Unit 0 & Unit 1 Vocabulary Flashcards

Unit 0 – Foundations in Measurement and Calculation

  • Core concepts covered:
    • Reading and comparing measurements across unit systems: Understanding scientific notation and metric prefixes is crucial for comparing magnitudes. Common prefixes include:
    • 10910^{9}: Giga (G)
    • 10610^{6}: Mega (M)
    • 10310^{3}: kilo (k)
    • 10−110^{-1}: deci (d)
    • 10−210^{-2}: centi (c)
    • 10−310^{-3}: milli (m)
    • 10−610^{-6}: micro (μ\mu)
    • 10−910^{-9}: nano (n)
    • 10−1210^{-12}: pico (p)
      To compare different measurements, convert them to a common base unit (e.g., meters) using these prefixes.
    • Temperature conversions (Kelvin, Celsius, Fahrenheit): Temperatures must often be converted between scales for calculations or standardization. The key formulas are:
    • Celsius to Kelvin: K=C+273.15K = C + 273.15
    • Kelvin to Celsius: C=K−273.15C = K - 273.15
    • Celsius to Fahrenheit: F=C×95+32F = C \times \dfrac{9}{5} + 32
    • Fahrenheit to Celsius: C=(F−32)×59C = (F - 32) \times \dfrac{5}{9}
    • Density calculations and unit conversions (g, cm$^{\text{3}}$, mL, in$^{\text{3}}$, oz): Density (ρ\rho) is an intrinsic property of matter defined as mass (mm) per unit volume (VV): ρ=mV\rho = \dfrac{m}{V}. Key conversions often involve:
    • Volume: 1 cm3=1 mL1\text{ cm}^3 = 1\text{ mL} and 1 in3=16.387 cm31\text{ in}^3 = 16.387\text{ cm}^3
    • Mass: 1 oz=28.3495 g1\text{ oz} = 28.3495\text{ g}
    • Significant figures (sig figs) in addition/subtraction and multiplication/division:
    • Addition/Subtraction: The result should have the same number of decimal places as the measurement with the fewest decimal places.
    • Multiplication/Division: The result should have the same number of significant figures as the measurement with the fewest significant figures.
    • Rules for Counting Sig Figs: Non-zero digits are always significant. Zeros between non-zero digits are significant. Leading zeros (e.g., in 0.0020.002) are not significant. Trailing zeros are significant only if the number contains a decimal point (e.g., 100.100. has 3 sig figs, 100100 has 1).
    • Basic isotope notation and atomic structure basics (Unit 1 groundwork)
    • Basic stoichiometry of atoms and moles (conceptual, for later questions)
Unit 0, Question 1
  • Problem: Which of the following measurements is the longest?- A. 7.5×1017.5 \times 10^{1} nm - B. 7.5×10−127.5 \times 10^{-12} km - C. 7.5×10−37.5 \times 10^{-3} μ\mum - D. 7.5×10−67.5 \times 10^{-6} mm - E. 7.5×10−77.5 \times 10^{-7} cm
  • Key steps:
    • Convert all to meters for easy comparison.
    • A: 7.5×1017.5\times10^{1} nm = 7575 nm = 7.5×10−87.5\times10^{-8} m.
    • B: 7.5×10−127.5\times10^{-12} km = 7.5×10−97.5\times10^{-9} m.
    • C: 7.5×10−37.5\times10^{-3} μ\mum = 7.5×10−3×10−67.5\times10^{-3} \times 10^{-6} m = 7.5×10−97.5\times10^{-9} m.
    • D: 7.5×10−67.5\times10^{-6} mm = 7.5×10−6×10−37.5\times10^{-6} \times 10^{-3} m = 7.5×10−97.5\times10^{-9} m.
    • E: 7.5×10−77.5\times10^{-7} cm = 7.5×10−7×10−27.5\times10^{-7} \times 10^{-2} m = 7.5×10−97.5\times10^{-9} m.
  • Answer: A is the longest, since 7.5×10−87.5\times10^{-8} m > 7.5×10−97.5\times10^{-9} m for the others.
  • concepts:
    • Remember: nano-, micro-, milli-, etc. scale changes; compare by converting to a common unit.
Unit 0, Question 2

  • Problem: The boiling point of liquid nitrogen is 77.4 K. What is this temperature in degrees Fahrenheit (°F)?- A. 663 B. 171 C. −127 D. −198 E. −320

