During a titration lab, a student splashes concentrated sulfuric acid onto her forearm. The nearest sink is across the room, but a drench hose and safety shower are 2 meters away. What should the teacher do FIRST?

Flush the arm with running water for at least 15 minutes, removing soaked clothing

Apply a sodium bicarbonate paste to neutralize the acid before rinsing the arm

Blot the acid off with dry paper towels and then cover the area with burn ointment

Send the student to the nurse with a partner while the teacher contains the spill

Experiment [A] (M) [B] (M) Initial rate (M/s) 1 0.10 0.10 2.0 × 10⁻³ 2 0.20 0.10 8.0 × 10⁻³ 3 0.10 0.30 6.0 × 10⁻³ Students collect the initial-rate data shown for the reaction A + B → C at constant temperature. Which rate law and rate constant are consistent with the data?

rate = k[A]²[B], k = 2.0 M⁻²·s⁻¹

rate = k[A][B], k = 0.20 M⁻¹·s⁻¹

rate = k[A]²[B]², k = 20 M⁻³·s⁻¹

rate = k[A][B]², k = 2.0 M⁻²·s⁻¹

A teacher receives a new oxidizing chemical and wants to know which glove material and type of eye and face protection the manufacturer recommends. Which section of the Safety Data Sheet should she consult?

Section 8, exposure controls and personal protection

Section 2, hazard classes, pictograms, and signal word

Section 7, precautions for safe handling and storage

Section 11, toxicological and health-effect information

A student builds a galvanic cell from a zinc electrode in 1.0 M Zn²⁺ and a copper electrode in 1.0 M Cu²⁺. Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, which statement about the cell is correct?

Zinc is the anode, E°cell = +1.10 V, and electrons flow from Zn to Cu

Copper is the anode, E°cell = +1.10 V, and electrons flow from Cu to Zn

Zinc is the anode, E°cell = +0.42 V, and electrons flow from Zn to Cu

Zinc is the cathode, E°cell = −0.42 V, and electrons flow from Cu to Zn

A student examines an igneous rock containing large, well-formed feldspar crystals scattered through a very fine-grained gray matrix. Which cooling history BEST explains this texture?

Slow cooling at depth, followed by eruption or rise and then rapid cooling

Rapid cooling at the surface, followed by slow cooling as lava flowed downhill

Uniform slow cooling deep underground, with feldspar crystallizing first

Rapid quenching in water, which formed crystals only around gas bubbles

TExES Science 7-12 (236)

Free TExES Science 7-12 (236) Practice Questions

10 sample questions for the TExES Science 7-12 (236), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.

Sections covered on the exam: Scientific Inquiry and Processes, Physics, Chemistry, Cell Structure and Processes, Heredity and Evolution of Life, Diversity of Life, Interdependence of Life and Environmental Systems, Earth's History and Structure and Function of Earth Systems, Solar System and the Universe, Science Learning, Instruction and Assessment.

