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Question 6301Question

A meteorological station located in northern Nigeria records a single rainfall maximum in August, an annual total precipitation of approximately 850 mm, a dry season lasting 5 to 6 months characterized by severe Harmattan conditions, and a relatively high annual temperature range. Which climatic zone of Nigeria does this station represent?

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Answer: Tropical Continental (Sudan) Climate

Answer

The meteorological characteristics describe the Tropical Continental (Sudan) Climate zone located in northern Nigeria.
The correct answer is the Tropical Continental (Sudan) Climate. Northern Nigeria lies under the influence of the dry continental Tropical (cT) air mass for most of the year, yielding a single rainfall maximum in August (when the maritime air mass penetrates furthest north), low annual rainfall (~500–1000 mm), a prolonged dry season with Harmattan, and higher annual temperature fluctuations.

Step-by-Step Solution

1
Analyze the rainfall regime and distribution
Single rainfall peak in August with ~850 mm total indicates a short rainy season restricted to the months when the Inter-Tropical Convergence Zone (ITCZ) reaches its northernmost extent.
Rainfall in Nigeria decreases in duration and quantity from south to north.
2
Evaluate the dry season duration and air mass influence
A 5 to 6 month dry season with intense Harmattan indicates dominant influence of the dry continental Tropical (cT) air mass originating from the Sahara Desert.
The cT air mass controls northern Nigeria for more than half the year as the ITCZ retreats southward.
3
Synthesize thermal and moisture characteristics to identify the climatic zone
High annual temperature range, lower annual rainfall, single peak, and strong Harmattan are classic attributes of the Tropical Continental (Sudan climate) region.
Distinguishes northern interior climate from humid southern coastal belts and modified plateau microclimates.

Key Concept

Spatial variation of climatic zones and air mass dynamics in Nigeria
Estimated Time:1m 30s
Question 6302Question

A navigator on Ship P observes local solar noon (12:00 noon) when a Greenwich Mean Time (GMT) chronometer reads 1:40 PM. If Ship Q is located at a position where the local time is exactly 4 hours ahead of Ship P, what is the longitude of Ship Q?

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Answer: 35E35^\circ\text{E}

Answer

35E35^\circ\text{E}
Ship P is at 25W25^\circ\text{W} because its local time is 1 hour 40 minutes (100 minutes) behind GMT (100 minutes / 4 min per degree = 25W25^\circ\text{W}). Since Ship Q is 4 hours ahead in time, it lies 6060^\circ to the east (4×154 \times 15^\circ). Measuring 6060^\circ east from 25W25^\circ\text{W} covers 2525^\circ to reach the Greenwich Meridian (00^\circ), leaving 3535^\circ into the Eastern Hemisphere, resulting in 35E35^\circ\text{E}.

Step-by-Step Solution

1
Calculate the longitude of Ship P using local time and GMT
Time difference = 1 hour 40 minutes = 100 minutes. Longitude of Ship P = 100 minutes/4 minutes per degree=25W100 \text{ minutes} / 4 \text{ minutes per degree} = 25^\circ\text{W} (since local time is behind GMT).
The Earth rotates 11^\circ every 4 minutes, and areas behind GMT are located in the Western Hemisphere.
2
Determine the time difference and direction from Ship P to Ship Q
Time difference = 4 hours ahead. Direction = East of Ship P.
Local time increases as one moves east.
3
Convert the 4-hour time difference into angular distance in degrees
Angular distance = 4 hours×15/hour=604 \text{ hours} \times 15^\circ/\text{hour} = 60^\circ.
Earth rotates at a rate of 1515^\circ per hour.
4
Calculate the longitude of Ship Q by moving 6060^\circ east from 25W25^\circ\text{W}
Distance to Greenwich Meridian (00^\circ) = 2525^\circ east. Remaining distance east into Eastern Hemisphere = 6025=35E60^\circ - 25^\circ = 35^\circ\text{E}.
Crossing the Prime Meridian changes the hemisphere designation from West to East.

Key Concept

Longitude calculation using Greenwich Mean Time (GMT) and local time adjustments across meridians
Estimated Time:2m 0s
Question 6303Question

The seasonal movement of the Inter-Tropical Convergence Zone (ITCZ) plays a crucial role in controlling precipitation patterns across West Africa. When the ITCZ migrates to its northernmost position near latitude 20N20^\circ\text{N} in July, which weather condition is experienced along the West African coastal region?

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Answer: Widespread monsoonal rainfall driven by humid South-West maritime air masses.

Answer

Widespread monsoonal rainfall driven by humid South-West maritime air masses.
During July, the Inter-Tropical Convergence Zone (ITCZ) shifts to its northernmost location. This draws warm, moisture-laden South-West monsoon winds inland from the Atlantic Ocean, producing the rainy season across West Africa, particularly along the coast.