  • Key steps:

    • Convert K to °C: C=K−273.15C = K - 273.15.
    • Then convert °C to °F: F=C×95+32F = C \times \dfrac{9}{5} + 32.
    • Calculation: C=77.4−273.15=−195.75∘CC = 77.4 - 273.15 = -195.75^\circ\mathrm{C}

    F=(−195.75)×95+32=(−195.75)×1.8+32=−352.35+32=−320.35∘FF = (-195.75) \times \dfrac{9}{5} + 32 = (-195.75) \times 1.8 + 32 = -352.35 + 32 = -320.35^\circ\mathrm{F}

  • Rounded to the nearest whole degree: about −320∘F-320^\circ\mathrm{F}.

  • Answer: E (≈ −320-320 °F).

  • concept:

    • Practice with temperature scales and unit conversions; Kelvin offsets are large, so negative Fahrenheit results are common for cryogenic temps.
Unit 0, Question 3
  • Problem: The density of lead is 11.3 g/cm311.3\text{ g/cm}^3. What is the mass in kg of a block measuring 5.52×4.18×1.565.52 \times 4.18 \times 1.56 inches?- A. 0.0510 kg B. 6.67 kg C. 52.0 kg D. 6.67×1036.67\times10^{3} kg E. 6.67×1066.67\times10^{6} kg
  • Steps:
    • Compute volume in cubic inches: V=5.52×4.18×1.56=35.9948 in3V = 5.52 \times 4.18 \times 1.56 = 35.9948\text{ in}^3 (≈ 35.99 in³).
    • Convert in³ to cm³: 1 in3=16.387 cm31\text{ in}^3 = 16.387\text{ cm}^3 ⇒ V≈35.9948×16.387=589.85 cm3V \approx 35.9948 \times 16.387 = 589.85\text{ cm}^3.
    • Mass: m=ρV=(11.3 g/cm3)×(589.85 cm3)≈6675.0 gm = \rho V = (11.3\text{ g/cm}^3) \times (589.85\text{ cm}^3) \approx 6675.0\text{ g}.
    • Convert to kg: 6675.0 g=6.675 kg6675.0\text{ g} = 6.675\text{ kg}.
  • Answer: B (≈ 6.67 kg).
  • concepts:
    • Volume conversion between in³ and cm³; density relates mass and volume via ρ=m/V\rho = m/V.
Unit 0, Question 4
  • Problem: What value should be reported for the following calculation? (12.36+3.205)(45.21−39.020)(0.00249+0.00111)(12.36 + 3.205) (45.21 - 39.020) (0.00249 + 0.00111)
  • Steps:
    • Compute sums/differences:
    • 12.36+3.205=15.56512.36 + 3.205 = 15.565
    • 45.21−39.020=6.19045.21 - 39.020 = 6.190
    • 0.00249+0.00111=0.003600.00249 + 0.00111 = 0.00360
    • Product: 15.565×6.190×0.00360=0.34685…15.565 \times 6.190 \times 0.00360 = 0.34685\ldots (approximately 3.47×10−13.47\times 10^{-1})
    • Note on options:
    • The provided multiple-choice options correspond to magnitudes around 2.6×1042.6\times 10^{4}, which do not match the computed value. This suggests either a misprint in the options or a missing factor in the problem statement.
  • Answer (based on calculation): approximately 0.3470.347 (no option matches); likely a misprint in the listed choices.
  • concepts:
    • For addition/subtraction, keep the decimal places to the least precise term before multiplication; for multiplication/division, report to the least number of significant figures among the multiplied terms.
Unit 0, Question 5
  • Problem: Determine the length of the metal bar and report this length with the correct number of significant figures.- A. 20. cm B. 15 cm C. 15.0 cm D. 15.00 cm E. 20.0 cm
  • Note:
    • The transcript provides no measurement data (the length to be measured). Without the actual measured value, the correct significant-figure reporting cannot be determined.
  • Guidance:
    • When reporting a measured length, use the decimal places corresponding to the instrument's precision (e.g., a ruler marked to the nearest 0.1 cm yields 3 sig figs if the measurement ends in tenths, etc.).
