CHMM Practice Questions (2026) — Free Sample Questions with Answers

Answer: C

1. Process definition (68.3): interconnected vessels, plus separate vessels located so they could be involved in a potential release, are one process.
2. Chlorine room: 2 manifolded containers × 2,000 lb = 4,000 lb; stored full container in the same room = 2,000 lb → 6,000 lb.
3. Pool building, one-half mile away: 4 × 150 lb = 600 lb, a separate process (also below 2,500 lb on its own).
4. Compare: 6,000 lb > 2,500 lb TQ → the chlorine room process is covered by the RMP rule.

2,000 lb: Treats each container as its own process; interconnected vessels and nearby containers that could be involved in a release form one process.

4,000 lb: Counts only the manifolded containers and omits the full container stored in the same room.

6,000 lb: Correct. The two connected containers and the co-located full container are one process: 3 × 2,000 = 6,000 lb, which exceeds 2,500 lb (40 CFR 68.3, 68.115(a)).

6,600 lb: Adds the pool cylinders half a mile away, which are a separate process that could not be involved in the same release.

RMP applicability is decided process by process: sum the maximum quantity in all interconnected and co-located vessels and compare it with the 68.130 threshold quantity.

Answer: A

1. Corrosivity: aqueous waste, pH 1.7 ≤ 2 → corrosive (D002).
2. Ignitability: flash point 68 °C vs limit < 60 °C → not ignitable.
3. Toxicity, lead: use the TCLP result, 1.9 mg/L < 5.0 mg/L → not D008 (the total of 210 mg/kg only shows a TCLP test was worth running).
4. Toxicity, chromium: 3.6 mg/L < 5.0 mg/L → not D007.
5. Conclusion: the pH result alone makes the waste hazardous (D002).

pH of 1.7: Correct. An aqueous waste with pH ≤ 2 is corrosive (D002); 1.7 is below 2 (40 CFR 261.22(a)(1)).

flash point of 68 °C: 68 °C (154 °F) is above the 60 °C (140 °F) ignitability limit, so the waste is not D001.

total lead of 210 mg/kg: Toxicity is judged on the TCLP extract, not the total result; the measured TCLP lead (1.9 mg/L) is below 5.0 mg/L.

TCLP chromium of 3.6 mg/L: 3.6 mg/L is below the 5.0 mg/L regulatory level, so chromium does not make the waste D007.

Compare each result with its own characteristic limit: pH ≤ 2 or ≥ 12.5 for aqueous wastes, flash point < 60 °C, and TCLP extract (not total) concentrations against the 261.24 levels.

Answer: D

DANGEROUS WHEN WET, FLAMMABLE and CORROSIVE placards: The Table 2 aggregate (800 lb) is under 1,001 lb, so FLAMMABLE and CORROSIVE placards are not required.

No placards, because the total load of 840 lb is below 1,001 lb: The 1,001-lb exception applies only to Table 2 materials; Table 1 materials such as 4.3 are placarded in any quantity.

A DANGEROUS placard in place of all three placards: DANGEROUS may substitute only for Table 2 placards, never for a Table 1 placard.

DANGEROUS WHEN WET placards only: Correct. Division 4.3 is a Table 1 material, placarded in any quantity (172.504(a), (e)). The Table 2 materials total 380 + 420 = 800 lb, below 1,001 lb, so they need no placards (172.504(c)(1)).

Placard Table 1 materials in any quantity; count only Table 2 materials toward the 1,001-lb aggregate exception.

Answer: C

Entry 1: The proper shipping name precedes the ID number; 172.202(b) requires the ID number first.

Entry 2: The packing group (PG II) is missing from the basic description.

Entry 3: Correct. The sequence is ID number, proper shipping name, class, packing group (172.202(b)), and "RQ" may be entered before or after the basic description (172.203(c)(2)). The shipping name already identifies the hazardous substance.

Entry 4: An n.o.s. entry needs the technical names in parentheses, e.g. (Toluene, Xylene), under 172.203(k).

Check every entry for the full sequence (ID, PSN, class, PG), technical names on n.o.s. entries, and "RQ" for hazardous substances.

