Tüm alıştırma soruları

13931 soru

Soru 3961Soru

Complete the statement regarding the choice of indicator and the expected endpoint color change when titrating aqueous ammonia with hydrochloric acid.

Aşağıdaki boşlukları doldurun

When titrating aqueous ammonia against hydrochloric acid, the appropriate indicator is , which changes from yellow in alkaline solution to at the equivalence point.
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Cevap

The appropriate indicator is methyl orange, which changes from yellow to red (or pink) at the equivalence point.
In a titration between a strong acid (HCl\text{HCl}) and a weak base (NH3\text{NH}_3), ammonium chloride (NH4Cl\text{NH}_4\text{Cl}) is formed. The hydrolysis of NH4+\text{NH}_4^+ ions makes the solution acidic at the equivalence point (pH<7\text{pH} < 7). Methyl orange is the correct indicator because its transition range (pH 3.14.4\text{pH } 3.1 - 4.4) coincides with this acidic equivalence point, turning red (or pink) when the endpoint is reached.

Adım Adım Çözüm

1
Analyze the nature of the reactants
Hydrochloric acid (HCl\text{HCl}) is a strong acid, and aqueous ammonia (NH3(aq)\text{NH}_3(aq)) is a weak base.
The relative strengths of the acid and base determine the pH at the equivalence point.
2
Determine the pH range at the equivalence point
The salt formed (NH4Cl\text{NH}_4\text{Cl}) undergoes hydrolysis, producing an acidic solution with an equivalence point at pH<7\text{pH} < 7 (typically between pH 3\text{pH } 3 and 66).
Salt hydrolysis of a strong acid and weak base yields ammonium ions (NH4+\text{NH}_4^+) which hydrolyze to produce H3O+\text{H}_3\text{O}^+ ions.
3
Select the suitable indicator and recall its color change
Methyl orange functions in the pH\text{pH} range of 3.14.43.1 - 4.4, changing from yellow in basic solution to red (or pink) at the acidic endpoint.
An indicator must undergo a distinct color transition within the steep pH\text{pH} region of the titration curve near the equivalence point.

Anahtar Kavram

Indicator Selection and Endpoint Determination in Volumetric Analysis
Tahmini Süre:1m 0s
Soru 3962Soru

Which of the following reagents reacts with propanal to produce a silver mirror, but shows no reaction with propanone?

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Cevap: Ammoniacal silver nitrate solution

Cevap

Ammoniacal silver nitrate solution
Ammoniacal silver nitrate solution (Tollen's reagent) acts as a mild oxidizing agent. It oxidizes propanal to propanoate ions while Ag+Ag^+ ions are reduced to elemental silver, depositing as a silver mirror on the container walls. Propanone resists oxidation by mild oxidizing agents and yields no silver deposit.

Adım Adım Çözüm

1
Identify the functional groups of the given compounds
Propanal (CH3CH2CHOCH_3CH_2CHO) is an alkanal (aldehyde), whereas propanone (CH3COCH3CH_3COCH_3) is an alkanone (ketone).
Alkanals possess a terminal carbonyl group with a hydrogen atom that is easily oxidized, while alkanones lack this hydrogen atom.
2
Determine the specific reagent that yields a silver mirror
Tollen's reagent (ammoniacal silver nitrate solution, [Ag(NH3)2]+[Ag(NH_3)_2]^+) is reduced by alkanals to form metallic silver (Ag(s)Ag(s)), which deposits on the glass wall as a silver mirror.
Alkanones cannot be easily oxidized by mild oxidizing agents like Tollen's reagent under standard conditions.

Anahtar Kavram

Distinction between alkanals and alkanones using Tollen's reagent
Soru 3963Soru

A biology student recorded four different representations for the scientific name of the housefly in a laboratory notebook. Which of the following options strictly adheres to the standard rules of Linnaean binomial nomenclature?

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Cevap: *Musca domestica*

Cevap

The format *Musca domestica*, where the genus name begins with a capital letter and the specific epithet is entirely in lowercase.
The correct answer properly applies Linnaean conventions by placing the genus name first with an initial capital letter (*Musca*), followed by the specific epithet entirely in lowercase (*domestica*), with both terms italicized.

