hess's Law.

  • N2(g) + 2O2(g) -> 2NO2(g) ΔH˚ = +67.6 kJ
  • 2NO(g) + O2(g) -> 2NO2(g) ΔH˚ = -113.2 kJ
  • Reverse the second equation:
    • 2NO2(g) -> 2NO(g) + O2(g) ΔH˚ = +113.2 kJ
  • Add the first equation and reversed second equation:
    • N2(g) + 2O2(g) -> 2NO2(g) ΔH˚ = +67.6 kJ
    • 2NO2(g) -> 2NO(g) + O2(g) ΔH˚ = +113.2 kJ
    • N2(g) + O2(g) -> 2NO(g) ΔH˚ = 67.6 + 113.2 = 180.8 kJ
Question 6

Metal and water calorimetry.

  • 5.00 g metal heated to 100 ºC, plunged into 100 g water at 24.0 ºC, final temperature 28.0ºC.
  • (a) Joules absorbed by water:
    • q = m * c * ΔT
    • q = 100 g * 4.184 J/g.ºC * (28.0 - 24.0)ºC = 1673.6 J
  • (b) Joules lost by metal:
    • The heat lost by the metal is equal to the heat absorbed by the water, so q = -1673.6 J
  • (c) Heat capacity of the metal:
    • q = m * c * ΔT
    • -1673.6 J = 5.00 g * c * (28.0 - 100)ºC
    • -1673.6 J = 5.00 g * c * (-72)ºC
    • c = -1673.6 J / (5.00 g * -72 ºC) = 4.65 J/g.ºC