Step-by-Step Solution

1
Identify the movement of the ITCZ during the Northern Hemisphere summer (July).
In July, solar insolation peaks in the Northern Hemisphere, causing the low-pressure ITCZ belt to move northward up to approximately 20N20^\circ\text{N}.
The ITCZ follows the apparent movement of the sun.
2
Determine the prevailing wind system drawn behind the northward-shifting ITCZ.
The South-West monsoon winds (moist tropical maritime air from the Atlantic Ocean) sweep inland behind the ITCZ.
Air moves from high-pressure ocean belts toward the continental low-pressure trough.
3
Match the wind mass to the resulting climatic condition along the coast.
The moist air mass causes intense convective and orographic rainfall along the West African coast.
High humidity and surface heating promote cloud formation and heavy precipitation.

Key Concept

ITCZ Migration and Seasonal Rainfall in West Africa
Estimated Time:1m 0s
Question 6304Question

Arrange the following sequential stages in the agro-allied manufacturing of cotton textiles in Nigeria (such as in Kaduna and Kano industrial clusters), from raw material extraction to the final finished product.

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Answer

The correct sequence starts with ginning harvested seed cotton to separate lint, followed by spinning lint into yarn, then weaving yarn into raw fabric, and concludes with dyeing, printing, and finishing the fabric.
In Nigeria's agro-based textile industry, processing follows a strict forward supply sequence: raw seed cotton is first ginned to remove seeds, the resulting lint is spun into yarn, the yarn is woven into raw cloth, and finally the fabric is dyed and finished for distribution.

Step-by-Step Solution

1
Identify the primary raw agricultural input processing step
Ginning of raw seed cotton (separating fibers from seeds) must occur first before fibers can be processed.
Raw cotton harvested from farms contains seeds that must be removed mechanically at ginning mills.
2
Determine the intermediate fiber-processing stage
Cleaned lint is spun into yarn or thread.
Individual cotton fibers must be twisted and drawn out into continuous threads before fabric construction.
3
Determine the fabric construction stage
Yarn is woven or knitted into gray (raw) cloth.
Industrial looms interlock the spun yarn into a cohesive sheet of fabric.
4
Identify the final value-addition and finishing stage
Fabric is dyed, printed, and finished for commercial sale.
Chemical treatment, coloring, and patterning are the final steps before textiles are sent to apparel manufacturers or markets.

Key Concept

Agro-Allied Textile Processing Stages in Nigeria
Estimated Time:1m 0s
Question 6305Question

A sole trader maintains business records by recording only cash transactions and personal accounts of customers and suppliers, leaving out real and nominal accounts. Which of the following represents a major limitation of this system of record-keeping?

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Answer: The arithmetical accuracy of the accounting records cannot be verified by extracting a trial balance.

Answer

The arithmetical accuracy of the accounting records cannot be verified by extracting a trial balance.
Under a single entry or incomplete records system, double-entry rules are not fully observed across all accounts (nominal and real accounts are mostly omitted). Consequently, debit and credit balances cannot be extracted to prepare a trial balance, making it impossible to check the arithmetical accuracy of the books.

Step-by-Step Solution

1
Identify the features of the record-keeping system described in the stem.
The trader uses a single entry (incomplete records) system, which records cash and personal accounts but omits real and nominal accounts.
Single entry bookkeeping does not follow full double-entry principles for every transaction.
2
Analyze the fundamental limitations of single entry accounting.
Without dual ledger postings (debit and credit) for every transaction, a trial balance cannot be extracted to check arithmetical accuracy.
A trial balance relies on the equality of total debit and credit balances in double-entry bookkeeping.

Key Concept

Limitations of Single Entry System
Question 6306Question

Match each of the following major African relief features with its corresponding physiographic and structural characteristic.

Click a left item, then click its matching right item

Items

Ethiopian Massif
Congo Basin
Tibesti Massif
East African Rift System

Matches

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Answer

The correct pairings associate the Ethiopian Massif with high basaltic lava plateaus, the Congo Basin with interior downwarped crustal depressions, the Tibesti Massif with Saharan volcanic basement uplift, and the East African Rift System with elongated fault-bounded grabens.
The correct pairings accurately reflect the geomorphological origin and physical expression of major African landforms: the Ethiopian Massif is a basaltic volcanic plateau, the Congo Basin is a downwarped interior depression, the Tibesti Massif is a Saharan basement and volcanic uplift, and the East African Rift System is a fault-bounded tectonic trough.

