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USAC Ciencias Químicas y Farmacia - fundamentos científicos study materials
USAC Ciencias Químicas y Farmacia - fundamentos científicos study resources

USAC Ciencias Químicas y Farmacia - fundamentos científicos practice questions

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Question 1Biología y razonamiento experimental

In chemical and clinical laboratory testing, what is the basis of the Benedict's or Fehling's test for reducing sugars?

  1. Iodine molecules insert into amylose helical coils to produce an intense blue-black complex
  2. Ninhydrin reacts with primary amino groups to form Ruhemann's purple
  3. Free aldehyde or hemiacetal groups in open-chain monosaccharides reduce cupric ions (Cu²⁺) to a brick-red precipitate of cuprous oxide (Cu₂O) Correct answer.
  4. Lipids react with osmium tetroxide to produce black metallic deposits

Explanation

Benedict's and Fehling's reagents contain alkaline copper(II) sulfate (Cu²⁺). Reducing sugars (all monosaccharides and disaccharides with a free hemiacetal group, such as maltose and lactose) can mutarotate to an open-chain form possessing a free aldehyde or alpha-hydroxyketone that reduces blue Cu²⁺ to an insoluble brick-red precipitate of copper(I) oxide (Cu₂O).

Question 2Biología y razonamiento experimental

Why do unsaturated fatty acids with cis double bonds (such as oleic acid) have lower melting points than saturated fatty acids of equal carbon chain length, remaining liquid oils at room temperature?

  1. Unsaturated fatty acids form strong ionic bonds with surrounding water molecules
  2. Double bonds increase the overall molecular weight, causing rapid thermal breakdown
  3. Unsaturated fatty acids lack a terminal carboxylic acid group
  4. Cis double bonds introduce rigid kinks in the hydrocarbon tail that prevent tight crystalline packing and reduce intermolecular van der Waals forces Correct answer.

Explanation

Saturated fatty acids have straight, flexible hydrocarbon chains that pack tightly together in an orderly crystalline lattice with extensive London dispersion forces, giving them high melting points (solid fats). Cis double bonds introduce rigid bends (kinks) in the chain that disrupt regular packing, decreasing dispersion attractions and dramatically lowering the melting point (liquid oils).

Question 3Biología y razonamiento experimental

Which level of protein structural organization remains intact when a globular enzyme is completely denatured by exposure to high temperature or extreme pH?

  1. Quaternary structure
  2. Primary structure Correct answer.
  3. Tertiary structure
  4. Secondary structure

Explanation

Protein denaturation disrupts non-covalent interactions (hydrogen bonds, ionic attractions, hydrophobic interactions) and disulfide bridges that stabilize secondary, tertiary, and quaternary conformations. Denaturation unfolds the protein into a random coil, but does not cleave the covalent peptide bonds of the primary structure (which requires enzymatic or chemical hydrolysis).

Question 4Biología y razonamiento experimental

What chemical distinction separates the nucleotide pentose sugar of RNA from that of DNA?

  1. RNA contains D-ribose with a hydroxyl group (-OH) on carbon 2', whereas DNA contains 2'-deoxyribose with a hydrogen atom (-H) on carbon 2' Correct answer.
  2. DNA contains a hexose sugar, whereas RNA contains a pentose sugar
  3. RNA contains a branched aldopentose, whereas DNA contains a cyclic ketopentose
  4. DNA contains ribitol with an open-chain structure

Explanation

Ribonucleic acid (RNA) contains D-ribose, which possesses a hydroxyl group (-OH) attached to carbon 2' (C2') of the furanose ring. Deoxyribonucleic acid (DNA) contains 2'-deoxy-D-ribose, which lacks this oxygen atom and has two hydrogen atoms (-H) attached to C2'. This lack of a 2'-OH group makes DNA significantly more resistant to alkaline chemical hydrolysis.

Question 5Biología y razonamiento experimental

Which of the following is NOT one of the three foundational tenets of classical Cell Theory (established by Schleiden, Schwann, and Virchow)?

  1. All living organisms are composed of one or more cells
  2. The cell is the most fundamental structural and functional unit of life
  3. All cells must possess a membrane-bound nucleus and mitochondria to be classified as alive Correct answer.
  4. All cells arise from pre-existing cells through cell division (Omnis cellula e cellula)

Explanation

The three classical postulates of cell theory are: 1) all organisms are made of cells, 2) the cell is the basic structural and functional unit of life, and 3) all cells arise from pre-existing living cells. Prokaryotic cells (bacteria and archaea) are fully living entities yet completely lack a membrane-bound nucleus and mitochondria.

