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Question 6661Question

A saturated solution of potassium chloride (KCl\text{KCl}) contains 14.9 g14.9\text{ g} of the salt dissolved in 100 g100\text{ g} of water at 25C25^\circ\text{C}. What is the solubility of potassium chloride in mol/dm3\text{mol/dm}^3 at this temperature?

(Take density of water = 1.0 g/cm31.0\text{ g/cm}^3, relative atomic masses: K=39\text{K} = 39, Cl=35.5\text{Cl} = 35.5)

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Answer: 2

Answer

The solubility of potassium chloride at 25C25^\circ\text{C} is 2.0 mol/dm32.0\text{ mol/dm}^3.
The correct calculated value is 2.0 mol/dm32.0\text{ mol/dm}^3. Molar mass of KCl=39+35.5=74.5 g/mol\text{KCl} = 39 + 35.5 = 74.5\text{ g/mol}. Number of moles of KCl=14.9 g74.5 g/mol=0.2 mol\text{KCl} = \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}. Volume of water solvent =100 g=0.1 dm3= 100\text{ g} = 0.1\text{ dm}^3. Therefore, solubility =0.2 mol0.1 dm3=2.0 mol/dm3= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3.

Step-by-Step Solution

1
Calculate the molar mass of KCl\text{KCl} and convert mass of solute to moles.
Molar mass =74.5 g/mol= 74.5\text{ g/mol}; Moles =14.9 g74.5 g/mol=0.2 mol= \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}.
Solubility in mol/dm3\text{mol/dm}^3 requires the quantity of solute to be expressed in moles rather than grams.
2
Convert the mass/volume of solvent into decimeters cubed (dm3\text{dm}^3).
Volume =100 g=100 cm3=0.1 dm3= 100\text{ g} = 100\text{ cm}^3 = 0.1\text{ dm}^3.
Molar solubility concentration is defined per 1.0 dm31.0\text{ dm}^3 of solution/solvent.
3
Compute the concentration in mol/dm3\text{mol/dm}^3 by dividing moles of solute by volume of solvent in dm3\text{dm}^3.
Solubility =0.2 mol0.1 dm3=2.0 mol/dm3= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3.
Solubility in molarity equal to total moles divided by total volume in dm3\text{dm}^3.

Key Concept

Calculating molar solubility in mol/dm3\text{mol/dm}^3 from solute mass and solvent volume.
Question 6662Question

Read the poetic excerpt below:

When storm clouds gather on the northern peak,
And ancient winds awaken from their sleep,
The mountain pines begin to bend and speak,
While valleys hold their breath in shadows deep.

Yet in the quiet heart of yonder grove,
A single candle burns against the night,
To whisper solace born of steadfast love,
And shield the weary traveler with its light.

Which of the following best describes the stanzaic structure and rhyme scheme of the excerpt?

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Answer: Two quatrains featuring an alternate abab cdcd rhyme scheme

Answer

Two quatrains featuring an alternate abab cdcd rhyme scheme
The poem is composed of two four-line stanzas (quatrains). Examining the end-rhymes shows that 'peak' rhymes with 'speak' (a) and 'sleep' rhymes with 'deep' (b) in the first stanza (abab). In the second stanza, 'grove' rhymes with 'love' (c) and 'night' rhymes with 'light' (d), yielding a cdcd pattern. Therefore, the excerpt consists of two quatrains in an abab cdcd rhyme scheme.

Step-by-Step Solution

1
Analyze stanzaic division
The excerpt consists of two stanzas, each containing four lines, making them quatrains.
A four-line stanza is structurally defined as a quatrain.
2
Map end-rhyme sounds for the first stanza
Line 1 (peak - a), Line 2 (sleep - b), Line 3 (speak - a), Line 4 (deep - b) -> abab
Lines 1 and 3 rhyme together, and Lines 2 and 4 rhyme together.
3
Map end-rhyme sounds for the second stanza
Line 5 (grove - c), Line 6 (night - d), Line 7 (love - c), Line 8 (light - d) -> cdcd
Lines 5 and 7 rhyme together, and Lines 6 and 8 rhyme together.

Key Concept

Stanzaic Classification and Rhyme Scheme Analysis
Estimated Time:1m 30s
Question 6663Question

A commercial firm based in Aba purchases crude palm oil in bulk from local producers in Imo State and resells it to soap manufacturing companies in Kaduna State. Which classification of trade best describes this firm's primary activity?

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Answer: Wholesale home trade

Answer

Wholesale home trade
Wholesale home trade is correct because the transaction involves purchasing goods in large quantities from local primary producers and distributing them to industrial users within the borders of the same country.