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1
Identify the two components of a scientific name according to binomial nomenclature rules.
The first name represents the Genus (generic name) and the second name represents the species (specific epithet).
Binomial nomenclature assigns every organism a two-part Latinized scientific name.
2
Apply standard capitalization rules to both parts of the binomial name.
The Genus name must start with a capital letter (*Musca*), whereas the species epithet must be written entirely in lowercase (*domestica*).
Linnaean formatting conventions require proper capitalization to distinguish taxonomy levels.
3
Verify sequence and typographic style.
The genus name must precede the specific epithet, and both terms should be italicized when typed (or underlined separately when handwritten).
Italicization or underlining indicates foreign (Latin) scientific nomenclature.

Anahtar Kavram

Rules of Linnaean Binomial Nomenclature
Tahmini Süre:1m 0s
Soru 3964Soru

Arrange the following major stages of cement manufacturing in the correct chronological sequence, from initial raw material handling to the final product stage.

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The correct sequence of cement production is: Extraction of limestone and clay, Crushing and blending raw meal, Heating in a rotary kiln to produce clinker, and Grinding clinker with gypsum.
Cement manufacturing begins with raw material extraction from quarries (limestone and clay). These raw materials are crushed and blended to form raw meal. The mixture is then calcined in a rotary kiln at temperatures up to 1450 °C to form clinker nodules. Finally, clinker is pulverized with a small quantity of gypsum (to retard flash setting) to produce commercial cement powder.

Adım Adım Çözüm

1
Identify raw material acquisition
Quarrying of limestone and clay occurs first.
Industrial processes must start with retrieving raw natural resources from the site.
2
Identify raw material preparation
Crushing and blending limestone and clay into raw meal occurs second.
The raw materials must be homogenized and reduced in size prior to high-temperature reaction.
3
Identify chemical transformation
Rotary kiln heating to form clinker occurs third.
High-temperature calcination converts calcium carbonate to calcium oxide, forming nodules of clinker.
4
Identify final processing and additive mixing
Grinding clinker with gypsum occurs last.
Gypsum is added at the final step to regulate the setting time of the finished cement.

Anahtar Kavram

Raw material processing sequence in cement chemical industry
Soru 3965Soru

Match each vertebrate group in Column I with its characteristic heart structure and circulatory pattern in Column II.

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Öğeler

Fish
Amphibians
Reptiles (excluding crocodiles)
Birds and Mammals

Eşleşmeler

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Cevap

Fish matches with Two-chambered heart with single circulation; Amphibians match with Three-chambered heart with a completely undivided ventricle; Reptiles match with Three-chambered heart with a partially divided ventricle; Birds and Mammals match with Four-chambered heart with complete double circulation.
Fish have a two-chambered heart with single circulation. Amphibians possess a three-chambered heart with two atria and an undivided ventricle. Non-crocodilian reptiles have a three-chambered heart featuring a partially divided ventricle. Birds and mammals possess a four-chambered heart providing complete separation of systemic and pulmonary circulation.

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1
Determine the heart structure of Fish.
Fish have 1 atrium and 1 ventricle operating in single circulation.
Deoxygenated blood enters the atrium, moves to the ventricle, and is pumped to the gills before flowing to the rest of the body.
2
Determine the heart structure of Amphibians.
Amphibians have 2 atria and 1 undivided ventricle.
Oxygenated blood from the lungs/skin and deoxygenated blood from the body enter separate atria but meet in a common ventricle.
3
Determine the heart structure of non-crocodilian Reptiles.
Reptiles have 2 atria and 1 ventricle partially divided by an incomplete septum.
The partial septum helps reduce the mixing of oxygenated and deoxygenated blood compared to amphibians.
4
Determine the heart structure of Birds and Mammals.
Birds and mammals have 2 atria and 2 completely separated ventricles.
Complete division prevents mixing of oxygenated and deoxygenated blood, supporting high metabolic rates.

Anahtar Kavram

Comparative anatomy of vertebrate hearts and circulatory pathways
Soru 3966Soru

In the industrial extraction of sodium metal using the Downs cell, calcium chloride (CaCl2\text{CaCl}_2) is added to molten sodium chloride (NaCl\text{NaCl}) primarily to:

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Cevap: lower the melting point of the sodium chloride electrolyte

Cevap

Calcium chloride is added to lower the melting point of the sodium chloride electrolyte.
Pure sodium chloride melts at 801C801^\circ\text{C}. Adding calcium chloride lowers the melting point of the mixture to about 600C600^\circ\text{C}, which conserves energy and minimizes evaporation of the liberated sodium metal during electrolysis in the Downs cell.