Question 6Biología y razonamiento experimental

Which of the following structures is found in eukaryotic cells but is strictly absent in prokaryotic cells?

  1. Plasma membrane
  2. Ribosomes
  3. Double-stranded DNA genome
  4. Nuclear envelope with pore complexes Correct answer.

Explanation

Prokaryotes lack a nuclear envelope; their genetic material is localized in an unenclosed region called the nucleoid. Eukaryotic cells possess a distinct, double-membrane nuclear envelope punctuated by nuclear pore complexes that encloses linear chromosomes.

Question 7Biología y razonamiento experimental

Which combination of cellular organelles and structures is unique to typical plant cells compared to animal cells?

  1. Centrosomes with centrioles, lysosomes, and glycogen granules
  2. Cellulose cell wall, plastids (chloroplasts), and a large central vacuole Correct answer.
  3. Mitochondria, rough endoplasmic reticulum, and Golgi apparatus
  4. Flagella, plasma membrane, and 80S ribosomes

Explanation

Plant cells are distinguished by possessing a rigid cellulose cell wall outside the plasma membrane, plastids (including photosynthetic chloroplasts), and a prominent central vacuole providing turgor pressure. Animal cells lack cell walls and chloroplasts, and typically possess centrioles and lysosomes.

Question 8Biología y razonamiento experimental

What role does cholesterol play within the phospholipid bilayer of eukaryotic animal cell membranes?

  1. It acts as a bidirectional membrane fluidity buffer, preventing solidification at low temperatures and restraining excessive fluidity at high temperatures Correct answer.
  2. It serves as a primary proton-pumping enzyme for active transport
  3. It forms the hydrophilic glycocalyx coating on the external cell surface
  4. It binds to ribosomal subunits to initiate translation

Explanation

Cholesterol is an amphipathic sterol intercalated between membrane phospholipids. At high physiological temperatures, its rigid steroid ring system restricts phospholipid hydrocarbon movement, reducing membrane fluidity. At low temperatures, it prevents close hydrocarbon packing, preventing crystallization and membrane freezing.

Question 9Biología y razonamiento experimental

When plant tissue is placed in a hypotonic aqueous solution, water enters the cells by osmosis until internal hydrostatic pressure balances the osmotic gradient. What is this rigid cellular state called?

  1. Plasmolysis
  2. Crenation
  3. Turgor (turgidity) Correct answer.
  4. Hemolysis

Explanation

In a hypotonic solution, water moves into the plant cell vacuole by osmosis. The swelling protoplast pushes firmly against the inelastic cell wall, generating wall pressure (turgor pressure). This turgid state is the normal healthy physiological state that provides structural support to herbaceous plants.

Question 10Biología y razonamiento experimental

Which cellular transport process involves the non-specific uptake of extracellular fluid and dissolved micro-solutes via small invaginating endocytic vesicles ('cellular drinking')?

  1. Phagocytosis
  2. Receptor-mediated endocytosis
  3. Exocytosis
  4. Pinocytosis Correct answer.

Explanation

Pinocytosis (cellular drinking) is a form of fluid-phase endocytosis in which the plasma membrane invaginates continuously to bring small droplets of extracellular fluid and dissolved solutes into the cell within microvesicles.

Question 11Biología y razonamiento experimental

How does a competitive enzyme inhibitor affect the kinetic parameters V_max (maximum velocity) and K_m (Michaelis-Menten constant) in an enzyme-catalyzed reaction?

  1. V_max decreases and K_m remains unchanged
  2. V_max remains unchanged and apparent K_m increases Correct answer.
  3. Both V_max and K_m decrease proportionally
  4. Both V_max and K_m increase proportionally

Explanation

A competitive inhibitor reversibly binds directly to the enzyme's active site, competing with the substrate. At sufficiently high substrate concentrations, the substrate outcompetes the inhibitor, so the maximum reaction rate (V_max) remains unchanged. However, a higher substrate concentration is required to achieve half-maximal velocity, meaning the apparent Michaelis constant (K_m) increases.

Question 12Biología y razonamiento experimental

During the glycolytic pathway in the cytoplasm, what is the net yield of energy-rich molecules produced from the catabolism of one molecule of glucose into two molecules of pyruvate?