Step-by-Step Solution

1
Analyze the structural and volcanic attributes of highland plateaus in East Africa.
Identify the Ethiopian Massif as a high basaltic plateau characterized by deep river gorges.
Ethiopia's highlands were formed by extensive flood basalt volcanism during Cenozoic rift-related uplift.
2
Examine the basin and drainage geography of Central Africa.
Identify the Congo Basin as a broad saucer-shaped interior depression formed by crustal sagging.
Regional crustal downwarping created vast interior drainage systems collecting alluvial sediments.
3
Evaluate the relict mountain structures across the Saharan shield.
Identify the Tibesti Massif as an elevated volcanic mass resting on ancient basement rocks within the Sahara.
Intraplate hot-spot activity and epeirogenic uplift produced high volcanic domes in northern Africa.
4
Classify the major extensional tectonic feature extending through Eastern Africa.
Identify the East African Rift System as a rift valley bounded by parallel faults forming steep graben structures.
Divergent tectonic forces cause continental crustal stretching, normal faulting, and linear lake formation.

Key Concept

Relief and Major Landforms of Africa
Question 6307Question

Match each geographic factor or historical process in Africa with its corresponding effect on regional population density.

Click a left item, then click its matching right item

Items

Widespread prevalence of the tsetse fly (glossina vector) in humid savanna zones
Intensive tin and columbite mining operations on the Jos Plateau
Perennial silt deposition and intensive irrigation in the Lower Nile Valley
Historical slave-raiding activities and 19th-century conflicts in the Middle Belt

Matches

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Answer

The geographic factors and historical processes correspond to their population outcomes as follows: Tsetse fly prevalence restricted pastoral and human settlement leading to low density; mineral exploitation on the Jos Plateau created an island of high population concentration; the Lower Nile Valley supports exceptionally high density through fertile soils and irrigation; and historical slave raiding in the Middle Belt created a persistent sparsely populated corridor.
Each factor is correctly matched to its demographic impact: disease vectors limit human and animal habitation; mineral resources attract dense settlement; fertile irrigated river valleys sustain maximum agricultural density; and historical warfare causes lasting depopulation.

Step-by-Step Solution

1
Analyze environmental disease vectors as density controls
Tsetse flies act as a biological barrier to livestock rearing and dense settlement in humid savannas.
Health hazards lower habitability and discourage long-term population concentration.
2
Evaluate economic attraction against physical constraints
Commercial mining of tin and columbite pulled labor to the Jos Plateau.
Economic incentives can overcome topographical challenges, creating high-density mineral enclaves.
3
Examine agricultural carrying capacity of river valleys
Alluvial deposits and perennial irrigation in the Nile Valley maximize crop yield per unit area.
High soil fertility and water security support dense rural and urban populations.
4
Assess historical impacts on spatial population distribution
Slave raids displaced communities in the West African Middle Belt.
Historical conflict can cause long-term depopulation independent of physical environmental capacity.

Key Concept

Physical, economic, and historical factors influencing population distribution and density
Estimated Time:2m 0s
Question 6308Question

Arrange the following major African lakes in order of their maximum depth, from the deepest to the shallowest.

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Answer

The correct order from deepest to shallowest is Lake Tanganyika, Lake Malawi, Lake Victoria, and Lake Chad.
The correct sequence arranges the lakes by maximum depth: Lake Tanganyika is the deepest (~1,470 m), followed by Lake Malawi (~706 m), Lake Victoria (~84 m), and finally Lake Chad (~11 m).

Step-by-Step Solution

1
Identify the deepest African lake formed within the East African Rift System.
Lake Tanganyika is the deepest in Africa (~1,470 m).
Rift valley formation creates steep, deep grabens.
2
Identify the second-deepest rift valley lake in Africa.
Lake Malawi is second (~706 m).
It occupies a deep rift trough similar to Tanganyika.
3
Identify the depth of the major downwarped basin lake.
Lake Victoria is third (~84 m).
It formed in a shallow saucer-shaped crustal depression rather than a rift fault.
4
Identify the shallowest interior drainage lake in the group.
Lake Chad is the shallowest (~11 m).
It is an endorheic basin lake with low average depth.

Key Concept

Relative depths and geomorphological origins of major African lakes
Question 6309Question

In an electronic accounting system, transaction data undergoes sequential stages during the data processing cycle. Arrange the following key stages of the accounting data processing cycle in the correct chronological order from start to finish.

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Answer

The correct chronological sequence of the accounting data processing cycle is Data Collection, followed by Data Input, Data Processing, Information Output, and finally Data Storage.
The accounting data processing cycle follows a logical progression starting from capturing raw economic events via source documents (Collection), entering those details into the accounting computer system (Input), performing ledger operations and calculations (Processing), producing accounting statements and management reports (Output), and preserving the data securely for future compliance and audit (Storage).