Adım Adım Çözüm

1
Identify the physical properties of pure sodium chloride
Pure NaCl\text{NaCl} has a high melting point of approximately 801C801^\circ\text{C}.
Maintaining such a high temperature industrially requires significant thermal energy.
2
Determine the effect of adding calcium chloride (CaCl2\text{CaCl}_2)
Mixing CaCl2\text{CaCl}_2 with NaCl\text{NaCl} forms an electrolytic mixture with a reduced melting point of about 600C600^\circ\text{C}.
Lowering the melting point makes the process economically viable and reduces heat loss and electrode wear.

Anahtar Kavram

Function of flux/admixture in the Downs process for sodium extraction
Tahmini Süre:45s
Soru 3967Soru

Consider the organic compound 3-methylbut-1-yne, which has the condensed structural formula HCCCH(CH3)2HC\equiv C-CH(CH_3)_2. How many carbon atoms in a single molecule of this compound are sp3sp^3 hybridized?

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Cevap: 3; three; 3 carbon atoms; 3 carbons

Cevap

There are 3 sp3sp^3 hybridized carbon atoms in one molecule of 3-methylbut-1-yne.
In 3-methylbut-1-yne (HCCCH(CH3)2HC\equiv C-CH(CH_3)_2), there are 5 total carbon atoms. The two terminal/alkyne carbons (C1C_1 and C2C_2) participate in a triple bond, giving them 2 σ\sigma bonds each and an spsp hybridization state. The central methine carbon (C3C_3) is bonded to four distinct atoms (C2C_2, HH, and two methyl carbons) via single σ\sigma bonds, making it sp3sp^3 hybridized. The two methyl group carbons (C4C_4 and C5C_5) are each single-bonded to three hydrogen atoms and C3C_3, making them sp3sp^3 hybridized as well. Thus, exactly 3 carbon atoms are sp3sp^3 hybridized.

Adım Adım Çözüm

1
Expand the condensed structural formula to identify every carbon atom and its bonding environment.
The expanded structure is HC1C2C3H(C4H3)(C5H3)H-C_1 \equiv C_2 - C_3H(C_4H_3)(C_5H_3), containing a total of 5 carbon atoms.
Expanding the formula clarifies the number of single (σ\sigma) and multiple bonds connected to each carbon atom.
2
Determine the hybridization state of the triply bonded carbon atoms (C1C_1 and C2C_2).
C1C_1 and C2C_2 are each involved in one triple bond and one single bond, forming 2 σ\sigma bonds and 2 π\pi bonds. Thus, both C1C_1 and C2C_2 are spsp hybridized.
A carbon atom with 2 steric domains (linear geometry) uses spsp hybrid orbitals.
3
Determine the hybridization state of the methine carbon atom (C3C_3) and the two methyl carbon atoms (C4C_4 and C5C_5).
C3C_3 forms four single σ\sigma bonds (one to C2C_2, one to HH, and two to methyl carbons). C4C_4 and C5C_5 each form four single σ\sigma bonds (one to C3C_3 and three to HH). Therefore, C3C_3, C4C_4, and C5C_5 are all sp3sp^3 hybridized.
A carbon atom bonded to 4 separate atoms via single σ\sigma bonds has 4 steric domains (tetrahedral geometry) and undergoes sp3sp^3 hybridization.
4
Count the total number of sp3sp^3 hybridized carbon atoms.
3 carbon atoms (C3C_3, C4C_4, and C5C_5) are sp3sp^3 hybridized.
Combining the results from steps 2 and 3 gives 2 spsp carbons and 3 sp3sp^3 carbons.

Anahtar Kavram

Identification of carbon hybridization states (sp3,sp2,spsp^3, sp^2, sp) in aliphatic molecules
Tahmini Süre:1m 30s
Soru 3968Soru

The table below shows the distribution of three distinct index fossils (XX, YY, and ZZ) found within four undisturbed sedimentary rock strata (Layer 1 being the lowermost and oldest, and Layer 4 being the uppermost and youngest):

Rock LayerFossils Present
Layer 4 (Top)Fossil ZZ
Layer 3Fossil YY, Fossil ZZ
Layer 2Fossil XX, Fossil YY
Layer 1 (Bottom)Fossil XX

Based on the law of superposition and paleontology principles, which of the following deductions regarding the evolutionary timeline of these organisms is correct?