  1. 2 ATP and 2 NADH Correct answer.
  2. 4 ATP and 2 NADH
  3. 2 ATP and 4 FADH₂
  4. 36 ATP and 6 CO₂

Explanation

Glycolysis consumes 2 ATP in its preparatory (investment) phase and synthesizes 4 ATP via substrate-level phosphorylation in its payoff phase, resulting in a net yield of 2 ATP per glucose molecule. Additionally, 2 molecules of NAD⁺ are reduced to 2 NADH, and 2 molecules of pyruvate are formed.

Question 13Biología y razonamiento experimental

In eukaryotic aerobic cellular respiration, where does the Krebs (citric acid) cycle take place?

  1. Cytosol
  2. Mitochondrial intermembrane space
  3. Mitochondrial matrix Correct answer.
  4. Inner mitochondrial cristae membrane

Explanation

The Krebs cycle takes place in the soluble mitochondrial matrix, where soluble enzymes (such as citrate synthase and isocitrate dehydrogenase) catalyze the stepwise oxidation of acetyl-CoA. (The exception is succinate dehydrogenase, which is bound to the inner mitochondrial membrane as Complex II).

Question 14Biología y razonamiento experimental

According to Peter Mitchell's chemiosmotic hypothesis, how does the mitochondrial electron transport chain drive the synthesis of ATP by ATP synthase?

  1. Electrons directly transfer high-energy phosphate groups to ADP molecules
  2. Molecular oxygen binds to ATP synthase to trigger substrate-level phosphorylation
  3. NADH molecules cross the inner membrane to deliver ATP directly into the cytoplasm
  4. Pumping of protons (H⁺) from the matrix into the intermembrane space creates an electrochemical proton-motive force that powers ATP synthase as protons flow back Correct answer.

Explanation

As electrons flow through Complexes I, III, and IV of the electron transport chain, free energy is used to pump protons (H⁺) from the mitochondrial matrix into the intermembrane space, establishing an electrochemical proton gradient (proton-motive force). The passive return of protons down this gradient through the F₀F₁ ATP synthase complex drives the rotary catalysis of ADP and P_i into ATP.

Question 15Biología y razonamiento experimental

In human skeletal muscle cells under strenuous anaerobic exercise, what is the fate of pyruvate produced during glycolysis?

  1. It is converted to ethanol and CO₂ by alcohol dehydrogenase
  2. It is reduced to lactate by lactate dehydrogenase, regenerating NAD⁺ to sustain glycolysis Correct answer.
  3. It enters the chloroplast stroma for the Calvin cycle
  4. It is immediately oxidized to acetyl-CoA in the cytosol

Explanation

Under anaerobic conditions, the mitochondrial electron transport chain cannot reoxidize NADH to NAD⁺ due to the absence of terminal oxygen. To prevent glycolysis from stalling, lactate dehydrogenase reduces pyruvate to lactate while oxidizing NADH back to NAD⁺, allowing continued substrate-level ATP generation.

Question 16Biología y razonamiento experimental

During the light-dependent reactions of oxygenic photosynthesis in the chloroplast thylakoid membranes, what process is directly responsible for releasing molecular oxygen (O₂)?

  1. The enzymatic photolysis (splitting) of water molecules at the oxygen-evolving complex of Photosystem II Correct answer.
  2. The fixation of carbon dioxide by RuBisCO in the stroma
  3. The reduction of NADP⁺ to NADPH by ferredoxin-NADP⁺ reductase
  4. The photorespiratory oxidation of glycolate in peroxisomes

Explanation

At Photosystem II (P680), light absorption drives the oxidation of water via the manganese-containing oxygen-evolving complex: 2 H₂O → 4 H⁺ + 4 e⁻ + O₂. The extracted electrons replace those excited in P680, protons contribute to the thylakoid lumen proton gradient, and molecular oxygen (O₂) is released as a byproduct.

Question 17Biología y razonamiento experimental

Which enzyme catalyzes the initial carbon fixation step of the Calvin cycle, combining CO₂ with the five-carbon acceptor ribulose-1,5-bisphosphate (RuBP) to produce 3-phosphoglycerate (3-PGA)?

  1. Phosphofructokinase-1 (PFK-1)
  2. ATP synthase
  3. Ribulose-1,5-bisphosphate carboxylase/oxygenase (RuBisCO) Correct answer.
  4. Pyruvate kinase

Explanation

RuBisCO (ribulose-1,5-bisphosphate carboxylase/oxygenase) is the primary carboxylating enzyme located in the chloroplast stroma. It fixes atmospheric CO₂ onto ribulose-1,5-bisphosphate (RuBP) to form an unstable 6-carbon intermediate that immediately cleaves into two molecules of 3-phosphoglycerate (3-PGA).