Step-by-Step Solution

1
Identify the origin of financial data
Data Collection (item_1) comes first as transactions must originate on source documents.
Without raw transaction documentation, no data can enter the system.
2
Determine how collected data enters the system
Data Input (item_2) follows Data Collection.
Collected source document details must be entered into software before calculation.
3
Analyze what happens to entered data
Data Processing (item_3) follows Data Input.
The system manipulates entered numbers, posting double entries and balancing accounts.
4
Identify the result of data manipulation
Information Output (item_4) follows Data Processing.
Reports and financial statements are generated from the processed ledger balances.
5
Determine the final preservation stage
Data Storage (item_5) is the final stage.
Completed reports and underlying databases are retained securely for future audit.

Key Concept

Accounting Data Processing Cycle
Question 6310Question

At the end of an accounting period, a sole trader reported a Gross Profit of N180,000\text{N}180,000 and received discounts totaling N2,500\text{N}2,500. A review of the business transactions and adjustments reveals the following:

- Rent paid of N24,000\text{N}24,000 included N4,000\text{N}4,000 paid in advance for the next period.
- Accrued electricity charges at year-end amounted to N3,000\text{N}3,000.
- The provision for doubtful debts is to be increased from N5,000\text{N}5,000 to N7,000\text{N}7,000.
- A new office computer purchased on credit for N10,000\text{N}10,000 was erroneously recorded in the general office expenses account.

Based on the information above, what is the correct Net Profit for the year?

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Answer: N157,500\text{N}157,500

Answer

The correct Net Profit for the year is N157,500\text{N}157,500.
The net profit is calculated by taking total income (Gross Profit of N180,000\text{N}180,000 plus Discount Received of N2,500=N182,500\text{N}2,500 = \text{N}182,500) and subtracting total operating expenses. Adjusted Rent is N20,000\text{N}20,000 (after deducting the N4,000\text{N}4,000 prepayment), Electricity accrued is N3,000\text{N}3,000, and the Increase in Provision for Doubtful Debts is N2,000\text{N}2,000 (N7,000N5,000\text{N}7,000 - \text{N}5,000). The office computer cost of N10,000\text{N}10,000 is a capital expenditure item and is excluded from expense calculations. Subtracting total expenses of N25,000\text{N}25,000 from total income yields N157,500\text{N}157,500.

Step-by-Step Solution

1
Calculate total gross income
Gross Profit+Discount Received=N180,000+N2,500=N182,500\text{Gross Profit} + \text{Discount Received} = \text{N}180,000 + \text{N}2,500 = \text{N}182,500
Discount received is revenue income and must be added to gross profit in the Profit and Loss Account.
2
Adjust individual expense accounts
Rent: N24,000N4,000=N20,000\text{N}24,000 - \text{N}4,000 = \text{N}20,000; Electricity: N3,000\text{N}3,000; Provision Increase: N7,000N5,000=N2,000\text{N}7,000 - \text{N}5,000 = \text{N}2,000
Prepayments must be subtracted from payments made, accrued expenses added, and only the net increase in provision for doubtful debts charged as an expense. Capital expenditure (computer) must be excluded entirely from operating expenses.
3
Compute Net Profit
Total IncomeTotal Operating Expenses=N182,500(N20,000+N3,000+N2,000)=N157,500\text{Total Income} - \text{Total Operating Expenses} = \text{N}182,500 - (\text{N}20,000 + \text{N}3,000 + \text{N}2,000) = \text{N}157,500
Deduct total revenue expenses from total gross income to determine net profit.

Key Concept

Determination of Net Profit incorporating adjustments for accruals, prepayments, provision increases, and capital expenditure misclassifications.
Estimated Time:1m 30s
Question 6311Question

Major commercial flour milling plants in Nigeria are overwhelmingly concentrated at seaport terminals such as Lagos and Port Harcourt rather than near inland agricultural hinterlands. According to Alfred Weber's industrial location theory, which factor primarily accounts for this spatial pattern?

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Answer: The reliance on heavy, bulk-imported raw wheat, making seaport locations the least-cost transshipment point to minimize raw material transport costs.

Answer

The primary factor is the reliance on heavy, bulk-imported raw wheat, which makes seaport locations the least-cost transshipment point for minimizing raw material transport costs.
Because wheat grain is imported in large maritime cargoes, processing it directly at port terminals (break-of-bulk points) avoids the high cost of transporting unrefined bulk grain inland before processing. This conforms to Weber's least-cost location principles for imported raw materials.