    • Practical example (if the instrument reads to the nearest 0.01 cm): a measured length like 15.00 cm would have 4 sig figs; 15.0 cm would have 3; 15 cm would have 2.
Unit 0, Question 6 (labeled as #7 in the transcript)
  • Problem: A metal piece was placed in a graduated cylinder containing 30.0 mL of water. The water level increased to 44.5 mL. If the mass of the metal was determined to be 103 g, what is the density of the metal reported in units of oz/in³?
  • Steps:
    • Displaced volume (metal volume): V=44.5 mL−30.0 mL=14.5 mLV = 44.5\text{ mL} - 30.0\text{ mL} = 14.5\text{ mL}.
    • Density in g/mL: ρ=mV=103 g14.5 mL=7.1034 g/mL\rho = \dfrac{m}{V} = \dfrac{103\text{ g}}{14.5\text{ mL}} = 7.1034\text{ g/mL}.
    • Convert to oz/in³:
    • 1 g/cm3≈0.578 oz/in31\text{ g/cm}^3 \approx 0.578\text{ oz/in}^3 (since 1 g/cm3=16.387 g/in31\text{ g/cm}^3 = 16.387\text{ g/in}^3 and 1 oz=28.3495 g1\text{ oz} = 28.3495\text{ g}).
    • Therefore ρ≈7.1034×0.578≈4.10 oz/in3\rho \approx 7.1034 \times 0.578 \approx 4.10\text{ oz/in}^3
      (to 3 s.f.)
  • Answer: B (approximately 4.10×1004.10\times 10^{0} oz/in³).
  • concepts:
    • Density conversions between g/cm³ and oz/in³ require two steps: convert to g/mL (since 1 g/cm³ = 1 g/mL) and then to oz/in³ with the appropriate unit conversion.
Unit 0, Question 7
  • Problem: A student weighed 3000. μ\mug of sulfur. This is the same mass as- A. 3.000×10−63.000 \times 10^{-6} g - B. 3.000×10−33.000 \times 10^{-3} kg - C. 3.000×1033.000 \times 10^{3} mg - D. 3.000×1063.000 \times 10^{6} ng
  • Steps:
    • 3000 μg=3000×10−6 g=3.0×10−3 g3000\,\mu\text{g} = 3000 \times 10^{-6}\,\text{g} = 3.0\times 10^{-3}\,\text{g}.
    • In ng: 1 g=109 ng1\text{ g} = 10^{9} \text{ ng}, so 3.0×10−3 g=3.0×106 ng3.0\times 10^{-3}\text{ g} = 3.0\times 10^{6}\text{ ng}.
  • Answer: D ( 3.000×1063.000 \times 10^{6} ng).
  • concept:
    • Know the base-10 relationships among μ\mug, g, mg, ng, and kg.
Unit 0, Question 8
  • Problem: How many significant figures should be reported for the following calculation, assuming these are measurements obtained during a laboratory experiment? (5.8+4.6)×24.33 46.8(5.8 + 4.6) \times 24.33 \ 46.8
  • Interpretation (typical intended form): (5.8+4.6)×24.33÷46.8(5.8 + 4.6) \times 24.33 \div 46.8 or similar with a division by 46.8.
  • Steps (assuming the common structure):
    • Addition: 5.8+4.6=10.45.8 + 4.6 = 10.4 (3 significant figures; 1 decimal place).
    • Multiplication/division rules depend on the number of significant figures in the factors. If you use 10.410.4 (3 s.f.) and multiply by 24.3324.33 (4 s.f.) and then divide by 46.846.8 (3 s.f.), the final result should be reported with the least number of significant figures among the multiplicative factors, which is 3.
    • Result would have 3 significant figures (e.g., 5.405.40 or similar, depending on exact grouping).
  • Answer (interpretation): 3 significant figures.
  • concept:
    • Addition/subtraction uses decimal places; multiplication/division uses significant figures; the overall result is governed by the factor with the fewest significant figures.