Question 18Biología y razonamiento experimental

What is the primary genetic difference in outcomes between normal mitotic cell division and meiotic cell division?

  1. Mitosis produces four genetically unique haploid gametes, whereas meiosis produces two genetically identical diploid somatic cells
  2. Mitosis involves two consecutive nuclear divisions, whereas meiosis involves only one division
  3. Meiosis occurs exclusively in prokaryotes, whereas mitosis occurs in all living organisms
  4. Mitosis produces two genetically identical diploid daughter cells, whereas meiosis produces four genetically distinct haploid gametes Correct answer.

Explanation

Mitosis consists of a single equational division yielding two genetically identical diploid (2n) somatic daughter cells for growth and tissue repair. Meiosis consists of one round of DNA replication followed by two nuclear divisions (reductional Meiosis I and equational Meiosis II), producing four genetically non-identical haploid (n) gametes for sexual reproduction.

Question 19Biología y razonamiento experimental

During which specific sub-stage of Prophase I in meiosis does crossing over (genetic recombination via chiasmata formation between non-sister homologous chromatids) take place?

  1. Leptotene
  2. Pachytene Correct answer.
  3. Zygotene
  4. Diakinesis

Explanation

During prophase I of meiosis, homologous chromosomes pair closely (synapsis) in zygotene. In the subsequent pachytene stage, crossing over occurs: non-sister chromatids break and rejoin at chiasmata, exchanging reciprocal genetic segments and creating recombinant chromosomes with novel allele combinations.

Question 20Biología y razonamiento experimental

During DNA replication in living cells, which enzyme is responsible for catalyzing the formation of phosphodiester bonds to join discontinuous Okazaki fragments on the lagging strand?

  1. DNA Ligase Correct answer.
  2. DNA Helicase
  3. DNA Topoisomerase (Gyrase)
  4. RNA Primase

Explanation

DNA ligase catalyzes the ATP-dependent (or NAD⁺-dependent in bacteria) formation of a covalent phosphodiester bond between the 3'-hydroxyl end of one Okazaki fragment and the 5'-phosphate end of the adjacent fragment, sealing nicks in the sugar-phosphate backbone.

Question 21Biología y razonamiento experimental

In the process of transcription, RNA polymerase synthesizes a messenger RNA (mRNA) transcript by reading the DNA template strand in which direction, and synthesizing the new RNA strand in which direction?

  1. Reads template 5' to 3'; synthesizes RNA 3' to 5'
  2. Reads template 5' to 3'; synthesizes RNA 5' to 3'
  3. Reads template 3' to 5'; synthesizes RNA 5' to 3' Correct answer.
  4. Reads template 3' to 5'; synthesizes RNA 3' to 5'

Explanation

All nucleic acid polymerases (both DNA and RNA polymerases) add incoming nucleotide triphosphates exclusively to the free 3'-hydroxyl (-OH) group of the growing strand. Consequently, the synthesized RNA strand elongates in the 5' → 3' direction, while reading the antiparallel DNA template strand in the 3' → 5' direction.

Question 22Biología y razonamiento experimental

Which codon in the standard genetic code serves as the universal translation start (initiation) codon in eukaryotic cytoplasm, specifying the amino acid methionine?

  1. UAA
  2. UGA
  3. UAG
  4. AUG Correct answer.

Explanation

AUG is the universal start codon recognized by the initiator tRNA (tRNA_i^Met) on the small ribosomal subunit, establishing the open reading frame and coding for methionine in eukaryotes (and N-formylmethionine in prokaryotes).

Question 23Biología y razonamiento experimental

A single nucleotide substitution in a gene changes a codon from UAC (which encodes tyrosine) to UAA (a stop codon). This point mutation is classified as a:

  1. Silent mutation
  2. Nonsense mutation Correct answer.
  3. Missense mutation
  4. Frameshift mutation

Explanation

A nonsense mutation is a point mutation that converts a codon specifying an amino acid into a premature stop codon (UAA, UAG, or UGA). This results in premature termination of polypeptide translation, typically yielding a truncated and nonfunctional protein product.

