Chemistry Homework Comprehensive Grading Report

The question was left unanswered. The chemical symbol Pb originates from the Latin name plumbum, which designates the element Lead (atomic number 82).

The question was left blank. The symbol Au comes from the Latin word aurum, meaning Gold (atomic number 79).

In introductory chemistry curriculum contexts, pure table salt is sodium chloride (NaCl). It is a chemical compound consisting of sodium (Na⁺) and chlorine (Cl⁻) ions chemically bonded in a fixed, constant 1:1 stoichiometric ratio. It is classified as a Compound (C). (Note: On Sheet 2, the student correctly classified table salt as C, indicating earlier hesitation).

Stainless steel is an alloy—a solid solution primarily made of iron, carbon, and chromium. Because alloys are physical blends of different elements that are not chemically combined in fixed stoichiometric proportions, stainless steel is a homogeneous mixture (M), not an element.

Omission. The chemical symbol for tin is Sn (from Latin stannum).

Omission. The chemical symbol for sodium is Na (from Latin natrium).

Omission. The chemical symbol for silver is Ag (from Latin argentum).

Omission. The chemical symbol for carbon is C.

Statement (a) is true because by definition in chemistry, all solutions are homogeneous mixtures. Statement (d) is false because a compound is composed of two or more different elements chemically united. Therefore, a compound can exist as molecules (e.g., H₂O) or ionic crystal lattices, but never as individual atoms. An atom always represents a single element.

Error on 4b (tap water): While pure distilled water is a compound (H₂O), tap water is not pure; it contains dissolved minerals (calcium, magnesium, chloride), disinfectants (chlorine), and gases. Therefore, tap water is a homogeneous Mixture (M).

The student responded with "e" (implying element). This is fundamentally incorrect. If a liquid evaporates completely without residue, it could be:

• A pure liquid element (e.g., liquid bromine, Br₂).
• A pure liquid compound (e.g., pure water, H₂O, or pure ethanol, C₂H₅OH).
• A homogeneous mixture of volatile liquids (e.g., a mixture of ethanol and water).

Because volatile compounds and volatile liquid mixtures also evaporate completely without leaving a solid residue, evaporation alone cannot distinguish between them. The correct answer is No, you cannot determine it.

Error on 7b (magnesium): The student wrote lowercase mg. In chemical notation (IUPAC standards), element symbols MUST start with a capitalized first letter followed by a lowercase second letter: Mg. The all-lowercase symbol mg is universally recognized as the abbreviation for milligrams (a unit of mass), not an element symbol.

Boiling (a) is a physical phase change, and pouring (d) is mechanical transport. Adding oxygen to liquid gasoline without ignition (b) merely dissolves oxygen gas in the hydrocarbon solvent, which is a physical mixture. Gasoline burns (c) is combustion: hydrocarbons chemically react with oxygen at high temperatures to form new chemical substances (CO₂ and H₂O) and release energy. Thus, c is the only chemical change.

Error on 2d (Soap mixed with water): Dissolving soap in water is a Physical change (P). The soap molecules disperse and aggregate into spherical micelles through intermolecular forces without forming or breaking covalent chemical bonds. Soap molecules retain their chemical identity and can be recovered unaltered by evaporating the water.

Potassium (K) and iron (Fe) are metallic conductors. Helium (He) is a noble gas insulator. Silicon (Si) is a metalloid with an intermediate energy bandgap (~1.1 eV) between conductors and insulators, which defines a semiconductor. Correct choice is c.

Critical Errors on 4a and 4c:
• 4a: Distillation is a physical separation method based on boiling point differences. Separating the orange liquid into a yellow liquid and red solid without breaking chemical bonds means the orange liquid was a Mixture (M), and distillation is a Physical change (P). The student wrote C, C.
• 4b: Decomposition into elements (metal and halogen gas) breaks chemical bonds, so the starting solid was a Compound (C) and the process is a Chemical change (C) (Student: correct).
• 4c: Adding sugar to tea forms a solution, which is a Mixture (M) (Student: correct). However, dissolving sugar is a Physical change (P), not a chemical change (Student incorrectly wrote C).