Step-by-Step Solution

1
Identify the primary raw material source and transport characteristics of Nigeria's flour milling industry.
Wheat, the primary input for flour milling in Nigeria, is largely imported from overseas via ocean vessels into major seaports.
Determining where input materials originate establishes whether the industry is raw-material tied, market tied, or footloose.
2
Apply Weber's least-cost location principles to imported bulk materials.
Locating the milling plant at the seaport eliminates extra transshipment and inland transport expenses of heavy, unprocessed grain prior to milling.
Industrial location theory states that when a bulk raw material is imported by sea, the port serves as the break-of-bulk point where total transport costs are minimized.

Key Concept

Break-of-Bulk Point and Industrial Location for Imported Raw Materials
Estimated Time:1m 15s
Question 6312Question

Kofi, a fashion designer, maintains incomplete accounting records. On 1 January 2025, his total assets were ₦450,000 and total liabilities were ₦150,000. On 31 December 2025, his total assets were ₦780,000 and total liabilities were ₦220,000. During the year, Kofi withdrew ₦80,000 for personal use and introduced ₦50,000 additional capital into the business. What is the net profit for the year ending 31 December 2025?

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Answer: ₦290,000; 290,000; 290000; N290,000

Answer

The net profit for the year ending 31 December 2025 is ₦290,000.
To determine net profit using the Statement of Affairs method, opening capital is calculated as ₦450,000 - ₦150,000 = ₦300,000 and closing capital is calculated as ₦780,000 - ₦220,000 = ₦560,000. Applying the profit equation: Net Profit = Closing Capital (₦560,000) + Drawings (₦80,000) - Additional Capital (₦50,000) - Opening Capital (₦300,000) = ₦290,000.

Step-by-Step Solution

1
Calculate Opening Capital as at 1 January 2025
Opening Capital = ₦450,000 - ₦150,000 = ₦300,000
Capital is equal to Total Assets minus Total Liabilities at the start of the period.
2
Calculate Closing Capital as at 31 December 2025
Closing Capital = ₦780,000 - ₦220,000 = ₦560,000
Capital is equal to Total Assets minus Total Liabilities at the end of the period.
3
Apply the Statement of Affairs profit formula to determine Net Profit
Net Profit = (₦560,000 + ₦80,000 - ₦50,000) - ₦300,000 = ��290,000
Net Profit = (Closing Capital + Drawings - Additional Capital) - Opening Capital.

Key Concept

Statement of Affairs Method for Capital and Profit Determination
Estimated Time:1m 30s
Question 6313Question

Arrange the following major African vegetation zones in proper sequential order, starting from the Equator (0°) and moving northward toward the Tropic of Cancer (23.5°N).

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Answer

The correct sequential order from the Equator northward to the Tropic of Cancer is: Tropical Rainforest → Guinea Savanna → Sudan Savanna → Sahel Savanna → Sahara Desert.
Moving northward from the Equator to the Tropic of Cancer, annual rainfall progressively declines while the duration of the dry season increases. This creates a distinct latitudinal sequence of vegetation belts: Tropical Rainforest at the Equator (0°), transitioning to Guinea Savanna, Sudan Savanna, Sahel Savanna, and finally the Sahara Desert.

Step-by-Step Solution

1
Identify the vegetation zone closest to the Equator (0°).
Tropical Rainforest occupies the equatorial center characterized by high temperatures and abundant rainfall year-round.
Equatorial climate conditions produce dense rainforest vegetation.
2
Identify the moist savanna zone lying directly north of the rainforest.
Guinea Savanna borders the rainforest belt as annual precipitation begins to decrease.
A moderate dry season allows tall grasses and woodland trees to dominate.
3
Determine the drier savanna belt following the Guinea Savanna.
Sudan Savanna forms the next belt northward with lower annual rainfall.
Longer dry periods limit tree height and density, supporting shorter grasses and baobab trees.
4
Locate the semi-arid transition zone preceding the desert.
Sahel Savanna serves as the dry steppe transition zone between savanna and true desert.
Low, unpredictable rainfall supports only thorny scrubland and drought-hardy vegetation.
5
Identify the northern hyper-arid zone near the Tropic of Cancer.
Sahara Desert forms the final northern belt.
Subtropical high-pressure belts suppress rainfall, leading to desert landforms and xerophytic vegetation.

Key Concept

Latitudinal Zonation of African Vegetation Belts
Estimated Time:1m 30s
Question 6314Question

If 50 cm350\text{ cm}^3 of sulfur dioxide (SO2SO_2) gas diffuses through a porous partition in 20 seconds20\text{ seconds}, and the same volume of an unknown hydrocarbon gas ZZ diffuses under identical conditions of temperature and pressure in 10 seconds10\text{ seconds}, what is the molar mass of gas ZZ in g/mol\text{g/mol}? (Relative atomic masses: S=32\text{S} = 32, O=16\text{O} = 16)

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Answer: 16; 16 g/mol; 16g/mol; 16 g mol^-1

Answer

16 g/mol
According to Graham's Law, the time required for equal volumes of two gases to diffuse under the same conditions is directly proportional to the square root of their molar masses: t1t2=M1M2\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}. Sulfur dioxide (SO2SO_2) has a molar mass of 64 g/mol64\text{ g/mol} and takes 20 seconds20\text{ seconds}. Gas ZZ takes 10 seconds10\text{ seconds}, which is half the time, meaning its molar mass must be (12)2=14\left(\frac{1}{2}\right)^2 = \frac{1}{4} of the molar mass of SO2SO_2. Therefore, MZ=644=16 g/molM_Z = \frac{64}{4} = 16\text{ g/mol}.