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Cevap: Fossil XX represents the oldest ancestral organism, while Fossil ZZ evolved most recently in the geological timeline.

Cevap

Fossil XX represents the oldest ancestral organism, while Fossil ZZ evolved most recently in the geological timeline.
The law of superposition dictates that in undisturbed sedimentary rock sequences, the deepest layer (Layer 1) is the oldest and the uppermost layer (Layer 4) is the youngest. Because Fossil XX is found in Layer 1, it represents the earliest organism in the record. Fossil ZZ, occurring in Layer 4, represents the most recently evolved organism.

Adım Adım Çözüm

1
Analyze the rock strata sequence using the Law of Superposition.
Layer 1 (bottom) is the oldest sedimentary deposit, followed by Layer 2, Layer 3, and Layer 4 (top, youngest).
In undisturbed sedimentary rock layers, deeper layers are deposited first and are older than overlying layers.
2
Map the occurrence of each fossil to its corresponding geological timeframe.
Fossil XX is present in Layers 1 and 2 (oldest timeframe); Fossil YY is present in Layers 2 and 3 (intermediate timeframe); Fossil ZZ is present in Layers 3 and 4 (youngest timeframe).
Fossils preserved in specific strata indicate the geological period during which those organisms lived.
3
Deduce the relative age and evolutionary succession of the organisms.
The evolutionary chronological order from oldest to newest is Fossil XX \rightarrow Fossil YY \rightarrow Fossil ZZ.
Sequential appearance of index fossils across rock strata reflects the chronological order of biological evolution over time.

Anahtar Kavram

Evidence for Evolution: Paleontology and Fossil Records
Tahmini Süre:2m 0s
Soru 3969Soru

The flipper of a whale and the wing of a bat possess a similar internal arrangement of bones derived from a common ancestral pentadactyl structure, despite being adapted for different functional roles. Which of the following terms best describes these anatomical features?

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Cevap: Homologous structures

Cevap

Homologous structures
Homologous structures are anatomical features in different species that share a common fundamental structure and developmental origin because they were inherited from a common ancestor. The pentadactyl limb layout found in whale flippers, bat wings, and human arms is a classic example of homology demonstrating evolutionary divergence.

Adım Adım Çözüm

1
Analyze the anatomical characteristics described in the stem.
The whale flipper and bat wing share a common underlying skeletal plan (pentadactyl limb structure) derived from a common ancestor, but carry out different functions (swimming vs. flying).
Structures with shared evolutionary origin and basic structural framework are classified as homologous.
2
Distinguish between homologous and analogous structures.
Homology indicates divergent evolution from a common ancestor (same origin, different function), whereas analogy indicates convergent evolution (different origin, similar function).
Understanding structural origins clarifies why pentadactyl limbs in mammals are homologous.

Anahtar Kavram

Homologous vs Analogous Anatomical Evidence for Evolution
Soru 3970Soru

Match each homologous series of organic compounds listed on the left with its correct general molecular formula on the right.

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Öğeler

Alkanals (Aldehydes)
Alkanoic acids (Carboxylic acids)
Alkynes
Alkanols (Alcohols)

Eşleşmeler

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Cevap

Alkanals match with CnH2nOC_n H_{2n}O; Alkanoic acids match with CnH2nO2C_n H_{2n}O_2; Alkynes match with CnH2n2C_n H_{2n-2}; Alkanols match with CnH2n+2OC_n H_{2n+2}O.
Each class of organic compound is defined by its characteristic functional group and degree of unsaturation. Alkanols are saturated single-oxygen compounds (CnH2n+2OC_n H_{2n+2}O), Alkanals feature one carbonyl double bond (CnH2nOC_n H_{2n}O), Alkanoic acids feature a carboxyl group (CnH2nO2C_n H_{2n}O_2), and Alkynes contain a triple bond reducing hydrogen content to CnH2n2C_n H_{2n-2}.