Conceptual Misidentification of Reactants:
Let us calculate the mass change ($\Delta m = m_{\text{after}} - m_{\text{before}}$) for each substance:
• Substance A: 0 - 18 = -18 g (Mass decreased $\rightarrow$ Reactant)
• Substance C: 2 - 2 = 0 g (Mass unchanged $\rightarrow$ Catalyst or spectator)
• Substance D: 12 - 32 = -20 g (Mass decreased $\rightarrow$ Reactant)
• By Conservation of Mass, the mass increase for substance B is: $\Delta m_B = -(-18 + 0 - 20) = \mathbf{+38\text{ g}}$ (Mass increased $\rightarrow$ Product). Final mass of B = 1 + 38 = 39 g.
The two reactants that reacted are A and D (NOT A and B). While the student computed the numerical ratio 18:20 = 9:10 correctly, they mistakenly labeled it A:B. Substance B is the reaction product! The correct ratio is A : D = 9 : 10.

Significant Figure Rule Violation: The explicit terminal decimal point in 30400000. indicates that all zeros are significant digits (8 significant figures). When converting to scientific notation, the number of significant figures must be conserved. Writing 3.04 × 10⁷ discards 5 significant figures. The correct form is 3.0400000 × 10⁷.

Significant Figure Rule Violation: Trailing zeros to the right of the decimal point in a measured quantity are significant. Therefore, 0.000045600 has 5 significant figures (4, 5, 6, and both trailing zeros). Writing 4.56 × 10⁻⁵ drops two significant figures. The correct scientific notation is 4.5600 × 10⁻⁵.

The question was left unanswered. The governing law stating that mass is neither created nor destroyed during chemical reactions is the Law of Conservation of Mass (formulated by Antoine Lavoisier).

Exact Number Concept: "Metersticks" is a countable discrete physical object, not an inexact quantity read from a graduated scale. Counted numbers and defined quantities have no uncertainty and possess an infinite number of significant figures (∞). The student treated it as a measured quantity with 2 sig figs.

Major Misunderstanding of Instructions:
The student converted numbers into scientific notation but completely failed to perform the required rounding to three significant figures for items a, b, d, e, and f!
• In a (87.073 m), the student wrote 8.7073 × 10¹ (kept 5 sig figs). It must be 8.71 × 10¹ m.
• In b (4.3621 × 10⁸ m), student wrote 4.3621 × 10⁸ (kept 5 sig figs). Must be 4.36 × 10⁸ m.
• In d (9009 kg), student wrote 9.009 × 10³ (kept 4 sig figs). Must be 9.01 × 10³ kg.
• In e (1.7777 × 10⁻³ m), student wrote 1.7777 × 10⁻³ (kept 5 sig figs). Must be 1.78 × 10⁻³ m.
• In f (629.55 m), student wrote 6.2955 × 10² (kept 5 sig figs). Must round up to 6.30 × 10² m.
• In g (7700. m³), student wrote 7.7 × 10³ (under-rounded to 2 sig figs). 3 sig figs requires 7.70 × 10³ m³.
• In i (dozen = 12), student wrote 1.2 × 10¹ (only 2 sig figs). 3 sig figs requires 1.20 × 10¹.

Items a, b, c, d, and f are correctly calculated.
• Error on 3e [(3.40 - 2.355) × 6.7209]: First, subtraction: 3.40 - 2.355 = 1.045. Since 3.40 has 2 decimal places, this intermediate difference is significant to the hundredths place (3 sig figs). Multiplying while carrying full precision: 1.045 × 6.7209 = 7.02334 $\rightarrow$ rounded to 3 sig figs = 7.02. The student got 6.99 because they prematurely truncated 1.045 to 1.04 before multiplying (1.04 × 6.7209 = 6.9897 $\rightarrow$ 6.99). Truncating intermediate steps leads to compounded rounding errors.
• Error on 3g [313.0 - 1.2 × 10³]: 1.2 × 10³ = 1200 has its last significant digit in the hundreds place (±100). In addition/subtraction, the result is limited by the least precise decimal position (the hundreds place). 313.0 - 1200 = -887. Rounding -887 to the hundreds place yields -900 or -9 × 10². The student wrote -887 without applying significant figure rounding.

Misconception regarding Scale Reading: Reading an analogue scale requires human visual interpolation between graduation lines. Fluctuations in eye level (parallax) and subjective estimation between markings vary randomly above and below the line from trial to trial. Therefore, scale reading is a classic source of random error. The statement is True (T).