Exam 0.5 Practice – Quick Reference Notes

Exam Logistics

  • Time: Practice quiz = 25 minutes; real quizzes = 50 minutes.
  • Materials: non-programmable calculator; #2 pencils or dark pens; drinks allowed; devices stored away; seating to avoid same-color exam.
  • Bubble sheet: sign honor code; print neatly; include name, student ID; color coding (Green = A, Yellow = B).
  • Version note: For Version A #6 and Version B #5, ignore option F.

Quick Reference Formulas

  • Percent by mass (sugar in cola)
    • Given sugar% = 10.5% ⇒ msugar = 0.105 × mcola
    • mcola = msugar / 0.105
    • Example: 50 g sugar → m_cola ≈ \frac{50}{0.105} ≈ 476 g = 0.476 kg
    • Formula: m<em>sugar=0.105 m</em>cola⇒m<em>cola=m</em>sugar0.105m<em>{sugar} = 0.105\, m</em>{cola} \Rightarrow m<em>{cola} = \frac{m</em>{sugar}}{0.105}
  • Area conversion (cm² to ft²)
    • 1 ft = 30.48 cm ⇒ 1 ft² = (30.48)² cm² ≈ 929.03 cm²
    • A<em>ft2=A</em>cm2(30.48)2≈Acm2929.03A<em>{ft^2} = \frac{A</em>{cm^2}}{(30.48)^2} \approx \frac{A_{cm^2}}{929.03}
    • Example: 65000 cm² ≈ 69.97 ft²
  • Chlorate ion and naming
    • Chlorate ion: ClO3−\mathrm{ClO_3^-}
    • Naming conventions (examples):
    • H₂S → hydrogen sulfide
    • PO₃³⁻ → phosphite
    • BF₃ → boron trifluoride
    • VCl₂ → vanadium(II) dichloride
    • CO₂ → carbon dioxide
  • Pure substances vs mixtures (concept)
    • Pure substances: single element or compound with fixed composition.
    • Mixtures: contain more than one substance.
  • Significant figures (quick rules)
    • Nonzero digits are significant.
    • Zeros between nonzero digits are significant.
    • Leading zeros are not significant.
    • Trailing zeros are significant only if a decimal point or explicit notation is present.
    • Examples for 4 sig figs:
    • 0.003000, 3.000 × 10³, 8.004, 0.0008004
    • Example (ambiguous): 300 (unclear without decimal or scientific notation; could be 1–3 sig figs)
  • Density relation
    • Density: d=mVd = \frac{m}{V}
    • Mass from density: m=d⋅Vm = d \cdot V
    • Example: if $d = 3.50\ \mathrm{g/cm^3}$ and $V = 1\ \mathrm{cm^3}$, then $m = 3.50\ \mathrm{g}$

Quick Notes on Problem-Solving Contexts

  • For problems with small procedural steps (percent by mass, unit conversions), write the core relation and plug in values.
  • Always check units and convert to compatible units before calculating.
  • When in doubt about sig figs, default to the rule of preserving digits up to the last reported figure in the problem statement.

Reference for Exam Setup (Reminders)

  • Bring only allowed tools; keep others hidden during the exam.
  • Complete the bubble sheet with name, ID, and honor code; ensure legibility for grading.
  • If a question references a diagram or image (e.g., pure substances), rely on the stated concepts unless images are provided in your version.