Step-by-Step Solution

1
Analyze the geographical scope of the trade transaction
The trade occurs between Aba (Abia State), Imo State, and Kaduna State, all within Nigeria.
Trade conducted entirely within the boundaries of a single nation is classified as home (domestic) trade.
2
Determine the scale and stage of distribution
The firm purchases in bulk from primary producers and sells to industrial manufacturers.
Buying in bulk from producers and reselling to manufacturers or retailers defines wholesale trade rather than retail trade.
3
Synthesize the trade classification
Combining domestic scope and bulk distribution yields wholesale home trade.
Home trade is divided into wholesale trade and retail trade; bulk distribution internally falls under wholesale home trade.

Key Concept

Classification of Home Trade into Wholesale and Retail Trade
Question 6664Question

A sample of oxygen gas is collected over water at 27C27^\circ\text{C} and a total pressure of 755 mmHg755\text{ mmHg}. If the saturated vapor pressure of water at 27C27^\circ\text{C} is 25 mmHg25\text{ mmHg}, what is the partial pressure of the dry oxygen gas in mmHg\text{mmHg}?

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Answer: 730

Answer

The partial pressure of the dry oxygen gas is 730 mmHg730\text{ mmHg}.
According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. When a gas is collected over water, the gas absorbs water vapor, so Ptotal=Pdry gas+Pwater vaporP_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapor}}. To find the pressure of the dry oxygen, the water vapor pressure (25 mmHg25\text{ mmHg}) must be subtracted from the total barometric pressure (755 mmHg755\text{ mmHg}), giving 730 mmHg730\text{ mmHg}.

Step-by-Step Solution

1
Identify Dalton's Law equation for a gas collected over water
Ptotal=Pdry gas+PwaterP_{\text{total}} = P_{\text{dry gas}} + P_{\text{water}}
When a gas is collected over water, it becomes saturated with water vapor. The total observed pressure is the sum of the partial pressure of the dry gas and the vapor pressure of water (aqueous tension).
2
Subtract the aqueous tension from the total pressure
Pdry gas=755 mmHg25 mmHg=730 mmHgP_{\text{dry gas}} = 755\text{ mmHg} - 25\text{ mmHg} = 730\text{ mmHg}
Isolating Pdry gasP_{\text{dry gas}} gives the pressure exerted purely by the collected oxygen gas.

Key Concept

Dalton's Law of Partial Pressures and Collection of Gas over Water
Question 6665Question

A public limited company is legally permitted to commence business operations and exercise its borrowing powers immediately upon receiving its Certificate of Incorporation.

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Answer: False

Answer

The statement is False. A public limited company must obtain a Certificate of Trading in addition to its Certificate of Incorporation before it can lawfully begin operations.
The statement is false because a public limited company cannot commence business or exercise borrowing powers merely upon receiving a Certificate of Incorporation. It must fulfill statutory capital subscription conditions, file the appropriate prospectus or statement in lieu of prospectus, and obtain a Certificate of Trading.

Step-by-Step Solution

1
Examine the legal requirements for commencing business operations in corporate structures.
Incorporation brings the corporate entity into legal existence, but additional statutory prerequisites exist for public entities.
Distinguishing the legal requirements of public limited companies from private limited companies is essential.
2
Differentiate between the Certificate of Incorporation and the Certificate of Trading.
The Certificate of Incorporation brings the company into legal existence, whereas the Certificate of Trading authorizes a public limited company to commence business after fulfilling prospectus and capital subscription requirements.
Public limited companies invite public capital, requiring regulators to verify minimum subscription compliance before trading begins.

Key Concept

Commencement of Business and Statutory Documentation of Public Limited Companies
Question 6666Question

Match each atmospheric pollutant listed on the left with its primary health or environmental impact listed on the right.

Click a left item, then click its matching right item

Items

Carbon (II) oxide (CO\text{CO})
Sulfur (IV) oxide (SO2\text{SO}_2)
Chlorofluorocarbons (CFCs\text{CFCs})
Oxides of nitrogen (NOx\text{NO}_x)

Matches

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Answer

Carbon (II) oxide matches with binding to hemoglobin; Sulfur (IV) oxide matches with acid rain formation; Chlorofluorocarbons match with ozone layer depletion; Oxides of nitrogen match with photochemical smog formation.
Each atmospheric pollutant has a distinct chemical property and mechanism of damage: Carbon (II) oxide binds tightly to blood hemoglobin; Sulfur (IV) oxide generates acidic rain upon dissolution in atmospheric water; Chlorofluorocarbons decompose UV light to yield chlorine free radicals that destroy ozone; and Oxides of nitrogen react under solar radiation to create photochemical smog.