Question 24Biología y razonamiento experimental

In a cross between two true-breeding snapdragon plants (Antirrhinum majus) where one parent has red flowers (CᴿCᴿ) and the other has white flowers (CᵂCᵂ), all F1 offspring produce pink flowers. When these F1 plants are self-pollinated, what phenotypic ratio is expected in the F2 generation?

  1. 1 red : 2 pink : 1 white Correct answer.
  2. 3 red : 1 white
  3. 9 red : 3 pink : 4 white
  4. All pink

Explanation

Flower color in snapdragons is a textbook example of incomplete dominance, where the heterozygote exhibits an intermediate phenotype. Selfing F1 pink plants (CᴿCᵂ × CᴿCᵂ) yields the genotypic ratio 1 CᴿCᴿ : 2 CᴿCᵂ : 1 CᵂCᵂ, which directly reflects the phenotypic ratio of 1 red (25%) : 2 pink (50%) : 1 white (25%).

Question 25Biología y razonamiento experimental

According to the Law of Independent Assortment (Mendel's Second Law), what phenotypic ratio is expected from a dihybrid cross of two heterozygous individuals (AaBb × AaBb), assuming complete dominance and unlinked autosomal loci?

  1. 1:2:1
  2. 3:1
  3. 9:3:3:1 Correct answer.
  4. 1:1:1:1

Explanation

When two unlinked gene pairs assort independently during gametogenesis, each monohybrid cross produces a 3:1 phenotypic ratio. Multiplying the independent probabilities [(3/4 A_ + 1/4 aa) × (3/4 B_ + 1/4 bb)] yields the classic Mendelian dihybrid phenotypic ratio of 9 (A_B_) : 3 (A_bb) : 3 (aaB_) : 1 (aabb).

Question 26Química fundamental

Which laboratory separation technique is most appropriate for separating a homogeneous mixture of two miscible liquids that have significantly different boiling points (e.g., water and ethanol)?

  1. Gravity filtration
  2. Centrifugation
  3. Liquid-liquid decantation
  4. Fractional or simple distillation Correct answer.

Explanation

Distillation exploits differences in the boiling points (volatilities) of miscible liquids in a homogeneous solution. The component with the lower boiling point vaporizes first, passes through a condenser, and is collected as distillate, leaving the less volatile component behind.

Question 27Química fundamental

What is the primary operational difference between an extensive property and an intensive property of a chemical system?

  1. Intensive properties change when a substance undergoes a physical phase change, whereas extensive properties never change
  2. Extensive properties depend on the quantity of matter present, whereas intensive properties are independent of sample size Correct answer.
  3. Extensive properties can be measured without altering the chemical identity, whereas intensive properties require a chemical reaction
  4. Intensive properties are always additive when combining two identical systems

Explanation

Extensive properties (such as mass, volume, total enthalpy, and heat capacity) depend directly on the amount or size of matter in the sample. Intensive properties (such as density, boiling point, refractive index, and temperature) are intrinsic to the substance and remain unchanged regardless of sample size.

Question 28Química fundamental

Which mixture is classified as a colloidal dispersion exhibiting the Tyndall effect rather than a true homogeneous solution?

  1. Aqueous sodium chloride solution (0.9% saline)
  2. Aqueous hydrochloric acid (1.0 M HCl)
  3. Whole milk Correct answer.
  4. Ethanol dissolved in water (70% rubbing alcohol)

Explanation

Whole milk is an emulsion/colloid composed of microscopic fat globules and casein protein micelles (1 to 1000 nm in diameter) suspended in an aqueous medium. These dispersed particles are large enough to scatter visible light (the Tyndall effect), making the mixture cloudy and preventing it from being a true molecular solution.

Question 29Química fundamental

In thin-layer chromatography (TLC) of pharmaceutical alkaloids using a polar silica gel stationary phase and a nonpolar organic mobile phase, which compound will display the highest retention factor (R_f)?

  1. The most polar compound, because it binds most strongly to the polar stationary phase
  2. The compound with the largest molecular weight, regardless of polarity
  3. The compound that forms the strongest ionic bonds with the mobile solvent
  4. The least polar (most nonpolar) compound, because it interacts preferentially with the mobile phase Correct answer.

Explanation

In normal-phase chromatography, the stationary phase is polar (silica gel) and the mobile phase is relatively nonpolar. Nonpolar compounds interact weakly with the polar silica and partition preferentially into the mobile phase, traveling further up the plate and yielding a higher R_f value (distance traveled by solute / distance traveled by solvent front).