Answer: C

1. Aboveground containers ≥ 55 gal: diesel 550 gal; tote 275 gal; drums 8 × 55 = 440 gal.
2. Exclude: 5-gal pails (< 55 gal) and the completely buried tank subject to part 280.
3. Aggregate: 550 + 275 + 440 = 1,265 gal.
4. Compare: SPCC applies only above 1,320 gal aboveground; 1,265 gal ≤ 1,320 gal.
5. Conclusion: not subject to SPCC on aboveground capacity (buried capacity is 0 gal after the part-280 exclusion).

3,265 gal; subject: Adds the buried part-280 tank to the aboveground total; it is excluded and is a separate buried-capacity test anyway.

1,465 gal; subject: Counts the 5-gal pails, which are below the 55-gal counting threshold.

1,265 gal; not subject: Correct. Count only aboveground containers ≥ 55 gal: 550 + 275 + 8 × 55 = 1,265 gal, which is not more than 1,320 gal (40 CFR 112.1(d)(2)(ii)).

3,465 gal; subject: Counts both the pails and the excluded buried tank.

SPCC applies above 1,320 gal aggregate aboveground capacity, counting only containers of 55 gal or more by shell capacity, or above 42,000 gal completely buried capacity excluding part-280 USTs.

Answer: D

1. Givens: total lead = 150 mg/kg; waste is 100% solid; TCLP limit = 5.0 mg/L; extraction ratio 20:1.
2. Theoretical maximum leachate = total ÷ 20 = 150 ÷ 20 = 7.5 mg/L (assumes all lead leaches).
3. Compare: 7.5 mg/L > 5.0 mg/L, so the limit could be exceeded.
4. Total analysis cannot rule out D008, and the theoretical value does not prove failure.
5. Conclusion: run the TCLP to determine whether the waste is D008.

The waste is D008 because its total lead of 150 mg/kg exceeds 5.0 mg/L.: A total concentration in mg/kg cannot be compared directly with a leachate limit in mg/L. The 20:1 dilution of the extraction must be accounted for.

The waste is not D008 because its theoretical maximum leachate is 1.5 mg/L.: This divides by 100 instead of 20. The 20:1 extraction ratio gives a theoretical maximum of 7.5 mg/L.

The waste is D008 because its theoretical maximum leachate of 7.5 mg/L exceeds 5.0 mg/L.: The 20× calculation assumes 100% of the lead leaches. It can rule a waste out but cannot prove it fails; only an actual TCLP result above 5.0 mg/L makes it D008.

The TCLP must be run because the 7.5 mg/L theoretical maximum could exceed 5.0 mg/L.: Correct. 150 ÷ 20 = 7.5 mg/L, which is above 5.0 mg/L, so total analysis cannot rule out the characteristic. Method 1311 §1.2 lets total analysis replace the TCLP only when the limit could not possibly be exceeded.

For a 100% solid waste, total ÷ 20 below the TCLP limit rules the characteristic out; a result above the limit means the TCLP must be run.

Answer: C

1. Containers with free liquids: 40 × 55 gal = 2,200 gal, plus the 330-gal tote = 2,530 gal. The 10 solids drums are excluded because they hold no free liquids.
2. 10% of the liquid-container volume: 0.10 × 2,530 gal = 253 gal.
3. Volume of the largest container: 330 gal (the tote).
4. Take the greater value: 330 gal > 253 gal.
5. Minimum containment capacity = 330 gal.

253 gal: This is 10% of the liquid-container volume, but the rule takes the greater of that figure or the largest container, and the 330-gal tote is larger.

308 gal: This takes 10% of all containers, including the solids drums. Containers holding no free liquids are left out of the calculation, and the largest-container test still governs.

330 gal: Correct. 40 CFR 264.175(b)(3) requires the greater of 10% of the volume of containers holding free liquids (253 gal) or the volume of the largest container (330 gal).

363 gal: This applies a 110% rule of thumb (330 × 1.1) from fire-code or company practice. The RCRA container standard sets the minimum at the greater of 10% of total volume or the largest container.

Permitted container storage containment must hold the greater of 10% of the volume of containers with free liquids or the volume of the largest container; containers without free liquids are not counted.