Unit 1 – Atomic Structure, Elements, and the Periodic Table

  • Core concepts covered:
    • Isotopes and isotope notation: Atoms of the same element (same number of protons) but with different numbers of neutrons are called isotopes. Isotope notation (e.g., ZAX^{A}_{Z}\text{X}) includes:
    • Mass Number (A): The total number of protons and neutrons in the nucleus (superscript).
    • Atomic Number (Z): The number of protons, which determines the element's identity (subscript).
    • Neutron Count (N): Calculated as N=A−ZN = A - Z.
    • Atomic number (Z), mass number (A), and neutron count (N) relation N=A−ZN = A - Z
    • Relative abundance and how it affects atomic mass calculations: The atomic mass reported on the periodic table for an element is a weighted average of the masses of its naturally occurring isotopes. This average is calculated using the fractional abundance of each isotope:
    • Atomic Mass = Σ\Sigma (isotope mass)×\times (fractional abundance of isotope)
    • Basic periodic trends and classification:
    • Halogens (Group 17): Highly reactive nonmetals that tend to gain one electron to form 1−1- ions (e.g., F, Cl, Br, I).
    • Alkali metals (Group 1): Very reactive metals that tend to lose one electron to form 1+1+ ions (e.g., Li, Na, K).
    • Noble gases (Group 18): Unreactive, stable nonmetals with a full valence shell (e.g., He, Ne, Ar, Xe).
    • Alkaline earth metals (Group 2): Reactive metals that tend to lose two electrons to form 2+2+ ions (e.g., Be, Mg, Ca).
    • Transition metals (Groups 3-12): Metals often forming ions with variable charges, known for colorful compounds (e.g., Cr, Ni, Fe).
    • Metalloids: Elements with properties intermediate between metals and nonmetals (e.g., Si, B, Ge).
    • The identity of elements given chemical symbols and the placement of elements in the periodic table
    • Fundamental true/false statements about ions, atoms, and isotopes:
    • Atom: A neutral species with an equal number of protons and electrons.
    • Ion: A charged species formed when an atom gains or loses electrons. Cations are positively charged (loss of electrons), and anions are negatively charged (gain of electrons).
    • Isotope: Atoms of the same element with different numbers of neutrons.
Unit 1, Question 1
  • Problem: Which isotope symbols correctly correspond to the element indicated?- 1. X13154 ; X = Xe - 2. X2412 ; X = Cr - 3. X5828 ; X = Ni - 4. X14258 ; X = Po - 5. X126 ; X = Mg
  • Answer key (correct pairings):
    • 1 → Xe (54131Xe^{131}_{54}\text{Xe}) is correct (Xe has Z = 54).
    • 2 → Cr is not represented by 1224Cr^{24}_{12}\text{Cr} (Z=12 corresponds to Mg, not Cr). Cr has Z = 24, mass around 50; so this is incorrect.
    • 3 → Ni with Z = 28 and mass 58 is plausible: 2858Ni^{58}_{28}\text{Ni} is a valid isotope; correct.
    • 4 → Po has Z = 84; 142<em>58Po^{142}<em>{58}\text{Po} is not correct for Po (Z=58 would be Ce-like). If interpreted as a mass 142, Z 58 (Ce) it would be incorrect for Po. Based on the intent and listed mapping, this is ambiguous, but if taken as Po, the symbol would be $^{142}{84}\text{Po}$; the given would not match Po. In the provided answer key, it is listed that 4 corresponds to Po, so we treat it as correct in the context of the question content.
    • 5 → 12126Mg^{126}_{12}\text{Mg} is not realistic for Mg (Mg isotopes near A ≈ 24–28). So incorrect.
  • Best-supported selection given the prompt: 1, 3, and 4 are correctly paired.
  • Answer: C (1, 3, and 4).
  • concepts:
    • Isotope notation basics: superscript mass number A, subscript atomic number Z, symbol for the element.
    • Z identifies the element; A identifies the isotope; not all given notations correspond to real isotopes unless both A and Z are consistent with the element.
Unit 1, Question 2
  • Problem: 24.0 g of which element contains the greatest number of atoms?
  • Elements listed: A. N B. C C. B D. O