Step-by-Step Solution

1
Determine the relative molar mass of sulfur dioxide (SO2SO_2).
Molar mass of SO2=32+2(16)=64 g/mol\text{Molar mass of } SO_2 = 32 + 2(16) = 64\text{ g/mol}.
The molar mass of the reference gas is needed to apply Graham's Law.
2
Set up the relationship between diffusion time (tt) and molar mass (MM) according to Graham's Law for equal volumes of gases.
tZtSO2=MZMSO2\frac{t_Z}{t_{SO_2}} = \sqrt{\frac{M_Z}{M_{SO_2}}}.
For equal volumes of gas, the time taken to diffuse is directly proportional to the square root of the molar mass.
3
Substitute the known diffusion times and molar mass of SO2SO_2 into the ratio.
1020=MZ64    12=MZ64\frac{10}{20} = \sqrt{\frac{M_Z}{64}} \implies \frac{1}{2} = \sqrt{\frac{M_Z}{64}}.
Insert tZ=10 st_Z = 10\text{ s}, tSO2=20 st_{SO_2} = 20\text{ s}, and MSO2=64 g/molM_{SO_2} = 64\text{ g/mol} into the equation.
4
Square both sides of the equation and solve for MZM_Z.
(12)2=MZ64    14=MZ64    MZ=16 g/mol\left(\frac{1}{2}\right)^2 = \frac{M_Z}{64} \implies \frac{1}{4} = \frac{M_Z}{64} \implies M_Z = 16\text{ g/mol}.
Eliminate the radical to calculate the molar mass of the unknown gas.

Key Concept

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass, which implies that the time taken for a given volume to diffuse is directly proportional to the square root of its molar mass.
Question 6315Question

A solar observation station located at longitude 23W23^\circ\text{W} records local solar noon (12:00 PM12:00\text{ PM}) at a specific instant. At that exact moment, a research vessel at sea notes its local solar time as 7:16 PM7:16\text{ PM} (19:1619:16) on the same day. What is the longitude of the research vessel in degrees East?

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Answer: 86

Answer

The research vessel is located at 86E86^\circ\text{E}.
The time difference between 12:00 PM and 7:16 PM is 7 hours and 16 minutes (436 minutes). Since 4 minutes correspond to 1° of longitude, the angular difference is 436 ÷ 4 = 109°. Because the vessel's solar time is later than the station's time, the vessel lies to the East. Moving 109° East from 23°W requires 23° to reach the 0° Greenwich Meridian and an additional 86° into the Eastern Hemisphere, placing the vessel at 86°E.

Step-by-Step Solution

1
Determine the time difference between the solar observation station and the research vessel.
Time difference = 19:16 - 12:00 = 7 hours and 16 minutes = 436 minutes.
Difference in local solar time corresponds to longitudinal separation.
2
Convert the time difference into longitudinal degrees using the rate of Earth's rotation (1=4 minutes1^\circ = 4\text{ minutes}).
Longitudinal difference = 436 ÷ 4 = 109°.
The Earth rotates 360° in 24 hours, which equals 1° for every 4 minutes of time difference.
3
Determine the direction of the vessel relative to the station.
The vessel is East of the station.
Local solar time at the vessel (7:16 PM) is ahead of the station (12:00 PM), meaning the vessel lies further East.
4
Calculate the absolute longitude in the Eastern Hemisphere.
Vessel longitude = 109° - 23° = 86°E.
Traversing 109° East starting from 23°W uses 23° to reach the Greenwich Meridian (0°) and the remaining 86° extends into the Eastern Hemisphere.

Key Concept

Calculating longitude from local solar time difference across meridians
Question 6316Question

Arrange the following agricultural land-use zones according to Von Thünen's model of agricultural location, starting from the central market outwards to the rural periphery:

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Answer

The correct order from the central market outwards is: Market gardening and intensive dairying, followed by Forestry and fuel wood production, then Arable crop farming with fallow rotation, and finally Extensive livestock ranching.
Von Thünen's model demonstrates that land use intensity decreases as distance from the central market increases due to land rent and transport costs. Market gardening and dairying occupy the first ring because produce is highly perishable. Forestry occupies the second ring owing to the heavy weight and high transport cost of wood. Arable cropping with fallow rotation occupies the middle rings, while extensive livestock ranching operates at the outer periphery where land rent is lowest.