Question 1 · Scientific Inquiry and Processes
During a titration lab, a student splashes concentrated sulfuric acid onto her forearm. The nearest sink is across the room, but a drench hose and safety shower are 2 meters away. What should the teacher do FIRST?
- A.Apply a sodium bicarbonate paste to neutralize the acid before rinsing the arm
- B.Flush the arm with running water for at least 15 minutes, removing soaked clothing
- C.Blot the acid off with dry paper towels and then cover the area with burn ointment
- D.Send the student to the nurse with a partner while the teacher contains the spill
Show answer and explanation
Answer: B. Flush the arm with running water for at least 15 minutes, removing soaked clothing
Immediate, prolonged flushing with water (at least 15 minutes) while removing contaminated clothing is the standard first response to a corrosive splash on skin. Neutralizing on the skin with a base wastes time and the exothermic reaction can deepen the injury, and delaying rinsing to walk to the nurse lets the acid keep damaging tissue.
Question 2 · Physics
A 12 V battery is connected to a 4.0 Ω resistor in series with a parallel pair of resistors, 6.0 Ω and 12 Ω. What is the current through the 6.0 Ω resistor?
- A.0.50 A
- B.1.5 A
- C.2.0 A
- D.1.0 A
Show answer and explanation
Answer: D. 1.0 A
The parallel pair has an equivalent resistance of (6.0 × 12)/(6.0 + 12) = 4.0 Ω, so the total is 8.0 Ω and the battery current is 1.5 A. The voltage across the parallel pair is 1.5 A × 4.0 Ω = 6.0 V, so the 6.0 Ω branch carries 1.0 A. The 2.0 A answer wrongly places the full 12 V across the 6.0 Ω resistor.
Question 3 · Chemistry
Experiment [A] (M) [B] (M) Initial rate (M/s)
1 0.10 0.10 2.0 × 10⁻³
2 0.20 0.10 8.0 × 10⁻³
3 0.10 0.30 6.0 × 10⁻³
Students collect the initial-rate data shown for the reaction A + B → C at constant temperature. Which rate law and rate constant are consistent with the data?
- A.rate = k[A][B], k = 0.20 M⁻¹·s⁻¹
- B.rate = k[A]²[B], k = 2.0 M⁻²·s⁻¹
- C.rate = k[A]²[B]², k = 20 M⁻³·s⁻¹
- D.rate = k[A][B]², k = 2.0 M⁻²·s⁻¹
Show answer and explanation
Answer: B. rate = k[A]²[B], k = 2.0 M⁻²·s⁻¹
Doubling [A] at constant [B] (experiments 1 and 2) quadruples the rate, so the reaction is second order in A; tripling [B] at constant [A] (experiments 1 and 3) triples the rate, so it is first order in B. Then k = (2.0 × 10⁻³ M/s)/[(0.10 M)²(0.10 M)] = 2.0 M⁻²·s⁻¹. Writing rate = k[A][B] from the balanced equation's coefficients is a common error, since reaction orders must be determined experimentally.
Question 4 · Heredity and Evolution of Life
In a plant, genes A, B, and C assort independently, and each shows complete dominance. Two plants with genotype AaBbCc are crossed. What fraction of the offspring will show the dominant A phenotype, the recessive b phenotype, and the dominant C phenotype?
- A.3/64
- B.9/64
- C.27/64
- D.3/16
Show answer and explanation
Answer: B. 9/64
Each gene is treated as its own monohybrid cross: P(A_) = 3/4, P(bb) = 1/4, and P(C_) = 3/4. Multiplying the independent probabilities gives 3/4 × 1/4 × 3/4 = 9/64. The value 27/64 is the probability of all three dominant phenotypes, which results from using 3/4 for the b gene instead of 1/4.
Question 5 · Earth's History and Structure and Function of Earth Systems
Test Result
Luster Metallic
Color Brass yellow
Streak Greenish-black
Glass plate Sample scratches glass
Crystal habit Cubic crystals
A student tests an unknown mineral sample and records the results shown. Which mineral is the sample MOST likely to be?
- A.Pyrite
- B.Native gold
- C.Chalcopyrite
- D.Native copper
Show answer and explanation
Answer: A. Pyrite
Pyrite is brass-yellow and metallic, leaves a greenish-black streak, and has a hardness of 6–6.5, so it scratches glass (about 5.5). Gold has a golden-yellow streak and a hardness under 3, and chalcopyrite, though it also has a dark streak, is too soft (3.5–4) to scratch glass.
Question 6 · Scientific Inquiry and Processes
A teacher receives a new oxidizing chemical and wants to know which glove material and type of eye and face protection the manufacturer recommends. Which section of the Safety Data Sheet should she consult?
- A.Section 8, exposure controls and personal protection
