This area carries 7–11 questions. Until 2020 the area was Chemistry, with only basic biochemistry; the current outline adds microbiology, molecular biology and bioprocessing.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
questions on the exam7–11
problems on this page15
of 17 knowledge areas05
Checked
What the specification covers
The outline lists five subtopics. In our words:
Inorganic chemistry: concentration units, acids and bases, pH and pK, redox, solubility products, electrochemistry, the periodic table.
Biochemistry and molecular biology: the cell, metabolic cycles (glycolysis, Krebs, Calvin), enzymes and proteins, genetics, transcription and translation.
Bioprocessing: fermentation, aerobic and anaerobic biological treatment, nutrient removal.
15 chemistry and biology problems
0 / 15 answered · 0 correct
Question 1 of 15 · Chemistry and Biology · 7–11 items
In an analytical determination using UV-Vis spectrophotometry, a diluted sample exhibits an absorbance of 0.75 at a wavelength with a molar absorptivity of 15000 L/(mol·cm) and a 1-cm path length. If the sample was diluted by a factor of 10 prior to measurement, calculate the original concentration of the absorbing compound.
Given
Absorbance A = 0.75
Molar absorptivity ε = 15000 L/(mol·cm)
Path length b = 1 cm
Dilution factor = 10
Find
Original concentration (M)
Solution
A = εbc → cdiluted = 0.75/(15,000 L/(mol·cm) × 1 cm) = 5.0 × 10−5 M
coriginal = 10 × 5.0 × 10−5 M = 5.0 × 10−4 M (Beer-Lambert law)
c = A/(εb) = 0.75/(15,000 × 1) = 5.0 × 10−5 M in the diluted sample; × 10 for the dilution gives 5.0 × 10−4 M.
8.3 × 10−5 M treats 0.75 as a transmittance and converts it with −log T; the reading is already an absorbance.
5.0 × 10−5 M is the concentration of the diluted solution in the cuvette; it still has to be multiplied by 10.
5.0 × 10−6 M divides by the dilution factor instead of multiplying; the original sample is more concentrated than the diluted one.
Question 2 of 15 · Chemistry and Biology · 7–11 items
For the gas-phase reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), the equilibrium constant Kc is 280 at 727°C. Calculating with the ideal gas constant R = 0.082057 L atm mol−1 K−1, what is the value of Kp at this temperature?
0.0337 uses R = 8.314 J/(mol·K); with Kc in mol/L and Kp in atm, R must be 0.08206 L·atm/(mol·K).
Question 3 of 15 · Chemistry and Biology · 7–11 items
Calculate the molarity of a solution prepared by dissolving 5.85 g of NaCl in water to make 250 mL of solution. (Molecular weight of NaCl = 58.5 g/mol)
Given
Mass of NaCl = 5.85 g
Solution volume V = 250 mL
Molecular weight of NaCl = 58.5 g/mol
Find
Molarity (M)
Solution
n = 5.85 g/58.5 g/mol = 0.100 mol
M = n/V = 0.100 mol/0.250 L = 0.40 M
0.00040 M divides the moles by 250 instead of 0.250 L, i.e. it leaves the volume in milliliters.
0.025 M multiplies the moles by the volume instead of dividing.
0.10 M is the number of moles of NaCl; molarity is moles per liter of solution.
n = 5.85 g/58.5 g/mol = 0.100 mol; M = 0.100 mol/0.250 L = 0.40 M.
Question 4 of 15 · Chemistry and Biology · 7–11 items
A buffer solution is prepared by mixing equal volumes of 0.1 M acetic acid (CH3COOH) and 0.1 M sodium acetate (CH3COONa). If Ka of acetic acid is 1.74 × 10−5, what is the pH of this buffer?
Chlorine is produced by oxidation (2Cl− → Cl2 + 2e−), which happens at the anode, not the cathode.
NaOH forms in brine (aqueous) electrolysis, where water is reduced to H2 and OH−; molten NaCl contains no water.