Adım Adım Çözüm

1
Determine the general formula for aliphatic monohydric alkanols.
Alkanols consist of an alkyl group (CnH2n+1C_n H_{2n+1}) bonded to a hydroxyl group (OH-\text{OH}), giving the molecular formula CnH2n+2OC_n H_{2n+2}O.
Saturated aliphatic monohydric alcohols have the maximum possible hydrogen-to-carbon ratio for oxygenated single-bonded species.
2
Determine the general formula for alkanals.
Alkanals contain a terminal carbonyl group (CHO-\text{CHO}), introducing one double bond (C=O\text{C=O}) which reduces the hydrogen atom count by two compared to alkanols, giving CnH2nOC_n H_{2n}O.
Each site of unsaturation (such as a π\pi-bond) decreases the hydrogen atom count by two.
3
Determine the general formula for alkanoic acids.
Alkanoic acids contain a carboxyl group (COOH-\text{COOH}), giving two oxygen atoms and one carbon-oxygen double bond, yielding CnH2nO2C_n H_{2n}O_2.
Carboxylic acids are functional group isomers of esters and share the general formula CnH2nO2C_n H_{2n}O_2.
4
Determine the general formula for alkynes.
Alkynes possess one carbon-carbon triple bond (two π\pi-bonds), reducing the hydrogen count by four relative to alkanes (CnH2n+2C_n H_{2n+2}), resulting in CnH2n2C_n H_{2n-2}.
A triple bond accounts for two degrees of unsaturation.

Anahtar Kavram

General Molecular Formulas of Organic Homologous Series
Tahmini Süre:1m 15s
Soru 3971Soru

At the end of the financial year, a trader's inventory has a cost price of N45,000\text{N}45,000 and a net realizable value of N40,000\text{N}40,000. In accordance with the prudence concept, at what value should the closing stock be recorded in the final accounts?

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Cevap: N40,000\text{N}40,000

Cevap

The closing stock should be recorded at N40,000\text{N}40,000, which is the lower of cost and net realizable value.
Under the prudence concept and standard accounting rules (IAS 2 / SAS 4), closing stock must be recorded at the lower of cost and net realizable value. Since the net realizable value (N40,000\text{N}40,000) is lower than the cost price (N45,000\text{N}45,000), the inventory is valued at N40,000\text{N}40,000.

Adım Adım Çözüm

1
Identify the given inventory values
Cost price = N45,000\text{N}45,000; Net Realizable Value (NRV) = N40,000\text{N}40,000.
Both figures are required to apply the valuation rule.
2
Apply the inventory valuation rule based on the prudence concept
Lower value = N40,000\text{N}40,000.
Accounting standards dictate that inventory must be valued at the lower of cost and net realizable value to avoid overstating assets and profit.

Anahtar Kavram

Lower of Cost and Net Realizable Value (Prudence Concept)
Tahmini Süre:45s
Soru 3972Soru

Which of the following accounting entries is correct when a business purchases office equipment on credit from Ade & Co.?

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Cevap: Debit Equipment Account, Credit Ade & Co. Account

Cevap

Debit Equipment Account, Credit Ade & Co. Account
When office equipment is purchased on credit, the business acquires a non-current asset (which must be debited) and incurs a financial obligation/liability to the supplier (which must be credited).

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1
Identify the nature of the accounts involved in the transaction.
Equipment Account is a Real/Asset account, and Ade & Co. Account is a Personal/Liability account.
Office equipment is a non-current asset being acquired, while Ade & Co. is a trade creditor.
2
Apply the double entry rules of bookkeeping.
Debit Equipment Account (increase in asset) and Credit Ade & Co. Account (increase in liability).
The rule states: Debit what comes in (or increase in assets) and Credit the giver (or increase in liabilities).

Anahtar Kavram

Double Entry Rules for Credit Purchases of Non-Current Assets
Soru 3973Soru

Chidi, an auto mechanics workshop owner, maintains incomplete accounting records. The details of his assets and liabilities at the beginning and end of the year 2025 are presented in the table below:

Item1 January 2025 (₦)31 December 2025 (₦)
Equipment (Cost)500,000500,000
Inventory180,000210,000
Trade Debtors120,000150,000
Bank Balance60,00045,000
Rent Accrued10,000
Rent Prepaid15,000
Trade Creditors110,00095,000
Electricity Accrued8,000

Additional information for the year ended 31 December 2025:
1. Equipment is to be depreciated at 10%10\% per annum on cost.
2. A provision for doubtful debts of 5%5\% is to be created on trade debtors at year-end.
3. Chidi introduced additional capital of ₦40,000 into the business during the year.
4. He withdrew ₦3,000 cash monthly for personal use and took workshop spare parts valued at ₦14,000 for private use.

What is Chidi's net profit or loss for the year ended 31 December 2025?