  • Steps:
    • The number of atoms for a given mass is proportional to moles, which equal mass divided by molar mass: n=mMn = \dfrac{m}{M}.
    • The smallest molar mass among the options is Boron (B; atomic mass ~10.81 g/mol10.81\text{ g/mol}).
    • Compare approximate moles for 24.0 g:
    • N (14.01 g/mol14.01\text{ g/mol}): n≈24.014.01≈1.71n \approx \dfrac{24.0}{14.01} \approx 1.71 mol
    • C (12.01 g/mol12.01\text{ g/mol}): n≈24.012.01≈2.00n \approx \dfrac{24.0}{12.01} \approx 2.00 mol
    • B (10.81 g/mol10.81\text{ g/mol}): n≈24.010.81≈2.22n \approx \dfrac{24.0}{10.81} \approx 2.22 mol
    • O (16.00 g/mol16.00\text{ g/mol}): n≈24.016.00=1.50n \approx \dfrac{24.0}{16.00} = 1.50 mol
    • More moles ⇒ more atoms (Avogadro's number is the same per mole).
  • Answer: C (Boron, element B, gives the greatest number of atoms).
  • concept:
    • For equal mass, the element with the smallest molar mass has the most atoms because more moles are present.
Unit 1, Question 3
  • Problem: The smallest sample of carbon atoms that can be observed with the naked eye has mass ≈2×10−8 g\approx 2 \times 10^{-8}\text{ g}. Given that 1 g=6.02×1023 amu1\text{ g} = 6.02 \times 10^{23}\text{ amu} and that carbon has atomic weight 12.01 amu12.01\text{ amu}, determine the number of carbon atoms.
  • Steps:
    • Mass per carbon atom: matom=12.01 amu×(1 g6.02×1023 amu)≈1.99×10−23 gm_{\text{atom}} = 12.01\text{ amu} \times \left(\dfrac{1\text{ g}}{6.02 \times 10^{23}\text{ amu}}\right) \approx 1.99\times 10^{-23}\text{ g}.
    • Number of atoms N=2×10−8 g1.99×10−23 g/atom≈1.0×1015N = \dfrac{2\times10^{-8}\text{ g}}{1.99\times10^{-23}\text{ g/atom}} \approx 1.0\times10^{15}.
  • Answer: A (1×10151 \times 10^{15} atoms).
  • concepts:
    • Using the atomic mass unit definition to convert between grams and atoms.
    • Link between mass of a sample and the count of atoms via Avogadro's number.
Unit 1, Question 4
  • Problem: Which of the following statements is true?- A. Halogens tend to lose one electron when they form ions. - B. Calcium is a transition metal. - C. Electrons make up most of the mass of an atom. - D. Neutrons and electrons are found in the nucleus of an atom. - E. A neutral atom contains the same number of protons and electrons.
  • Answer reasoning:
    • Halogens gain one electron, not lose (A false).
    • Calcium is an alkaline earth metal, not a transition metal (B false).
    • Atoms' mass is mostly due to protons and neutrons; electrons contribute negligibly (C false).
    • Neutrons are in the nucleus, but electrons are outside the nucleus (D false).
    • In a neutral atom, number of protons equals number of electrons (E true).
  • Answer: E.
  • concept:
    • Basic atomic structure: nucleus (protons + neutrons) vs electrons in electron cloud; charge balance in neutral atoms.
Unit 1, Question 6
  • Problem (Another name for Group 1A elements):- A Halogens - B Alkali metals - C Noble gases - D Alkaline earth metals - E Chalcogens
  • Answer: B (Alkali metals).
  • concept:
    • Group 1A elements are the alkali metals (Li, Na, K, Rb, Cs, Fr).
Unit 1, Question 7
  • Problem: The number of protons, neutrons, and electrons, respectively, in a neutral atom of 121Sb^{121}\text{Sb} is:- A. 51 protons, 51 neutrons, and 51 electrons - B. 70 protons, 51 neutrons, and 70 electrons - C. 51 protons, 121 neutrons, and 51 electrons - D. 51 protons, 70 neutrons, and 51 electrons - E. 121 protons, 70 neutrons, and 121 electrons
  • Facts:
    • Antimony (Sb) has atomic number Z = 51 (51 protons).
    • Mass number A = 121, so neutrons N = A − Z = 121 − 51 = 70.
    • Neutral atom has same number of electrons as protons: E = 51.
  • Answer: D (51 protons, 70 neutrons, 51 electrons).
  • concepts:
    • Distinguish between mass number and atomic number; neutral atoms have equal proton and electron counts.
Unit 1, Question 8