Answer: C

1. Given: LQG; the initial transporter accepted the waste on April 10; no signed copy from the designated facility.
2. Rule: 40 CFR 262.42(a)(1) — contact the transporter and/or designated facility at 45 days; 262.42(a)(2) — exception report at 60 days; both counted from acceptance by the initial transporter.
3. Count: April 10 → April 30 = 20 days; 25 more days → May 25 = 45 days.
4. Conclusion: contact by May 25; if the signed copy still has not arrived, file the exception report by June 9 (60 days).

April 30: 20 days is the current TSDF window to reconcile a significant manifest discrepancy (40 CFR 264.72), not the generator's follow-up deadline.

May 15: Uses the former 35-day contact period, which the current rule replaced with 45 days.

May 25: Correct. 40 CFR 262.42(a)(1): an LQG that has not received the signed copy within 45 days of acceptance by the initial transporter must contact the transporter and/or designated facility; April 10 + 45 days = May 25. The exception report follows at 60 days (June 9) under 262.42(a)(2).

June 9: June 9 is 60 days, the deadline to file the exception report, not the earlier contact step.

An LQG missing the signed manifest contacts the transporter and/or designated facility at 45 days and files an exception report at 60 days; both clocks start when the initial transporter accepts the waste.

Answer: B

1. Ammonia: Mon 06:00 to Tue 04:00 = 22 h; 55 + 50 = 105 lb ≥ 100 lb RQ → reportable.
2. Chlorine: Mon 16:00 to Wed 09:00 = 41 h; at most 6 lb in any 24 h < 10 lb RQ → not reportable.
3. Toluene: Wed 10:00 to Thu 14:00 = 28 h; at most 900 lb in any 24 h < 1,000 lb RQ → not reportable.
4. Conclusion: only the ammonia releases required immediate NRC notification.

Only the chlorine releases required notification, because they total 12 lb: The two chlorine releases are 41 hours apart, so no 24-hour period holds more than 6 lb.

Only the ammonia releases required notification: Correct. The ammonia releases are 22 hours apart: 55 + 50 = 105 lb ≥ 100-lb RQ within 24 hours (40 CFR 302.6(a)). The chlorine releases are 41 hours apart and the toluene releases 28 hours apart, so neither pair reaches its RQ in any 24-hour period.

The ammonia and toluene releases required notification: The toluene releases are 28 hours apart; no 24-hour period holds 1,000 lb.

No release required notification, because no single event reached its RQ: Releases within any 24-hour period are added; the ammonia total of 105 lb in 22 hours reaches the RQ.

Sum releases of each substance over any rolling 24-hour period and compare the total with that substance's RQ.

Answer: D

First responder awareness level, with no minimum hours specified: Awareness-level employees only recognize a release and notify others. They take no containment action.

40 hours of off-site instruction with 3 days of supervised field experience: This is the general site worker requirement for cleanup operations under 1910.120(e), not emergency response under paragraph (q).

Hazardous materials technician level, at least 24 hours: Technicians approach the point of release to plug, patch or stop it. These employees deliberately stay back.

First responder operations level, at least 8 hours or demonstrated competency: Correct. 1910.120(q)(6)(ii) covers responders who act defensively from a safe distance to contain a release without stopping it, and requires at least 8 hours of training or sufficient experience to objectively demonstrate competency.

Match the (q)(6) level to the role: awareness (detect and notify), operations (defensive, 8 h), technician (stop the release, 24 h), specialist and incident commander (24 h).

Answer: C

1. Givens: inside diameter 12.0 ft, so radius r = 6.0 ft; level drop h = 14.0 − 11.5 = 2.5 ft; 1 ft³ = 7.48 gal.
2. Volume released = πr²h = 3.1416 × 36.0 ft² × 2.5 ft = 282.7 ft³.
3. Convert: 282.7 ft³ × 7.48 gal/ft³ = 2,115 gal.
4. Conclusion: about 2,115 gal was released.

283 gal: This is the volume in cubic feet (282.7 ft³); it was not converted to gallons.

705 gal: This uses the side-wall area (2πrh) instead of the cross-sectional area (πr²) times the level drop.

2,115 gal: Correct. V = πr²h = 3.1416 × (6.0 ft)² × 2.5 ft = 282.7 ft³; 282.7 × 7.48 = 2,115 gal.

8,460 gal: This uses the diameter (12 ft) in place of the radius, which overstates the volume by a factor of four.