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Cevap: ₦29,500 net profit

Cevap

₦29,500 net profit
The correct answer of ₦29,500 net profit is obtained by properly preparing both opening and closing Statements of Affairs with necessary asset valuation adjustments. Opening capital is ₦740,000 (₦860,000 assets minus ₦120,000 liabilities). Closing capital after 10% depreciation on equipment (₦50,000) and 5% provision on debtors (₦7,500) is ₦759,500 (₦862,500 assets minus ₦103,000 liabilities). Applying the formula Profit = Closing Capital + Total Drawings (₦50,000) - Additional Capital (₦40,000) - Opening Capital (₦740,000) yields ₦29,500.

Adım Adım Çözüm

1
Calculate Opening Capital at 1 January 2025
Total Opening Assets = ₦500,000 (Equipment) + ₦180,000 (Inventory) + ₦120,000 (Debtors) + ₦60,000 (Bank) = ₦860,000.
Total Opening Liabilities = ₦110,000 (Creditors) + ₦10,000 (Rent Accrued) = ₦120,000.
Opening Capital = ₦860,000 - ₦120,000 = ₦740,000.
Opening capital is derived from the opening Statement of Affairs by deducting opening liabilities from opening assets.
2
Calculate Adjusted Closing Capital at 31 December 2025
Equipment Net = ₦500,000 - (10% × ₦500,000) = ₦450,000.
Debtors Net = ₦150,000 - (5% × ₦150,000) = ₦142,500.
Total Adjusted Closing Assets = ₦450,000 + ₦210,000 + ₦142,500 + ₦45,000 + ₦15,000 (Prepaid Rent) = ₦862,500.
Total Closing Liabilities = ₦95,000 (Creditors) + ₦8,000 (Electricity Accrued) = ₦103,000.
Closing Capital = ₦862,500 - ₦103,000 = ₦759,500.
Closing capital must reflect all year-end adjustments including non-current asset depreciation, provisions, accruals, and prepayments.
3
Determine Total Owner Drawings
Cash Drawings = 12 × ₦3,000 = ₦36,000.
Goods Drawings = ₦14,000.
Total Drawings = ₦36,000 + ₦14,000 = ₦50,000.
Drawings comprise both monetary withdrawals and business inventory/assets taken for personal use.
4
Calculate Net Profit using the Capital Comparison Formula
Net Profit = (Closing Capital + Total Drawings - Additional Capital) - Opening Capital
Net Profit = (₦759,500 + ₦50,000 - ₦40,000) - ₦740,000 = ₦769,500 - ₦740,000 = ₦29,500.
Applying the standard Statement of Affairs profit equation gives the true operational profit for the financial year.

Anahtar Kavram

Statement of Affairs Method for Capital and Profit Determination
Soru 3974Soru

In the balance sheet of a sole trader, under which section should accrued expenses at the end of the accounting period be classified?

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Cevap: Current liabilities

Cevap

Current liabilities
Accrued expenses are amounts owed by the firm for goods or services already received during the financial year. Since these obligations are short-term and payable within 12 months, they are classified under current liabilities in the balance sheet.

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1
Identify the nature of accrued expenses
Accrued expenses are expenses incurred during the accounting period but not yet paid for by the balance sheet date.
Understanding the definition of an accrual determines whether it is an asset, liability, or capital item.
2
Determine the time horizon for settlement
These expenses are expected to be paid within the next short-term operating cycle (less than 12 months).
Obligations due within one year are classified as current rather than non-current.
3
Classify the item under the appropriate balance sheet heading
Classify accrued expenses as current liabilities.
Current liabilities group together short-term debts and obligations owed by the entity.

Anahtar Kavram

Balance Sheet Classification of Current Liabilities
Soru 3975Soru

Match each contour line pattern on a topographical map with the corresponding relief feature it represents.

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Öğeler

Concentric closed contours with values increasing towards the center
V-shaped contours pointing towards higher elevation (uphill)
Contour lines drawn very close together

Eşleşmeler

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Cevap

Concentric closed contours with increasing values inward match Hill top or hill peak; V-shaped contours pointing uphill match Valley or stream channel; Closely spaced contour lines match Steep slope or cliff.
Each contour pattern uniquely corresponds to a fundamental topographic feature according to standard map reading conventions.