At the cathode Na+ + e− → Na; with no water present, sodium ions are the only species to reduce.
There is no water or oxygen-bearing anion in molten NaCl, so no O2 can form at either electrode.
Question 7 of 15 · Chemistry and Biology · 7–11 items
In DNA replication, which enzyme is responsible for joining DNA fragments on the lagging strand?
Solution
Lagging strand: primase (RNA primer) → DNA polymerase III (Okazaki fragment) → DNA polymerase I (replaces primer) → DNA ligase (seals the nick).
DNA ligase forms the phosphodiester bond that seals the nicks between Okazaki fragments on the lagging strand.
Helicase unwinds the double helix at the replication fork; it does not seal strands.
Primase lays down the short RNA primers that start each Okazaki fragment; it does not join them.
DNA polymerase extends each fragment and replaces the primers, but the final nick is sealed by ligase.
Question 8 of 15 · Chemistry and Biology · 7–11 items
Which of the following is the correct IUPAC name for CH3–CH2–CH(OH)–CH3?
Solution
Longest chain = 4 C (butane); OH on C-2 counted from the nearer end → butan-2-ol (2-butanol), a secondary alcohol.
1-Butanol has the OH on the end carbon (CH3CH2CH2CH2OH); here it sits on an inner carbon.
2-Propanol has only three carbons; this chain has four.
Four-carbon chain with OH on carbon 2 when numbered from the nearer end: butan-2-ol (2-butanol).
Numbering must give the OH the lowest locant; counting from the other end makes 3 into 2, so 3-butanol is not a valid name.
Question 9 of 15 · Chemistry and Biology · 7–11 items
A sample of 3-methyl-2-butanol is oxidized with acidified potassium dichromate. What is the major organic product?
Solution
Secondary alcohol (OH on C-2) + Cr(VI) oxidant → ketone at the same carbon: 3-methyl-2-butanone. Primary alcohols give aldehydes/acids; tertiary alcohols resist oxidation.
This is the starting material; an acidified dichromate oxidant does react with a secondary alcohol.
3-Methyl-1-butanol is a primary-alcohol isomer of the starting material; oxidation changes the functional group, not the OH position.
A ketone carbon cannot also carry a methyl branch and stay neutral and four-bonded; the name describes an impossible structure.
3-Methyl-2-butanol is a secondary alcohol (OH on C-2), and Cr(VI) oxidizes secondary alcohols to ketones: 3-methyl-2-butanone.
Question 10 of 15 · Chemistry and Biology · 7–11 items
Potassium permanganate (KMnO4) is used to oxidize trichloroethylene (C2HCl3) in contaminated groundwater under acidic conditions. The unbalanced skeletal reaction is MnO4− + C2HCl3 → Mn2+ + CO2 + Cl−. When the overall redox reaction is balanced with the smallest integer coefficients in acidic solution, what is the coefficient of water (H2O), and on which side does it appear?
The count is right, but the 24 waters formed by permanganate reduction outweigh the 20 consumed by oxidation, so the net 4 are products.
20 H2O is the five-times oxidation half-reaction before it is combined with the reduction half; part of it cancels.
24 H2O is the six-times reduction half-reaction before canceling the 20 waters consumed on the other side.
Question 11 of 15 · Chemistry and Biology · 7–11 items
What is the primary function of enzymes in biochemical reactions?
Solution
Enzymes are biological catalysts: lower Ea by stabilizing the transition state; ΔG and K of the reaction are unchanged (Michaelis-Menten kinetics).
Enzymes supply no energy; endergonic steps in cells are driven by coupling to ATP hydrolysis.
Enzymes bind the substrate and stabilize the transition state, lowering Ea and raising the rate without being consumed.
A catalyst speeds forward and reverse reactions equally, so K is unchanged; only the approach to equilibrium is faster.
Energy conversion of that kind is not what a catalyst does; enzymes change the path, not the energy of the products.