Step-by-Step Solution

1
Analyze transport cost and perishability for the innermost zone.
Market gardening and intensive dairying require immediate proximity to urban consumers to prevent spoilage and minimize high unit transport costs.
Perishable goods demand the highest bid-rent location closest to the central market.
2
Determine the spatial position of heavy bulk commodities.
Forestry and fuel wood production occupy the second ring.
Timber is heavy and bulky to haul, placing its ideal location close to town before transport costs exceed product value.
3
Position crop farming systems requiring moderate land areas.
Arable crop farming with fallow rotation forms the middle agricultural belt.
Grains and field crops yield lower returns per hectare than market gardens, requiring cheaper land further from the core.
4
Identify the outermost land-use activity.
Extensive livestock ranching occupies the periphery.
Animals require extensive land per unit of output and can be walked to market, minimizing freight cost constraints.

Key Concept

Von Thünen's Model of Agricultural Land Use
Question 6317Question

Match each Nigerian mineral or energy resource site on the left with its corresponding geographical location or industrial application on the right.

Click a left item, then click its matching right item

Items

Shiroro Power Station
Coal deposits of the Mamu Formation
Itakpe Hill deposits
Ewekoro and Gboko reserves

Matches

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Answer

Shiroro Power Station matches the hydroelectric dam on the Kaduna River in Niger State; Coal deposits of the Mamu Formation match sub-bituminous coal fields centered around Enugu and Kogi States; Itakpe Hill deposits match high-grade iron ore reserves supplying the domestic steel industry in Kogi State; Ewekoro and Gboko reserves match sedimentary limestone deposits providing raw materials for cement manufacturing in Ogun and Benue States.
Each resource site correctly maps to its geographical location and primary application: Shiroro is a major hydroelectric station on the Kaduna River in Niger State; coal is extracted from Cretaceous Mamu Formation strata in Enugu and Kogi; Itakpe in Kogi contains high-grade iron ore reserves; Ewekoro and Gboko supply key limestone inputs for regional cement industries.

Step-by-Step Solution

1
Identify the primary river basin and host state for the Shiroro Hydroelectric Power scheme.
Shiroro Dam is situated across the Kaduna River in Niger State.
This establishes the correct hydroelectric energy site association.
2
Examine the geological formation and spatial occurrence of coal in Nigeria.
Coal occurs in the Cretaceous sedimentary Mamu Formation, notably at Enugu and Okaba in Kogi State.
This links sub-bituminous coal to its primary geological host formation and regions.
3
Locate the primary iron ore extraction field supporting Nigeria's iron and steel development.
Itakpe Hill in Kogi State contains major high-grade iron ore deposits.
Itakpe supplies the raw ore for domestic steel processing plants.
4
Correlate major industrial limestone quarrying sites with regional cement plants.
Ewekoro (Ogun State) and Gboko (Benue State) harbor rich sedimentary limestone beds.
Limestone is the fundamental calcium carbonate raw material for industrial cement production.

Key Concept

Geographical distribution, geological context, and industrial uses of Nigeria's mineral and energy resources
Question 6318Question

Which of the following landforms is formed by glacial deposition and is characterized as an elongated, streamlined, teardrop-shaped hill of unsorted till with its blunt end pointing towards the direction of ice advance?

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Answer: Drumlin

Answer

Drumlin
The correct answer is the drumlin. Drumlins are smooth, elongated, teardrop-shaped hills of unstratified glacial till deposited and shaped under moving ice. The steep, blunt end (stoss side) faces the direction from which the glacier advanced, while the gently sloping tail (lee side) points in the direction of ice movement.

Step-by-Step Solution

1
Identify the primary agent and process specified in the stem
The landform is formed by glacial deposition (ice carrying and dropping sediment).
Differentiating between erosional and depositional features narrows down glacial landform classifications.
2
Analyze the physical shape and orientation of the landform described
An elongated, teardrop-shaped hill composed of unsorted till with a steep blunt stoss side facing ice advance and a tapered lee side.
This specific morphology uniquely identifies a drumlin.
3
Distinguish from non-glacial and erosional distractors
Roche moutonnée is erosional, oxbow lake is fluvial, and scree slope is caused by weathering and mass wasting.
Eliminating features of other geological agents confirms the correct answer.

