Chemistry — Semester B Practice

Question 1 of 10

TEKS 12A-12C Easy Word

Which type of radioactive emission has NO mass and NO charge, consisting of high-energy electromagnetic radiation?

A Beta (β) particle

B Gamma (γ) ray

C Alpha (α) particle

D Neutron

Explanation

📌 Gamma rays are pure electromagnetic radiation (like light, but much higher energy) — no mass, no charge. Alpha particles are helium nuclei (mass 4, charge +2). Beta particles are high-speed electrons (mass ~0, charge −1). Neutrons have mass 1 and no charge but are usually not classified with the three main decay types. Gamma rays often accompany alpha or beta decay as the nucleus releases excess energy.

Question 2 of 10

TEKS 12A-12C Medium Calc Word

Carbon-14 has a half-life of about 5,730 years. If a fossil sample initially contained 100. g of carbon-14, approximately how many grams remain after 17,190 years?

A 50.0 g

B 12.5 g

C 6.25 g

D 25.0 g

Explanation

📌 17,190 years ÷ 5,730 years per half-life = 3 half-lives. After each half-life, the remaining amount halves. So 100 g → 50 g (1st) → 25 g (2nd) → 12.5 g (3rd). General formula: remaining = initial × (1/2)ⁿ, where n is the number of half-lives elapsed. This is the basis of radiocarbon dating for fossils up to ~50,000 years old.

Question 3 of 10

TEKS 10A-10D Easy Word

Which statement about an ideal gas is TRUE according to the kinetic molecular theory?

A Ideal gas particles move randomly and collide elastically

B Ideal gas particles have significant volume

C Ideal gas particles slow down as temperature rises

D Ideal gas particles attract each other strongly

Explanation

📌 Kinetic Molecular Theory assumes ideal gas particles (1) have negligible volume relative to the container, (2) experience no intermolecular forces, (3) move in random straight-line motion, (4) undergo elastic collisions (no energy lost), and (5) have average kinetic energy proportional to absolute temperature. Real gases deviate at high pressure and low temperature.

Question 4 of 10

TEKS 10A-10D Medium Calc Word

A sealed flexible container holds 4.0 L of helium gas at 200. K and 1.0 atm. The container is warmed until the gas temperature reaches 400. K, while the pressure remains at 1.0 atm. What is the new volume of the gas?

A 8.0 L

B 4.0 L

C 2.0 L

D 16 L

Explanation

📌 Charles's Law (constant pressure): V₁/T₁ = V₂/T₂. Rearranging: V₂ = V₁ × (T₂/T₁) = 4.0 L × (400 K / 200 K) = 8.0 L. Doubling the absolute temperature (in Kelvin) doubles the volume at constant pressure. Always use Kelvin — converting from °C would give wildly wrong answers.

Question 5 of 10

TEKS 11A-11D Easy

When ammonium nitrate (NH₄NO₃) is dissolved in water inside a beaker, the temperature of the solution drops noticeably and the beaker feels cold to the touch. Based on this observation, the dissolution of NH₄NO₃ is best classified as:

A Combustion

B Neutralization

C Exothermic

D Endothermic

Explanation

📌 When the surroundings (the beaker, your hand) get colder, the system absorbed heat from them → endothermic. The energy went INTO breaking apart the ionic lattice of NH₄NO₃ during dissolution. This is the principle behind instant cold packs — they use NH₄NO₃ dissolving in water for emergency cooling.

Question 6 of 10

TEKS 11A-11D Medium Calc Word

How much heat (in joules) is required to warm 50.0 g of water from 20.0 °C to 80.0 °C? (Specific heat of water = 4.18 J/(g·°C))

A 12,540 J

B 8,360 J

C 3,344 J

D 250.8 J

Explanation

📌 q = m × c × ΔT, where q is heat, m is mass, c is specific heat, and ΔT is temperature change. q = 50.0 g × 4.18 J/(g·°C) × (80.0 − 20.0) °C = 50.0 × 4.18 × 60.0 = 12,540 J. Always compute ΔT as final minus initial; positive ΔT means heat absorbed, negative means heat released.

Question 7 of 10

TEKS 9A-9D Easy Calc Word

A water sample contains 5 mg of dissolved chloride ions per liter of solution. This concentration is equivalent to:

A 50 ppm

B 500 ppm

C 0.5 ppm

D 5 ppm

Explanation

📌 For dilute aqueous solutions (density ≈ 1 g/mL), parts per million ≈ mg per kg of solvent ≈ mg per liter. So 5 mg/L ≈ 5 ppm. Pattern: ppm = (mass of solute in g ÷ mass of solution in g) × 10⁶. Used for trace concentrations like water-quality testing (e.g., 'fluoride at 1 ppm in tap water').

Question 8 of 10

TEKS 8A-8F Medium Calc Word

A student measures out 88.0 g of carbon dioxide (CO₂) for a stoichiometry experiment. How many moles of CO₂ does the sample contain? (Molar mass of CO₂ = 44.0 g/mol)

A 1.00 mol

B 2.00 mol

C 0.500 mol

D 3,872 mol

Explanation

📌 Moles = mass ÷ molar mass = 88.0 g ÷ 44.0 g/mol = 2.00 mol. The mole bridge: grams ↔ moles ↔ particles. To go from grams to moles, ALWAYS divide by molar mass. Multiplying (the common error in distractor D) gives an unreasonable answer.

Question 9 of 10

TEKS 9A-9D Medium Calc Word

What is the molarity of a solution that contains 0.500 mol of NaCl dissolved in enough water to make 250 mL of solution?

A 0.125 M

B 0.500 M

C 125 M

D 2.00 M

Explanation

📌 Molarity (M) = moles of solute ÷ liters of solution = 0.500 mol ÷ 0.250 L = 2.00 M. Common error: forgetting to convert mL to L. 250 mL = 0.250 L. Molarity tells you concentration; don't confuse with molality (moles per kg solvent).

Question 10 of 10

TEKS 8A-8F Easy Calc Word

How many moles of water are present in 36.0 g of H₂O? (Molar mass of H₂O = 18.0 g/mol)

A 2.00 mol

B 0.500 mol

C 1.00 mol

D 18.0 mol

Explanation

📌 Moles = mass ÷ molar mass = 36.0 g ÷ 18.0 g/mol = 2.00 mol. Repeat after me: grams → moles is ALWAYS division by molar mass. Multiplying produces an unreasonably huge answer (option D shows that error).

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