Question 12 of 15 · Chemistry and Biology · 7–11 items
Which process in industrial fermentation is used to separate and collect the desired product from the fermentation broth?
Solution
Upstream (strain, media, sterilization, inoculum) → fermentation → downstream (separation, purification, polishing) of the product from the broth.
Sterilizing the medium happens before inoculation, to keep contaminants out of the fermenter.
Downstream processing (cell separation, extraction, chromatography, crystallization, drying) recovers and purifies the product from the broth.
Batch cultivation is a way of running the fermentation itself, not of recovering the product.
Upstream processing covers strain, medium preparation and inoculum, everything before and up to the fermentation.
Question 13 of 15 · Chemistry and Biology · 7–11 items
In industrial fermentation, what is the primary difference between an aerobic and an anaerobic process?
Solution
Aerobic: O2 is the terminal electron acceptor (respiration, high biomass yield). Anaerobic: no O2; fermentation or other acceptors, lower energy and biomass yield (e.g. ethanol fermentation by yeast).
Both kinds of process run near neutral pH; pH does not separate them.
Different organisms dominate in each process, but that follows from the oxygen condition; it is not the defining difference.
Either process can run mesophilic or thermophilic; temperature is not the distinction.
Aerobic processes use O2 as the terminal electron acceptor; anaerobic ones run without it and yield less biomass and ATP per mole of substrate.
Question 14 of 15 · Chemistry and Biology · 7–11 items
Rank the following three organic compounds in order of decreasing water solubility at neutral pH: butanoic acid (CH3CH2CH2COOH), 1-butanol (CH3CH2CH2CH2OH) and n-butane (CH3CH2CH2CH3).
Solution
Polarity and hydrogen bonding set water solubility: –COOH (and –COO− at pH 7, pKa ≈ 4.8) > –OH > no polar group.
1-Butanol is only partly soluble (about 7 g per 100 mL), while butanoic acid is miscible and at pH 7 is mostly ionized as butanoate.
This puts the hydrocarbon above the carboxylic acid, the reverse of their polarity.
The carboxylic acid hydrogen-bonds as donor and acceptor and is largely ionized at pH 7 (pKa ≈ 4.8); the alcohol hydrogen-bonds less; the alkane barely dissolves.
This is the exact reverse: solubility in water rises with polar, hydrogen-bonding groups, not with their absence.
Butane has no polar group and is nearly insoluble; it cannot outrank an alcohol that hydrogen-bonds with water.
Question 15 of 15 · Chemistry and Biology · 7–11 items
Which of the following organic compounds is an aldehyde?
Solution
Aldehyde: R–CHO (carbonyl at a chain end, bonded to H). Ketone: R–CO–R′. Carboxylic acid: R–COOH. Alcohol: R–OH.
CH3COCH3 is acetone: its carbonyl sits between two carbons, which makes it a ketone.
CH3CH2OH is ethanol, an alcohol; it has no carbonyl group.
CH3CHO (acetaldehyde, ethanal) has a carbonyl on the end carbon bonded to a hydrogen, the aldehyde group –CHO.
CH3COOH is acetic acid; the carbonyl carbon also carries –OH, which makes it a carboxylic acid.
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Common traps
Molality per kilogram of solution. Molality is per kilogram of solvent; molarity is per liter of solution. The two converge only for dilute aqueous solutions.
Strong-acid pH for a weak acid. A weak acid dissociates partly; its pH comes from Ka, and the shortcut x ≈ √(Ka·C) holds only while x is small next to C.
Electrons left out of a redox balance. Balance atoms other than O and H, then O with water, H with H+, and charge with electrons; in base, neutralize the H+ at the end.
Growth units. A specific growth rate is per hour (or per day); a yield coefficient YX/S is mass of cells per mass of substrate. Mixing time bases misplaces every number.
Sources: Brown, LeMay and Bursten, Chemistry: The Central Science; Shuler and Kargi, Bioprocess Engineering: Basic Concepts.