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1
Identify the arrangement of concentric closed contours increasing inward.
Matched to Hill top or hill peak.
As height increases towards the middle of closed shapes, it shows a rising landform such as a hill summit.
2
Determine the landform associated with V-shaped contours pointing uphill.
Matched to Valley or stream channel.
V-shaped contour apexes pointing towards higher elevations signify drainage lines or valleys.
3
Examine the spacing of contour lines placed close together.
Matched to Steep slope or cliff.
Tight contour spacing indicates rapid elevation change over a small horizontal ground distance.

Anahtar Kavram

Relief representation and contour pattern identification
Soru 3976Soru

Kofi & Sons Enterprises operates two departments, Department P and Department Q. At the end of the trading period, total Rent and Rates incurred amounted to 200,000\text{₦}200,000. The records provide the following information:

- Floor area occupied: Department P = 1,200 m21,200\text{ m}^2; Department Q = 800 m2800\text{ m}^2
- Sales turnover: Department P = 300,000\text{₦}300,000; Department Q = 700,000\text{₦}700,000

What is the correct amount of Rent and Rates to be apportioned to Department P?

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Cevap: ₦120,000

Cevap

The correct amount of Rent and Rates to be apportioned to Department P is ₦120,000.
Rent and Rates are space-based expenses and must be apportioned based on floor area occupied. Department P occupies 1,200 m21,200\text{ m}^2 out of a total 2,000 m22,000\text{ m}^2, which is 35\frac{3}{5} or 60%60\%. Multiplying 60%60\% by total rent of 200,000\text{₦}200,000 yields 120,000\text{₦}120,000.

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1
Identify the appropriate basis for apportioning Rent and Rates.
Rent and Rates should be apportioned according to floor area occupied by each department.
Rent is a space-related indirect expense, so floor area provides an equitable basis of apportionment.
2
Calculate total floor area and Department P's share ratio.
Total floor area = 1,200 m2+800 m2=2,000 m21,200\text{ m}^2 + 800\text{ m}^2 = 2,000\text{ m}^2. Department P's ratio = 1,2002,000=35\frac{1,200}{2,000} = \frac{3}{5}.
Determining the proportion of space utilized by Department P relative to the total space.
3
Calculate Department P's share of total Rent and Rates.
Department P's Rent = 35×200,000=120,000\frac{3}{5} \times \text{₦}200,000 = \text{₦}120,000.
Multiplying the total expense by Department P's floor area proportion.

Anahtar Kavram

Apportionment of Indirect Departmental Expenses
Tahmini Süre:1m 30s
Soru 3977Soru

At 31st December 2025, a sole trader's trial balance showed Trade Debtors of ₦50,000. Additional bad debts of ₦2,000 are to be written off, and a provision for doubtful debts is to be created at 5% on the remaining trade debtors. What is the net trade debtors figure to be presented in the Statement of Financial Position?

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Cevap: 45600

Cevap

The net trade debtors figure to be presented in the Statement of Financial Position is ₦45,600.
To calculate net trade debtors for the Statement of Financial Position, first subtract the ₦2,000 additional bad debts from the gross debtors of ₦50,000 to get ₦48,000 adjusted debtors. Next, calculate 5% of ₦48,000, which gives a provision for doubtful debts of ₦2,400. Subtracting ₦2,400 from ₦48,000 yields the net debtors figure of ₦45,600.

Adım Adım Çözüm

1
Deduct the additional bad debts from gross trade debtors
Adjusted Debtors = ₦50,000 - ₦2,000 = ₦48,000
Bad debts identified at year-end must be written off against gross debtors first before calculating the required percentage provision.
2
Calculate the 5% provision for doubtful debts on the adjusted debtors balance
Provision = 5% × ₦48,000 = ₦2,400
The provision for doubtful debts is estimated based on the net collectible debtors.
3
Deduct the provision for doubtful debts from the adjusted debtors balance
Net Debtors = ₦48,000 - ₦2,400 = ₦45,600
Net debtors are reported in the Statement of Financial Position after subtracting the provision for doubtful debts.

Anahtar Kavram

Adjustment for bad debts written off and provision for doubtful debts in final accounts
Soru 3978Soru

On a topographical map drawn to a Representative Fraction (RF) scale of 1:80,0001 : 80,000, the measured distance along a proposed drainage canal between two agricultural communities is 17.5 cm17.5\text{ cm}. What is the actual ground distance of the canal in kilometers?