Step-by-Step Solution

1
Analyze the toxicological pathway of Carbon (II) oxide.
Carbon (II) oxide has a high affinity for hemoglobin, forming carboxyhemoglobin.
It interferes directly with oxygen transport in mammals.
2
Analyze atmospheric sulfur chemistry.
Sulfur (IV) oxide gas dissolves in rainwater forming trioxosulfate (IV) acid / tetraoxosulfate (VI) acid.
This acid rain degrades building materials and stonework.
3
Analyze stratospheric halide chemistry.
Chlorofluorocarbons release chlorine atoms under UV radiation.
Free chlorine atoms catalytically destroy protective ozone.
4
Analyze tropospheric nitrogen chemistry.
Oxides of nitrogen participate in secondary photochemical atmospheric reactions.
Interaction with hydrocarbons in sunlight forms haze-like photochemical smog.

Key Concept

Air Pollutants and Their Environmental Consequences
Estimated Time:1m 30s
Question 6667Question

A block and tackle system consisting of 55 pulleys is used to raise a load of 200 N200\text{ N} through a vertical height of 4 m4\text{ m}. If the efficiency of the system is 80%80\%, what is the work done against friction, in joules, during the lifting process?

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Answer: 200

Answer

The work done against friction during the lifting process is 200 J200\text{ J}.
The useful work output is achieved by lifting the 200 N200\text{ N} load through a vertical height of 4 m4\text{ m}, yielding Wout=200×4=800 JW_{\text{out}} = 200 \times 4 = 800\text{ J}. Given an efficiency of 80%80\% (0.800.80), the total work input required from the effort force is Win=8000.80=1000 JW_{\text{in}} = \frac{800}{0.80} = 1000\text{ J}. The energy lost to overcome frictional resistance in the pulleys equals the total work input minus useful work output: 1000 J800 J=200 J1000\text{ J} - 800\text{ J} = 200\text{ J}.

Step-by-Step Solution

1
Calculate useful work output
Wout=800 JW_{\text{out}} = 800\text{ J}
Useful work output is the energy required to raise the load through the specified height (Wout=Load×heightW_{\text{out}} = \text{Load} \times \text{height}).
2
Determine total work input
Win=1000 JW_{\text{in}} = 1000\text{ J}
The total work input is calculated from the efficiency formula: Efficiency=WoutWin\text{Efficiency} = \frac{W_{\text{out}}}{W_{\text{in}}}.
3
Compute work done against friction
Wfriction=200 JW_{\text{friction}} = 200\text{ J}
The work lost overcoming friction is the difference between total work input and useful work output (WinWoutW_{\text{in}} - W_{\text{out}}).

Key Concept

Work and Efficiency in Pulley Systems
Estimated Time:1m 30s
Question 6668Question

A longitudinal mechanical wave propagates through a gas at a speed of 340 m/s340\text{ m/s}. If a particle in the gas completes 5050 full oscillations in 0.10 s0.10\text{ s}, what is the distance between a compression and the immediate next rarefaction?

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Answer: 0.34 m0.34\text{ m}

Answer

The distance between a compression and the adjacent rarefaction is 0.34 m0.34\text{ m}.
The frequency of the particle oscillation is f=50/0.10=500 Hzf = 50 / 0.10 = 500\text{ Hz}. Using the wave equation v=fλv = f\lambda, the wavelength λ=340/500=0.68 m\lambda = 340 / 500 = 0.68\text{ m}. Because a compression and the consecutive rarefaction represent half of a full wave cycle, the distance between them is λ/2=0.34 m\lambda / 2 = 0.34\text{ m}.

Step-by-Step Solution

1
Calculate the frequency of oscillation of the wave
f=Number of oscillationsTime interval=500.10 s=500 Hzf = \frac{\text{Number of oscillations}}{\text{Time interval}} = \frac{50}{0.10\text{ s}} = 500\text{ Hz}
Frequency is defined as the number of complete vibrations per unit time.
2
Calculate the wavelength using the wave equation
λ=vf=340 m/s500 Hz=0.68 m\lambda = \frac{v}{f} = \frac{340\text{ m/s}}{500\text{ Hz}} = 0.68\text{ m}
The fundamental wave equation relates speed, frequency, and wavelength.
3
Determine the distance between consecutive compression and rarefaction
d=λ2=0.68 m2=0.34 md = \frac{\lambda}{2} = \frac{0.68\text{ m}}{2} = 0.34\text{ m}
In a longitudinal wave, a compression and the nearest rarefaction are separated by half a wavelength.