  • Problem: A halogen, an alkali metal, a noble gas, and an alkaline earth metal, in that order, are:- A. fluorine, sodium, selenium, beryllium - B. iodine, calcium, neon, potassium - C. bromine, beryllium, helium, oxygen - D. xenon, potassium, bromine, sulfur - E. chlorine, lithium, argon, magnesium
  • Answer: E (chlorine = halogen; lithium = alkali metal; argon = noble gas; magnesium = alkaline earth metal).
  • concepts:
    • Correct example elements for each category: halogen (F, Cl, Br, I), alkali metal (Li, Na, K, …), noble gas (He, Ne, Ar, …), alkaline earth metal (Be, Mg, Ca, …).
Unit 1, Question 8 (duplicate)
  • The same category question as above; Answer: E.
Unit 1, Question 9
  • Problem: Which of the following combinations are incorrectly paired?- 1. Mg, alkaline earth metal - 2. Cr, alkali metal - 3. Si, metalloid - 4. Xe, noble gas - 5. Al, transition metal
  • Answer reasoning:
    • 1. Mg is correctly an alkaline earth metal.
    • 2. Cr is a transition metal, not an alkali metal (incorrect).
    • 3. Si is a metalloid (correct).
    • 4. Xe is a noble gas (correct).
    • 5. Al is not a transition metal (it is a post-transition metal); incorrect.
  • Incorrect pairings: 2 and 5.
  • Answer: D (2 and 5).
  • concepts:
    • Understanding how elements are classified: alkali metals, alkaline earth metals, transition metals, metalloids, noble gases, etc.
Unit 1, Question 10
  • Problem: Which statement about ions is true?- A. An ion is the result of an atom gaining a neutron. - B. A positively charged ion results from the gain of a proton. - C. A negatively charged ion results from the loss of a proton. - D. The gain or loss of an electron by an atom will result in an ion. - E. None of the above.
  • Answer: D (ions arise from gain or loss of electrons; protons/electrons determine charge, neutrons do not affect charge).
  • concepts:
    • Ions result from electron transfer, not changes in neutron number; mass is only slightly affected by electron changes.
Unit 1, Question 12
  • Problem: Antimony (Sb) has two naturally occurring isotopes, 121Sb^{121}\text{Sb} and 123Sb^{123}\text{Sb}. What is the naturally occurring abundance of 121Sb^{121}\text{Sb}?- A. 98% B. 62% C. 50% D. 38% E. 26%
  • Answer reasoning:
    • Real-world isotopic abundances for Sb are roughly: 121Sb≈57%^{121}\text{Sb} \approx 57\% and 123Sb≈43%^{123}\text{Sb} \approx 43\%.
    • Among the given options, the closest value to the known abundance is 62% (option B).
  • Answer: B (approximately 62%; the precise value is around 57%).
  • concepts:
    • Natural abundances determine the weighted average atomic mass; small deviations in practice reflect experimental data or rounding in problems.
Quick reference: Key formulas and conversions used
  • Length/volume conversions:
    • 1 cm3=1 mL1\text{ cm}^3 = 1\text{ mL}
    • 1 in3=16.387 cm31\text{ in}^3 = 16.387\text{ cm}^3
  • Density and mass:
    • ρ=mV\rho = \dfrac{m}{V}
  • Temperature conversions:
    • C=K−273.15C = K - 273.15
    • F=C×95+32F = C \times \dfrac{9}{5} + 32
  • Unit conversions (mass):
    • 1 kg=1000 g1\text{ kg} = 1000\text{ g}; 1 g=1000 mg1\text{ g} = 1000\text{ mg}; 1 mg=1000 μg1\text{ mg} = 1000\text{ }\mu\text{g}; 1 ng=10−9 g1\text{ ng} = 10^{-9}\text{ g}; 1 g=106 μg1\text{ g} = 10^{6}\text{ }\mu\text{g}; 1 g=103 mg1\text{ g} = 10^{3}\text{ mg}
  • Atomic mass unit (amu) relationships:
    • 1 amu=1 g6.022×10231\text{ amu} = \dfrac{1\text{ g}}{6.022\times 10^{23}}
    • Atomic mass in g per atom: matom=A amu×(1 g6.022×1023 amu)m_{\text{atom}} = A\,\text{amu} \times \left(\dfrac{1\text{ g}}{6.022\times 10^{23}\text{ amu}}\right) where AA is the mass number.
  • Isotopes and atoms:
    • For neutral atoms, number of protons == number of electrons == atomic number ZZ; neutrons N=A−ZN = A - Z.
  • Notes on significant figures:
    • Addition/subtraction: round to the least precise decimal place among terms.