Russo Semester I Review problems 2024

Semester I Review Problems

Conversions & Calculations

• Nanometers to Meters: 1 x 10^9 nm = 1 m

• Atomic Mass Unit to Grams: 1 amu = 1.6606 x 10^-24 g

Problem 1: Speed Conversion

• Given: Average speed of hydrogen at 25ºC = 1770 m/s

• Convert m/s to km/h:

  • Formula: speed (km/h) = speed (m/s) x 3.6

  • Calculation: 1770 m/s x 3.6 = 6372 km/h

Problem 2: Rate of Evaporation

• Given: 1 drop of water = 1.70 x 10^21 molecules, evaporation rate = 1,830,000 molecules/second

• Calculation of time in seconds:

  • Total time = Total molecules / Rate of evaporation

  • Total time = (1.70 x 10^21 molecules) / (1,830,000 molecules/second)

  • Total time ≈ 9.3 x 10^12 seconds

• Convert seconds to years:

  • Total years ≈ (9.3 x 10^12 s) / (60s/min x 60min/h x 24h/day x 365.25 days/year) ≈ 295,000 years

Problem 3: Ionization Energy Calculation

• Given: Ionization energy for Na = 0.50 MJ/mol

  • 1000 kJ = 1 MJ, 1 mol = 6.022 x 10^23 atoms

• Calculation for 4.8 x 10^26 sodium atoms:

  • Energy required = (Number of atoms / Avogadro's number) x Ionization energy

  • Energy = (4.8 x 10^26 / 6.022 x 10^23) x (0.50 x 10^6 kJ)

  • Total energy ≈ 39,918,377 kJ

Problem 4: Density of Gallium

• Given: Density of gallium = 5.88 g/ml, vessel volume = 25.0 ml

• Calculation of mass:

  • Mass = density x volume

  • Mass = 5.88 g/ml x 25.0 ml = 147 g

Problem 5: Density of Sulfuric Acid

• Given: Density = 1.86 g/ml, mass = 97.6 g

• Calculation of volume:

  • Volume = mass / density

  • Volume = 97.6 g / 1.86 g/ml ≈ 52.3 ml

Problem 6: Mass of Cat After Eating

• Given: Mass of cat = 3.220 kg, mass of rat = 213.6 g, weight of cheese = 0.120 g

• Total mass:

  • Before consumption: Cat's mass = 3.220 kg + (213.6 g + 0.120 g converted to kg)

  • Total mass = 3.220 kg + 0.21372 kg = 3.43372 kg

Problem 7: Precision and Accuracy of Boiling Point

• Given: Experimental boiling point of octane = 124.1ºC, theoretical boiling point = 125.7ºC

• Determination:

  • The boiling point is close (124.1ºC vs. 125.7ºC) indicating the work is accurate but not precise.

Problem 8: Error and Percent Error

• Calculation:

  • Error = Experimental value - Theoretical value = 124.1 - 125.7 = -1.6ºC

  • Percent Error = |Error| / Theoretical value x 100 = |-1.6| / 125.7 x 100 ≈ 1.27%

Problem 9: Average Atomic Mass of Neon

• Isotopes:

  • 90.51% of 19.992 amu, 0.2700% of 20.994 amu, 9.22% of 21.991 amu

• Calculation:

  • Average Atomic Mass = (fraction of isotope) x (mass of isotope)

  • = (0.9051 x 19.992) + (0.0002700 x 20.994) + (0.0922 x 21.991)

  • ≈ 20.179 amu

Problem 10: Magnesium Isotope Mass

• Given: Magnesium isotope mass = 24.9858 amu

• Mass in grams:

  • 1 amu = 1.6606 x 10^-24 grams

  • Mass in grams ≈ 24.9858 x (1.6606 x 10^-24) = 4.1474 x 10^-23 g

• Mass of 2.4890 x 10^21 atoms:

  • Total mass = number of atoms x mass of one atom

  • Total mass = (2.4890 x 10^21)x(4.1474 x 10^-23 g) ≈ 0.103 g

Problem 11: Energy Gained by Water Sample

• Given: Mass = 40.8 g, initial T = 22.4ºC, final T = 35.3ºC, specific heat = 4.184 J/gºC