Question 30Química fundamental

A student measures a mass of 0.0520 g on an analytical balance. How many significant figures are recorded in this measurement?

  1. 2
  2. 3 Correct answer.
  3. 4
  4. 5

Explanation

In 0.0520 g, the leading zeros (0.0) serve only as decimal placeholders and are not significant. The digits '5' and '2', and the trailing zero following the '2' in a decimal number, are significant, yielding exactly 3 significant figures.

Question 31Química fundamental

Which phase change occurs with a release of heat (exothermic process) into the surroundings?

  1. Condensation of steam into liquid water Correct answer.
  2. Sublimation of dry ice
  3. Fusion (melting) of ice into liquid water
  4. Vaporization of liquid diethyl ether

Explanation

Condensation is an exothermic phase change in which gas molecules lose kinetic energy and form intermolecular attractions to become a liquid, releasing latent heat into the surroundings (ΔH < 0). Sublimation, fusion, and vaporization are all endothermic processes requiring an absorption of heat (ΔH > 0).

Question 32Química fundamental

A 4.00 L rigid cylinder holds 2.00 mol of nitrogen gas at 300 K. Using the ideal gas law PV = nRT with R = 0.0821 L·atm/(mol·K), what is the pressure inside the cylinder?

  1. 3.08 atm
  2. 24.6 atm
  3. 12.3 atm Correct answer.
  4. 49.3 atm

Explanation

Rearranging PV = nRT gives P = nRT / V. Substituting the values: P = (2.00 mol)(0.0821 L·atm/mol·K)(300 K) / 4.00 L = 49.26 L·atm / 4.00 L = 12.3 atm. Temperature must always be expressed in kelvin in this equation, and the value of R already fixes the pressure unit as atmospheres.

Question 33Química fundamental

Which set of quantum numbers (n, l, m_l, m_s) is permissible for an electron in an atom?

  1. n = 2, l = 2, m_l = 0, m_s = +1/2
  2. n = 3, l = 1, m_l = -2, m_s = -1/2
  3. n = 3, l = 2, m_l = +1, m_s = +1
  4. n = 4, l = 0, m_l = 0, m_s = +1/2 Correct answer.

Explanation

Quantum number rules dictate: n must be a positive integer (n ≥ 1); l ranges from 0 to n - 1; m_l ranges from -l to +l; and m_s must be either +1/2 or -1/2. For n = 4, l = 0 (an s orbital), m_l = 0, and m_s = +1/2, all values are strictly valid.

Question 34Química fundamental

What fundamental principle states that no two electrons in the same atom can have an identical set of four quantum numbers?

  1. Hund's Rule of Maximum Multiplicity
  2. Pauli Exclusion Principle Correct answer.
  3. Heisenberg Uncertainty Principle
  4. Aufbau Principle

Explanation

The Pauli Exclusion Principle (Wolfgang Pauli, 1925) dictates that no two fermions (such as electrons) within an atom can possess the same quantum state, meaning they cannot share identical values for all four quantum numbers (n, l, m_l, m_s). Consequently, an individual orbital can hold a maximum of two electrons with opposing spins.

Question 35Química fundamental

Which of the following atoms or ions possesses the largest radius?

  1. O²⁻ (oxide anion) Correct answer.
  2. Na⁺ (sodium cation)
  3. Mg²⁺ (magnesium cation)
  4. F⁻ (fluoride anion)

Explanation

The species Na⁺ (Z = 11), Mg²⁺ (Z = 12), F⁻ (Z = 9), and O²⁻ (Z = 8) form an isoelectronic series with 10 electrons ([Ne] configuration). In an isoelectronic series, the radius decreases as atomic number (nuclear charge Z) increases. Oxide (O²⁻) has the lowest nuclear charge (Z = 8) pulling on the 10 electrons, giving it the largest ionic radius.

Question 36Química fundamental

Which halogen possesses the highest electron affinity (most negative enthalpy of electron gain)?

  1. Fluorine (F)
  2. Bromine (Br)
  3. Chlorine (Cl) Correct answer.
  4. Iodine (I)

Explanation

Although periodic trends generally show electron affinity increasing upward within a group, chlorine (Cl, -349 kJ/mol) has a higher electron affinity than fluorine (F, -328 kJ/mol). Fluorine's exceptionally small 2p orbital creates intense inter-electronic repulsions that diminish the net energy released when an additional electron is accepted.

Question 37Química fundamental

What is the total number of unpaired electrons in a ground-state gaseous phosphorus atom (P, Z = 15)?

