CHEM 1411 Practice Exam 1 Notes

A. Fills any container completely
- State of Matter: Gas
- Gases do not have a fixed volume or shape and will expand to fill any container they are placed in.
B. Assumes the shape of its container
- State of Matter: Liquid
- Liquids have a definite volume but take the shape of their container.
C. Is rigid and only slightly compressible
- State of Matter: Solid
- Solids have a definite shape and volume, and are not easily compressible.

A. Formation of Ice Cubes
- Type: Exothermic process
- Heat is released when water freezes into ice.
B. Condensation
- Type: Exothermic process
- Heat is released when water vapor condenses into liquid water.
C. Burning Crude Oil
- Type: Exothermic process
- Heat is released when crude oil undergoes combustion.

A. Oxygen
- Type: Pure Substance
- Oxygen (O₂) is a diatomic molecule and a pure substance.
B. Chocolate Chip Cookie
- Type: Heterogeneous Mixture
- Contains distinct components (chocolate chips, cookie dough) that can be observed.
C. Aluminum Foil
- Type: Pure Substance
- Aluminum foil is made of pure aluminum atoms without mixture.

A. 15892
- Standard Scientific Notation: $1.5892 imes 10^4$
- Significant Figures: 5
B. 0.015892
- Standard Scientific Notation: $1.5892 imes 10^{-2}$
- Significant Figures: 5
C. 0.1
- Standard Scientific Notation: $1.0 imes 10^{-1}$
- Significant Figures: 2

A. 1.070 L to gallons
- Conversion: $1.070 ext{ L} imes 0.264172 = 0.282 ext{ gallons}$
- Significant Figures: 3
B. 6.2 x 10^{-7} kg to short tons
- Conversion: $6.2 imes 10^{-7} ext{ kg} imes (1 ext{ short ton} / 907.185 ext{ kg}) \ = 6.84 imes 10^{-10} ext{ short tons}$
- Significant Figures: 2

Density of Sucrose: 1.65 g/cm³
Dimensions of Crystal: 2.20 mm x 1.36 mm x 1.23 mm
- Volume Calculation:
- Convert mm to cm: 2.20 mm = 0.220 cm, 1.36 mm = 0.136 cm, 1.23 mm = 0.123 cm
- $ext{Volume} = 0.220 ext{ cm} imes 0.136 ext{ cm} imes 0.123 ext{ cm} = 0.003836 ext{ cm}³$
- Mass Calculation using Density:
- $ext{Mass} = ext{Density} imes ext{Volume} = 1.65 ext{ g/cm}³ imes 0.003836 ext{ cm}³ = 0.0063 ext{ g}$

A. Mass of Carbon in 7.65g of CO₂
- Molar Mass of CO₂: Carbon (C) = 12.01 g/mol, Oxygen (O) = 16.00 g/mol
- Molar mass of CO₂ = 12.01 + 2(16.00) = 44.01 g/mol
- $ext{Mass of C} = rac{12.01 ext{ g/mol}}{44.01 ext{ g/mol}} imes 7.65 ext{ g} = 2.09 ext{ g}$
B. Number of Carbon Atoms in 7.65g CO₂
- Number of moles of CO₂ = $rac{7.65 ext{ g}}{44.01 ext{ g/mol}} = 0.1748 ext{ mol}$
- Number of moles of C in CO₂ = 0.1748 mol (1:1 ratio)
- Number of atoms = $0.1748 ext{ mol} imes 6.022 imes 10^{23} ext{ atoms/mol} = 1.05 imes 10^{23} ext{ atoms}$

Complete the following table:
- Atom/Ion: 90Sr
- Chemical Symbol: Sr
- Number of Protons: 38
- Number of Electrons: 38
- Number of Neutrons: 52
- Atom/Ion: 128Te²⁻
- Chemical Symbol: Te
- Number of Protons: 52
- Number of Electrons: 54
- Number of Neutrons: 76
- Atom/Ion: 41Sc³⁺
- Chemical Symbol: Sc
- Number of Protons: 21
- Number of Electrons: 18
- Number of Neutrons: 20
- Atom/Ion: 34S⁴⁺
- Chemical Symbol: S
- Number of Protons: 16
- Number of Electrons: 12
- Number of Neutrons: 18

Name or Write Balanced Formula for the following:
- A. PbO₂
- B. Al₂(SO₃)₂
- C. N₂O
- D. HClO₄
- E. H₂S(aq)
- F. Cu₂O
- G. Mg₃P₂
- H. Sn(NO₃)₄
- I. HNO₂
- J. HClO
- K. Cuprous Chlorite
- L. Lithium Carbonate
- M. Potassium Acetate
- N. Nitric Acid
- O. Sulfuric Acid
- P. Hydroiodic Acid
- Q. Dinitrogen Tetrasulfide
- R. Phosphorus Pentachloride
- S. Tin (IV) Iodide
- T. Ammonium Oxide

Given Composition: 68.3% lead, 10.6% sulfur, remainder oxygen
- Determine: Remainder for Oxygen = 100% - (68.3% + 10.6%) = 21.1%
- Convert to Moles:
- Moles of Pb = $rac{68.3 ext{ g}}{207.2 ext{ g/mol}} = 0.329 ext{ mol}$
- Moles of S = $rac{10.6 ext{ g}}{32.07 ext{ g/mol}} = 0.331 ext{ mol}$
- Moles of O = $rac{21.1 ext{ g}}{16.00 ext{ g/mol}} = 1.32 ext{ mol}$
Empirical Formula: PbS₁O₄
Molar Mass Approx.: 910 g
- Molecular Formula Calculation:
- Divide molar mass by empirical formula mass (lead = 207.2, sulfur = 32.07, oxygen = 16.00): ( ext{Empirical Formula Mass} = 207.2 + 32.07 + 64 = 303.27 ext{ g/mol} )
- Ratio = $rac{910}{303.27} <br>ightarrow ext{approximately } 3$
- So, $ext{Molecular Formula} = Pb₃S₃O₁₂$

A. Atomic Structure Before Rutherford's Experiment
- Model: Plum Pudding Model (proposed by J.J. Thomson)
- Atoms are a uniform positive sphere with negative electrons embedded within it.
- Expected Outcome: Deflection of alpha particles expected to be minimal – as particles were thought to be uniformly distributed.
B. Nuclear Atom Model
- Description: After the gold foil experiment, the concept of the nuclear atom was introduced.
- The nucleus is a dense core containing protons and neutrons, with electrons orbiting around it.
- Development:
- Outcome of the experiment: Some alpha particles were deflected at large angles, which showed that there is a concentrated mass (the nucleus) at the center of the atom, contradicting the plum pudding model.