• Calculation of energy:

  • Energy = mass x specific heat x (T_final - T_initial)

  • Energy = 40.8 g x 4.184 J/gºC x (35.3 - 22.4)ºC ≈ 2232.76 J

Problem 12: Mass of Heated Silver

• Given: Energy = 6050 J, initial temp = 17.3ºC, final temp = 38.5ºC, specific heat of silver = 0.233 J/gºC

• Calculation:

  • Mass = Energy / (specific heat x (T_final - T_initial))

  • Mass = 6050 J / (0.233 J/gºC x (38.5 - 17.3)ºC) ≈ 1278.26 g

Problem 13: Frequency of UV Light

• Given: Wavelength = 295 nm

• Calculation of frequency:

  • Speed of light (c) = 3 x 10^8 m/s, Frequency (f) = c / wavelength

  • Frequency = (3 x 10^8 m/s) / (295 x 10^-9 m) ≈ 1.02 x 10^{15} Hz

Problem 14: Frequency of Yellow Light

• Given: Energy of light = 3.37 x 10^-19 J

• Calculation of frequency:

  • Energy of photon = h x f, where h = 6.626 x 10^-34 J·s

  • Frequency = Energy / h = (3.37 x 10^-19 J) / (6.626 x 10^-34 J·s)

  • Frequency ≈ 5.08 x 10^{14} Hz

Problem 15: Energy of Green Light Photon

• Given: Wavelength = 550 nm

• Calculation:

  • Convert to meters and use Energy = h x f:

  • Energy = h x (speed of light / wavelength)

  • Energy = 6.626 x 10^-34 J·s x (3 x 10^8 m/s / 550 x 10^-9 m) ≈ 3.61 x 10^-19 J

Problem 16: Energy to Cool Air in Refrigerator

• Given: Mass of escaping air = 720 g, initial temp = 27.6ºC, final temp = 4.8ºC, specific heat of air = 1 J/gºC

• Energy Calculation:

  • Energy = mass x specific heat x (T_final - T_initial)

  • Energy = 720 g x 1 J/gºC x (4.8 - 27.6)ºC = 720 x (-22.8) ≈ -16,416 J

Classification and Properties

Problem 17: Classification of Substances

• a) Air: Mixture

• b) Table salt: Compound

• c) Milk: Mixture

• d) Mercury: Element

Problem 18: Homogeneous vs Heterogeneous

• a) Ranch dressing: Heterogeneous

• b) Oatmeal raisin cookie dough: Heterogeneous

• c) Gasoline: Homogeneous

• d) Seawater: Homogeneous

Problem 19: Phosphorus Properties

• a) Physical (exist as different forms)

• b) Physical (solid at 25°C)

• c) Physical (specific heat capacity)

• d) Chemical (burns in chlorine)

Problem 20: Discovery of Nucleus

• Discoverer: Ernest Rutherford

• Experiment: Gold foil experiment using alpha particles which led to the conclusion that atoms contain a small, dense nucleus.

Problem 21: Iodine Properties

• a) Formula: I2

• b) Protons & electrons in I2: 53 each, in iodide ion (I-): 54 protons, 53 electrons

• c) Atomic number for iodine: 53

• d) Neutrons: Mass number - Atomic number = 130 - 53 = 77

• e) Group/Period: Group 17, Period 5; Nonmetal

Problem 22: Atomic Radius Trend

• Trend: Increases down a group and decreases across a period.

• Cause: Increased electron shielding in heavy elements and increased nuclear charge.

Problem 23: Link between Atomic Radius and Ionization Energy

• Explanation: Larger atomic radii generally lead to a lower ionization energy, as outer electrons are further from the nucleus and easier to remove.