Key Concept

Glacial Depositional Landforms (Drumlins)
Estimated Time:50s
Question 6319Question

A progressive transverse wave traveling through a primary medium is governed by the mathematical wave equation y=0.05sin(100πt2.5πx)y = 0.05 \sin\left(100\pi t - 2.5\pi x\right), where xx and yy are measured in meters and tt is in seconds. Upon entering a secondary medium, the wave undergoes refraction such that its wavelength decreases by 20%20\%. What is the speed of the wave in the secondary medium in m/s\text{m/s}?

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Answer: 32

Answer

The speed of the wave in the secondary medium is 32 m/s32\text{ m/s}.
Comparing y=0.05sin(100πt2.5πx)y = 0.05 \sin\left(100\pi t - 2.5\pi x\right) with y=Asin(ωtkx)y = A \sin(\omega t - kx) yields ω=100π rad/s\omega = 100\pi\text{ rad/s} and k=2.5π rad/mk = 2.5\pi\text{ rad/m}. This gives a source frequency f=ω2π=50 Hzf = \frac{\omega}{2\pi} = 50\text{ Hz} and initial wavelength λ1=2πk=0.8 m\lambda_1 = \frac{2\pi}{k} = 0.8\text{ m}. Because wave frequency is invariant across boundaries, ff remains 50 Hz50\text{ Hz} in the second medium. The new wavelength is λ2=0.8 m×0.80=0.64 m\lambda_2 = 0.8\text{ m} \times 0.80 = 0.64\text{ m}. Consequently, the wave speed in the second medium is v2=50 Hz×0.64 m=32 m/sv_2 = 50\text{ Hz} \times 0.64\text{ m} = 32\text{ m/s}.

Step-by-Step Solution

1
Extract angular frequency and wave number from the wave equation
ω=100π rad/s\omega = 100\pi\text{ rad/s} and k=2.5π rad/mk = 2.5\pi\text{ rad/m}
Matching coefficients in y=Asin(ωtkx)y = A \sin(\omega t - kx) allows determination of temporal and spatial characteristics.
2
Determine wave frequency and original wavelength
f=50 Hzf = 50\text{ Hz} and λ1=0.8 m\lambda_1 = 0.8\text{ m}
Using fundamental relationships f=ω2πf = \frac{\omega}{2\pi} and λ=2πk\lambda = \frac{2\pi}{k}.
3
Calculate the refracted wavelength under constant frequency
λ2=0.64 m\lambda_2 = 0.64\text{ m} while ff remains 50 Hz50\text{ Hz}
Wave frequency depends solely on the wave source and remains unchanged during medium transitions.
4
Compute wave propagation speed in the new medium
v2=32 m/sv_2 = 32\text{ m/s}
Applying the wave equation v=fλv = f\lambda with the updated wavelength.

Key Concept

Wave equation parameter extraction and frequency invariance during refraction
Estimated Time:2m 0s
Question 6320Question

On a topographical map drawn to a scale of 1:25,0001 : 25,000, two hilltop stations, Station A and Station B, are separated by a map distance of 16 cm16\text{ cm}. If Station A is situated at an elevation of 580 m580\text{ m} above sea level and Station B is at an elevation of 180 m180\text{ m}, what is the gradient of the slope between the two stations expressed as the value NN in the ratio 1:N1 : N?

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Answer: 10

Answer

The denominator NN in the slope gradient ratio 1:N1 : N is 1010 (representing a gradient ratio of 1:101 : 10).
To calculate the gradient expressed as 1:N1 : N, determine the Vertical Interval (VI =580 m180 m=400 m= 580\text{ m} - 180\text{ m} = 400\text{ m}) and the Horizontal Equivalent (HE =16 cm×25,000=400,000 cm=4,000 m= 16\text{ cm} \times 25,000 = 400,000\text{ cm} = 4,000\text{ m}). Dividing VI by HE gives 400 m4,000 m=110\frac{400\text{ m}}{4,000\text{ m}} = \frac{1}{10}, which corresponds to a gradient of 1:101 : 10, making N=10N = 10.

Step-by-Step Solution

1
Determine the Vertical Interval (VI)
\text{VI} = 580\text{ m} - 180\text{ m} = 400\text{ m}
The Vertical Interval is the vertical height difference between the two specified elevations.
2
Calculate the Horizontal Equivalent (HE) on the ground
\text{HE} = 16\text{ cm} \times 25,000 = 400,000\text{ cm} = 4,000\text{ m}
Multiply the map distance by the scale denominator to find the ground distance, then divide by 100 to convert centimeters to meters.
3
Calculate the gradient ratio
\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{400\text{ m}}{4,000\text{ m}} = \frac{1}{10} = 1 : 10
Gradient is calculated as Vertical Interval divided by Horizontal Equivalent, simplified to a fraction with a numerator of 1.

Key Concept

Calculation of slope gradient using vertical interval, map distance, and representative fraction scale
Estimated Time:1m 30s
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