  1. 0
  2. 1
  3. 5
  4. 3 Correct answer.

Explanation

Phosphorus has the ground-state electron configuration 1s² 2s² 2p⁶ 3s² 3p³. According to Hund's rule, each of the three degenerate 3p orbitals (3p_x, 3p_y, 3p_z) is occupied by a single electron with parallel spins before any pairing occurs. Thus, phosphorus has 3 unpaired electrons.

Question 38Química fundamental

What type of chemical bond forms between two atoms when their difference in electronegativity (ΔEN) on the Pauling scale is approximately zero (e.g., in diatomic nitrogen, N₂)?

  1. Ionic bond
  2. Nonpolar (pure) covalent bond Correct answer.
  3. Polar covalent bond
  4. Coordinate dative bond

Explanation

When two identical nonmetal atoms bond (ΔEN = 0), they share the bonding electron pairs equally with no permanent dipole moment, creating a nonpolar (pure) covalent bond.

Question 39Química fundamental

According to Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular geometry (shape) of the water molecule (H₂O)?

  1. Bent (angular) Correct answer.
  2. Linear
  3. Trigonal planar
  4. Tetrahedral

Explanation

The central oxygen atom in H₂O has 4 electron domains (two bonding pairs with hydrogen and two nonbonding lone pairs), giving it a tetrahedral electron-pair geometry. Because lone pairs exert stronger electrostatic repulsion than bonding pairs, the bonded atoms are compressed into a bent (angular) molecular shape with a bond angle of approximately 104.5°.

Question 40Química fundamental

Which of the following chemical species represents an exception to the octet rule by possessing an expanded valence shell of more than eight electrons?

  1. Carbon dioxide (CO₂)
  2. Methane (CH₄)
  3. Sulfur hexafluoride (SF₆) Correct answer.
  4. Ammonia (NH₃)

Explanation

In sulfur hexafluoride (SF₆), sulfur is bonded to six fluorine atoms, sharing 12 valence electrons. Elements in Period 3 and beyond have accessible d orbitals that allow them to expand their valence shell beyond eight electrons (hypervalence).

Question 41Química fundamental

Which molecule has polar covalent bonds but has a net molecular dipole moment of zero (nonpolar molecule) due to symmetric cancellation?

  1. Water (H₂O)
  2. Sulfur dioxide (SO₂)
  3. Hydrochloric acid (HCl)
  4. Carbon tetrachloride (CCl₄) Correct answer.

Explanation

In CCl₄, each C-Cl bond is polar due to the electronegativity difference between carbon (2.5) and chlorine (3.0). However, the four dipole vectors point toward the vertices of a regular tetrahedron. Due to tetrahedral symmetry, the four bond dipole vectors cancel vectorially, resulting in a net dipole moment of μ = 0.

Question 42Química fundamental

Which intermolecular force is strongest in a pure sample of liquid ethanol (CH₃CH₂OH)?

  1. London dispersion forces
  2. Hydrogen bonding Correct answer.
  3. Ion-dipole interactions
  4. Metallic bonding

Explanation

Ethanol possesses a hydroxyl group (-OH) in which hydrogen is covalently bonded to highly electronegative oxygen. This allows ethanol molecules to form strong intermolecular hydrogen bonds, resulting in a substantially higher boiling point (78 °C) compared to dimethyl ether (-24 °C), its constitutional isomer.

Question 43Química fundamental

What is the empirical formula of an organic compound determined to contain 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass (atomic weights: C = 12.0, H = 1.0, O = 16.0 g/mol)?

  1. CH₂O Correct answer.
  2. CHO
  3. C₂H₄O
  4. C₆H₁₂O₆

Explanation

Assume a 100 g sample: Carbon: 40.0 g / 12.0 g/mol = 3.33 mol C Hydrogen: 6.7 g / 1.0 g/mol = 6.70 mol H Oxygen: 53.3 g / 16.0 g/mol = 3.33 mol O Divide each mole quantity by the smallest value (3.33): C = 3.33 / 3.33 = 1 H = 6.70 / 3.33 ≈ 2 O = 3.33 / 3.33 = 1 The empirical formula is CH₂O.

Question 44Química fundamental

How many moles of oxygen gas (O₂) are required to completely react with 4.0 moles of aluminum in the synthesis reaction: 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)?