- B.Section 2, hazard classes, pictograms, and signal word
- C.Section 7, precautions for safe handling and storage
- D.Section 11, toxicological and health-effect information
Show answer and explanation
Answer: A. Section 8, exposure controls and personal protection
Under the Globally Harmonized System format, Section 8 lists exposure limits, engineering controls, and specific personal protective equipment such as glove material and eye protection. Section 2 summarizes hazard classes and pictograms, Section 7 covers safe handling and storage conditions, and Section 11 describes health effects and toxicity data.
Question 7 · Physics
A student plucks a guitar string alongside a 440 Hz tuning fork and hears 4 beats per second. After slightly tightening the string, she hears 2 beats per second. What was the string's original frequency?
- A.438 Hz
- B.436 Hz
- C.442 Hz
- D.444 Hz
Show answer and explanation
Answer: B. 436 Hz
Four beats means the string was at 436 Hz or 444 Hz. Tightening raises the frequency, and the beats dropped to 2 per second, so the string moved closer to 440 Hz from below and began at 436 Hz. The 444 Hz answer would have produced more beats after tightening, not fewer.
Question 8 · Chemistry
A student builds a galvanic cell from a zinc electrode in 1.0 M Zn²⁺ and a copper electrode in 1.0 M Cu²⁺. Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, which statement about the cell is correct?
- A.Copper is the anode, E°cell = +1.10 V, and electrons flow from Cu to Zn
- B.Zinc is the anode, E°cell = +0.42 V, and electrons flow from Zn to Cu
- C.Zinc is the cathode, E°cell = −0.42 V, and electrons flow from Cu to Zn
- D.Zinc is the anode, E°cell = +1.10 V, and electrons flow from Zn to Cu
Show answer and explanation
Answer: D. Zinc is the anode, E°cell = +1.10 V, and electrons flow from Zn to Cu
Cu²⁺ has the higher reduction potential, so it is reduced at the copper cathode while zinc is oxidized at the anode, and electrons travel through the wire from Zn to Cu. E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V. The value 0.42 V comes from adding the two potentials as written instead of subtracting the anode potential.
Question 9 · Heredity and Evolution of Life
Cystic fibrosis is an autosomal recessive disorder. Two unaffected parents have one affected son and one unaffected daughter. The daughter later has a child with a man who is a known carrier. What is the probability that their first child will have cystic fibrosis?
- A.1/4
- B.1/8
- C.1/2
- D.1/6
Show answer and explanation
Answer: D. 1/6
Both parents must be carriers (Ff), so their unaffected daughter is either FF or Ff in a 1 : 2 ratio, giving her a 2/3 chance of being a carrier. If she is a carrier, a cross with a carrier gives a 1/4 chance of an affected child, so the overall probability is 2/3 × 1/4 = 1/6. Using 1/2 for her carrier chance gives 1/8, which forgets that her unaffected phenotype rules out the ff genotype.
Question 10 · Earth's History and Structure and Function of Earth Systems
A student examines an igneous rock containing large, well-formed feldspar crystals scattered through a very fine-grained gray matrix. Which cooling history BEST explains this texture?
- A.Rapid cooling at the surface, followed by slow cooling as lava flowed downhill
- B.Uniform slow cooling deep underground, with feldspar crystallizing first
- C.Rapid quenching in water, which formed crystals only around gas bubbles
- D.Slow cooling at depth, followed by eruption or rise and then rapid cooling
Show answer and explanation
Answer: D. Slow cooling at depth, followed by eruption or rise and then rapid cooling
A porphyritic texture records two stages: large crystals (phenocrysts) grew slowly in a magma chamber, and the remaining melt then cooled quickly after the magma rose or erupted, forming the fine matrix. Uniform slow cooling would produce only coarse crystals, and quenching in water produces glass rather than large phenocrysts.

Experiment	[A] (M)	[B] (M)	Initial rate (M/s)
1	0.10	0.10	2.0 × 10⁻³
2	0.20	0.10	8.0 × 10⁻³
3	0.10	0.30	6.0 × 10⁻³

Test	Result
Luster	Metallic
Color	Brass yellow
Streak	Greenish-black
Glass plate	Sample scratches glass
Crystal habit	Cubic crystals

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