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Cevap: 14.0 km14.0\text{ km}

Cevap

The actual ground distance of the drainage canal is 14.0 km14.0\text{ km}.
The option showing 14.0 km14.0\text{ km} is correct because multiplying the map distance (17.5 cm17.5\text{ cm}) by the RF scale factor (80,00080,000) gives a ground measurement of 1,400,000 cm1,400,000\text{ cm}. Converting to kilometers by dividing by 100,000 cm/km100,000\text{ cm/km} yields exactly 14.0 km14.0\text{ km}.

Adım Adım Çözüm

1
Identify the given map scale and distance
Representative Fraction scale = 1:80,0001 : 80,000; Map distance = 17.5 cm17.5\text{ cm}.
Establishing known values is essential for scale conversion.
2
Calculate the ground distance in centimeters
Ground distance in cm = 17.5 cm×80,000=1,400,000 cm17.5\text{ cm} \times 80,000 = 1,400,000\text{ cm}.
The RF denominator indicates that 1 cm1\text{ cm} on the map represents 80,000 cm80,000\text{ cm} on the ground.
3
Convert the ground distance from centimeters to kilometers
Ground distance in km = 1,400,000 cm100,000 cm/km=14.0 km\frac{1,400,000\text{ cm}}{100,000\text{ cm/km}} = 14.0\text{ km}.
Since 1 km=100,000 cm1\text{ km} = 100,000\text{ cm}, dividing by 100,000100,000 converts centimeters to kilometers.

Anahtar Kavram

Ground Distance Calculation using Representative Fraction (RF)
Tahmini Süre:1m 30s
Soru 3979Soru

A sole trader's trial balance extracts show Motor Vehicles at a cost of 800,000\text{₦}800,000. If depreciation is to be charged at 15%15\% per annum on cost, what is the depreciation expense to be debited to the Profit and Loss Account for the year?

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Cevap: ₦120,000

Cevap

₦120,000
Under the straight-line method, the annual depreciation expense is calculated by multiplying the specified depreciation rate by the asset's original cost: 15%×800,000=120,00015\% \times \text{₦}800,000 = \text{₦}120,000. This amount is debited to the Profit and Loss Account as an expense.

Adım Adım Çözüm

1
Identify the depreciation method and rates given in the problem statement.
Method: Straight-line (on cost), Cost = 800,000\text{₦}800,000, Rate = 15%15\%.
The straight-line method calculates depreciation as a fixed percentage of the asset's original cost.
2
Compute the annual depreciation expense to be charged to the Profit and Loss Account.
Depreciation=15%×800,000=120,000\text{Depreciation} = 15\% \times \text{₦}800,000 = \text{₦}120,000.
This is the annual charge reflecting the wear and tear of the fixed asset for the current period.

Anahtar Kavram

Straight-line method of depreciation adjustment in final accounts
Soru 3980Soru

Match each goodwill transaction or valuation method in partnership accounts with its corresponding accounting treatment or valuation rule.

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Öğeler

Raising goodwill in the partnership books
Writing off goodwill in the partnership books
Valuing goodwill using the Average Profit method

Eşleşmeler

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Cevap

Raising goodwill matches crediting old partners' capital accounts in the old profit-sharing ratio; Writing off goodwill matches debiting partners' capital accounts in the new profit-sharing ratio; Valuing goodwill using the Average Profit method matches multiplying average profits by the agreed number of years' purchase.
Each item correctly matches standard partnership accounting rules: raising goodwill credits old partners in their old profit-sharing ratio, writing off goodwill debits partners in their new profit-sharing ratio, and the average profit valuation method computes goodwill by multiplying average profit by the specified number of years' purchase.

Adım Adım Çözüm

1
Determine the double entry for raising goodwill in a partnership
Debit Goodwill Account and credit Old Partners' Capital Accounts in their old profit-sharing ratio.
Existing partners are credited for the goodwill accrued up to the date of reconstitution based on their historical profit share.
2
Determine the double entry for writing off goodwill
Debit all partners' capital accounts in the new profit-sharing ratio and credit Goodwill Account.
Writing off removes goodwill from the balance sheet while adjusting partners' capitals according to the new profit-sharing arrangement.
3
Determine the formula for the Average Profit valuation method
Goodwill = Average Annual Profit ×\times Number of Years' Purchase.
This method estimates expected future super-normal earnings based on past average performance.

Anahtar Kavram

Treatment and Valuation of Goodwill in Partnership Accounts
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