  1. 2.0 moles
  2. 4.0 moles
  3. 3.0 moles Correct answer.
  4. 6.0 moles

Explanation

From the balanced chemical equation, the stoichiometric mole ratio between Al and O₂ is 4 mol Al : 3 mol O₂. Therefore, 4.0 moles of Al require exactly 3.0 moles of O₂.

Question 45Química fundamental

If 10.0 g of hydrogen gas (H₂, molar mass 2.0 g/mol) reacts with 64.0 g of oxygen gas (O₂, molar mass 32.0 g/mol) according to 2 H₂ + O₂ → 2 H₂O, which is the limiting reactant and what is the theoretical yield of water (H₂O, molar mass 18.0 g/mol)?

  1. H₂ is limiting; theoretical yield is 90.0 g H₂O
  2. H₂ is limiting; theoretical yield is 72.0 g H₂O
  3. Neither is limiting; theoretical yield is 74.0 g H₂O
  4. O₂ is limiting; theoretical yield is 72.0 g H₂O Correct answer.

Explanation

Calculate moles of reactants: 10.0 g H₂ / 2.0 g/mol = 5.0 mol H₂; 64.0 g O₂ / 32.0 g/mol = 2.0 mol O₂. The reaction requires 2 mol H₂ per 1 mol O₂. For 2.0 mol O₂, 4.0 mol H₂ are required. Since 5.0 mol H₂ are present, H₂ is in excess and O₂ is the limiting reactant. Yield = 2.0 mol O₂ × (2 mol H₂O / 1 mol O₂) = 4.0 mol H₂O × 18.0 g/mol = 72.0 g H₂O.

Question 46Química fundamental

The chemical reaction Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g) is best classified as which reaction type?

  1. Synthesis (combination)
  2. Single displacement (substitution) Correct answer.
  3. Decomposition
  4. Double displacement (metathesis)

Explanation

In this reaction, elemental zinc displaces hydrogen from hydrochloric acid to form zinc chloride and hydrogen gas. Because one element replaces another element in a compound (A + BC → AC + B), it is a single displacement (single replacement) reaction, which is also an oxidation-reduction reaction.

Question 47Química fundamental

What is the molarity (M) of an aqueous solution prepared by dissolving 20.0 g of sodium hydroxide (NaOH, molar mass 40.0 g/mol) in sufficient distilled water to make 500 mL of solution?

  1. 1.00 M Correct answer.
  2. 0.50 M
  3. 2.00 M
  4. 4.00 M

Explanation

Calculate moles of solute: moles = 20.0 g / 40.0 g/mol = 0.50 mol NaOH. Convert volume to liters: 500 mL = 0.500 L. Calculate molarity: M = moles / liters = 0.50 mol / 0.500 L = 1.00 M.

Question 48Química fundamental

What volume of a 6.0 M stock hydrochloric acid (HCl) solution is required to prepare 300 mL of a 0.50 M HCl working solution?

  1. 30 mL
  2. 50 mL
  3. 25 mL Correct answer.
  4. 100 mL

Explanation

Apply the dilution equation: M₁V₁ = M₂V₂. Here, M₁ = 6.0 M, M₂ = 0.50 M, and V₂ = 300 mL. Solving for V₁: V₁ = (M₂ × V₂) / M₁ = (0.50 M × 300 mL) / 6.0 M = 150 / 6.0 = 25 mL.

Question 49Química fundamental

What is the pH of a 0.0010 M aqueous solution of hydrochloric acid (HCl), assuming complete dissociation?

  1. 1.0
  2. 2.0
  3. 11.0
  4. 3.0 Correct answer.

Explanation

HCl is a strong monoprotic acid that dissociates completely in water: HCl → H⁺ + Cl⁻. Therefore, [H⁺] = 0.0010 M = 1.0 × 10⁻³ M. The pH is calculated as: pH = -log₁₀[H⁺] = -log₁₀(1.0 × 10⁻³) = 3.0.

Question 50Química fundamental

According to the Brønsted-Lowry theory of acids and bases, how are an acid and a base defined?

  1. An acid produces OH⁻ in water; a base produces H⁺ in water
  2. An acid is a proton (H⁺) donor; a base is a proton (H⁺) acceptor Correct answer.
  3. An acid is an electron-pair donor; a base is an electron-pair acceptor
  4. An acid contains oxygen; a base contains a metal cation

Explanation

Under the Brønsted-Lowry definition (1923), an acid is any species capable of donating a proton (hydrogen ion, H⁺), while a base is any species capable of accepting a proton.