Problem 24: Naming Compounds

• a) N2H4: Hydrazine

• b) NaClO: Sodium hypochlorite

• c) BrF5: Bromine pentafluoride

• d) NCl3: Nitrogen trichloride

• e) Fe2(SiO3)3: Iron(III) silicate

• f) IF7: Iodine heptafluoride

• g) KHCO3: Potassium bicarbonate

Problem 25: Writing Formulas for Compounds

• a) Calcium bromide: CaBr2

• b) Aluminum sulfide: Al2S3

• c) Tin(IV) cyanide: Sn(CN)4

• d) Strontium acetate: Sr(C2H3O2)2

• e) Silver chloride: AgCl

• f) Nitrogen dioxide: NO2

Problem 26: Electron Configurations

• a) S: 1s2 2s2 2p6 3s2 3p4; [Ne] 3s2 3p4

• b) K: 1s2 2s2 2p6 3s2 3p6 4s1; [Ar] 4s1

• c) Ti: 1s2 2s2 2p6 3s2 3p6 4s2 3d2; [Ar] 4s2 3d2

• d) Cl: 1s2 2s2 2p6 3s2 3p5; [Ne] 3s2 3p5

• e) Sb: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3; [Kr] 5s2 4d10 5p3

• f) Ba: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2; [Xe] 6s2

Problem 27: Unpaired Electrons

• a) Mg: 2 unpaired electrons

• b) P: 3 unpaired electrons

• c) V: 3 unpaired electrons

Problem 28: Completed Table of Atomic Parameters

• Example for Ba:

  • Symbol: Ba

  • of neutrons: 81

  • of protons: 56

  • Atomic #: 56

  • Mass #: 137

Metalloids and Molecular Structures

Problem 29: Metalloids

• a) Aluminum: No

• b) Boron: Yes

• c) Bismuth: No

• d) Antimony: Yes

• e) Selenium: No

Problem 30: Lewis Structures

• a) CS2: Linear

• b) CN-: Linear

• c) SeO3: Bent

• d) H3O+: Bent

• e) TeF2: Bent

• f) PH3: Trigonal pyramidal

• g) PCl4+: Tetrahedral

• h) CO3-2: Trigonal planar

Problem 31: Molecular Geometry and Bond Angles

• Geometrics:

  • Linear for CS2 and CN- (180°)

  • Bent for SeO3, H3O+ (120° in a case, similar)

  • Trigonal pyramidal for PH3 (107°)

  • Tetrahedral for PCl4+ (109.5°)

  • Trigonal planar for CO3-2 (120°)

• Polarity:

  • Molecules with lone pairs are polar, linear ones without lone pairs are nonpolar.

Problem 32: Dissolving in Hexane

• Molecules likely to dissolve in hexane: Nonpolar and weakly polar molecules from Problem 30

Problem 33: Bond Order in Silicate Ion

• Bond Order Calculation: Average number of bonds for atoms in silicate ion structure.

Problem 34: Bond Comparison

• a) Silicate: Has longer bonds due to lower bond order.

• b) Silicon Dioxide: Has stronger bonds due to higher bond order.

Problem 35: Bond Types

• a) Hg-P: Polar Covalent

• b) P-S: Polar Covalent

• c) Si-F: Polar Covalent

• d) Mg-N: Ionic

• e) Fe-N: Polar Covalent

Problem 36: Polarity of SO2

• Diagrammatic Representation: Asymmetrical shape leads to dipole moment resulting in a polar molecule.

Problem 37: Experimental Accuracy

• Given Accepted Value: 0.865 g/ml

• Assessment of Data:

  • 0.887 g/ml: Precise but not accurate

  • 0.672 g/ml: Neither accurate nor precise

  • 1.036 g/ml: Accurate but not precise

Problem 38: Measurement Assessment

• Identifiable Error: Requires accompanying arrow data for assessment.

Problem 39: Classification of Substances

• a) Nonpolar, high melting point, conducts electricity: Metallic

• b) Conducts electricity in water, high melting: Ionic

• c) Insoluble, low melting: Molecular

Problem 40: Bond Type Impact on Solubility

• Explanation: Polar substances dissolve in polar solvents, nonpolar in nonpolar.

Problem 41: Capsaicin and Solubility

• Bond Type: Nonpolar due to preference for fats over water.

Problem 42: Soap Functionality

• Soap Structure: Two-sided - hydrophilic end attracts water; hydrophobic end attracts grease and oil.
  

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