For a vertical cylinder, the release equals πr² × the level drop; convert cubic feet to gallons with 7.48 gal/ft³.

Answer: A

Trichloroethene: Correct. SVE suits volatile compounds with high vapor pressure and weak sorption in permeable soil; TCE's vapor pressure (~70 mm Hg) and low Koc make it readily extractable.

Benzo(a)pyrene: Essentially nonvolatile and very strongly sorbed (very high Koc); SVE would remove almost none.

Lead: Metals do not volatilize; they call for excavation, stabilization or capping.

Pentachlorophenol: Its low vapor pressure and moderate sorption make it a poor SVE candidate; bioremediation or other treatment fits better.

Match technology to properties: SVE needs volatility and permeable unsaturated soil; high Koc and low vapor pressure point away from SVE.

Answer: C

1. Givens: capital $48,000; savings $12,000/yr disposal + $8,000/yr purchases; O&M $4,000/yr.
2. Gross annual savings = $12,000 + $8,000 = $20,000.
3. Net annual savings = $20,000 − $4,000 = $16,000.
4. Payback = capital ÷ net annual savings = $48,000 ÷ $16,000 = 3.0 years.

2.0 years: Adds the $4,000 O&M cost to the savings ($24,000) instead of subtracting it.

2.4 years: Ignores the $4,000 annual O&M cost ($48,000 ÷ $20,000).

3.0 years: Correct. Net annual savings = $12,000 + $8,000 − $4,000 = $16,000; payback = $48,000 ÷ $16,000 = 3.0 years.

4.0 years: Counts only the disposal savings ($48,000 ÷ $12,000) and omits the purchase savings.

Simple payback = installed capital cost ÷ net annual savings, where net savings subtracts the project's own operating and maintenance cost.

Answer: B

Ingestion of contaminated groundwater: With no wells and a public supply from outside the area, there is no exposure point for drinking the groundwater, so this pathway is incomplete.

Inhalation of indoor air from vapor intrusion: Correct. TCE is volatile; vapors can migrate from shallow groundwater through soil gas into basements, giving a source, transport mechanism, exposure point, route and receptor — a complete pathway.

Incidental ingestion of surface soil: Surface soil is clean and the source is 15 ft down, so soil is not an exposure medium.

Dermal contact with groundwater while bathing: Bathing water comes from the uncontaminated public supply; residents do not contact the plume.

A pathway is complete only if source, transport, exposure point, route and receptor are all present; for volatile groundwater plumes under buildings, vapor intrusion is often the complete one.

Answer: C

Signal words "Danger" and "Warning"; flame, corrosion and exclamation mark pictograms: If "Danger" appears, "Warning" must not appear.

Signal word "Danger"; flame and corrosion pictograms only: Misapplies the precedence rule: the corrosive pictogram removes the exclamation mark only where it is used for skin or eye irritation, not acute toxicity.

Signal word "Danger"; flame, corrosion and exclamation mark pictograms: Correct. "Danger" (from skin corrosion) displaces "Warning" (C.2.1.1). The exclamation mark stays because it is used for acute toxicity Category 4; the corrosive pictogram displaces it only for skin or eye irritation (C.2.1.3).

Signal word "Warning"; flame and exclamation mark pictograms: Skin Corrosion Category 1B requires "Danger" and the corrosion pictogram; the more severe hazard governs.

Use one signal word, "Danger" if any class requires it; drop the exclamation mark only when the skull and crossbones (acute toxicity) or the corrosion pictogram (irritation) covers the same hazard.

Answer: A

An emergency action plan meeting 29 CFR 1910.38: Correct. 1910.120(q)(1): employers who evacuate employees from the danger area and do not permit them to assist in handling the emergency are exempt from (q) if they provide an emergency action plan complying with 1910.38.

A site safety and health plan under 1910.120(b): SSHPs apply to cleanup operations at hazardous waste sites, not to emergency evacuation at a warehouse.

An emergency response plan under 1910.120(q), with hazmat technician training for all employees: Required only when employees will respond to the release; this policy prohibits that.

An emergency response program under 40 CFR 68: That is an EPA RMP requirement for covered processes, not the OSHA plan that governs employee evacuation.

If employees only evacuate and call outside responders, a 1910.38 emergency action plan suffices; a 1910.120(q) ERP is needed when employees respond.