Key Concept

Relation between frequency, wavelength, wave speed, and structural intervals in longitudinal waves
Question 6669Question

Match each chemical species containing a central transition metal or halogen atom with its corresponding systematic IUPAC name and central element oxidation number.

Click a left item, then click its matching right item

Items

K3[Fe(CN)6]\text{K}_3[\text{Fe}(\text{CN})_6]
K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7
NaIO4\text{NaIO}_4
[Co(NH3)5Cl]Cl2[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2

Matches

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Answer

K3[Fe(CN)6]\text{K}_3[\text{Fe}(\text{CN})_6] matches Potassium hexacyanoferrate(III), +3; K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 matches Potassium heptaoxodichromate(VI), +6; NaIO4\text{NaIO}_4 matches Sodium tetraoxoiodate(VII), +7; [Co(NH3)5Cl]Cl2[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 matches Pentaamminechlorocobalt(III) chloride, +3.
Each complex chemical formula correctly pairs with its systematic IUPAC designation based on assigning oxidation numbers to central atoms according to standard IUPAC rules.

Step-by-Step Solution

1
Determine the oxidation state of Fe in K3[Fe(CN)6]\text{K}_3[\text{Fe}(\text{CN})_6] and match its IUPAC name
Fe oxidation state is +3; IUPAC name is Potassium hexacyanoferrate(III)
Potassium is +1, cyanide ion CN\text{CN}^- is -1. Solving 3(1)+x+6(1)=03(1) + x + 6(-1) = 0 gives x=+3x = +3.
2
Determine the oxidation state of Cr in K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 and match its IUPAC name
Cr oxidation state is +6; IUPAC name is Potassium heptaoxodichromate(VI)
Potassium is +1, oxygen is -2. Solving 2(1)+2x+7(2)=02(1) + 2x + 7(-2) = 0 gives 2x=122x = 12, so x=+6x = +6.
3
Determine the oxidation state of I in NaIO4\text{NaIO}_4 and match its IUPAC name
I oxidation state is +7; IUPAC name is Sodium tetraoxoiodate(VII)
Sodium is +1, oxygen is -2. Solving 1(1)+x+4(2)=01(1) + x + 4(-2) = 0 gives x=+7x = +7.
4
Determine the oxidation state of Co in [Co(NH3)5Cl]Cl2[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 and match its IUPAC name
Co oxidation state is +3; IUPAC name is Pentaamminechlorocobalt(III) chloride
Ammonia NH3\text{NH}_3 is a neutral molecule (0 charge), while chloride Cl\text{Cl}^- has -1 charge. Solving x+5(0)+3(1)=0x + 5(0) + 3(-1) = 0 gives x=+3x = +3.

Key Concept

Calculation of oxidation numbers in complex salts, coordination complexes, and polyatomic oxo-compounds, and applying systematic IUPAC nomenclature rules.
Question 6670Question

A sample of pure ammonium trioxocarbonate(IV), (NH4)2CO3(\text{NH}_4)_2\text{CO}_3, is determined to contain 3.6×10243.6 \times 10^{24} hydrogen atoms. What is the total mass of oxygen present in this sample?

[Relative atomic masses: H=1\text{H} = 1, C=12\text{C} = 12, N=14\text{N} = 14, O=16\text{O} = 16; Avogadro's constant NA=6.0×1023 mol1N_A = 6.0 \times 10^{23} \text{ mol}^{-1}]

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Answer: 36 g36\text{ g}

Answer

36 g36\text{ g}
The correct answer is 36 g36\text{ g}. Converting 3.6×10243.6 \times 10^{24} hydrogen atoms using Avogadro's constant gives 6.0 moles6.0\text{ moles} of hydrogen atoms. Because each formula unit of (NH4)2CO3(\text{NH}_4)_2\text{CO}_3 contains 88 hydrogen atoms and 33 oxygen atoms, the mole ratio of O to H is 3:83:8, giving 2.25 moles2.25\text{ moles} of oxygen atoms. Multiplying 2.25 moles2.25\text{ moles} by the molar mass of oxygen (16 g/mol16\text{ g/mol}) yields 36 g36\text{ g}.

Step-by-Step Solution

1
Calculate the total number of moles of hydrogen atoms from the given particle count.
Moles of H atoms=3.6×10246.0×1023 mol1=6.0 moles\text{Moles of H atoms} = \frac{3.6 \times 10^{24}}{6.0 \times 10^{23} \text{ mol}^{-1}} = 6.0\text{ moles}.
Dividing particle count by Avogadro's constant gives the amount in moles.
2
Determine the number of hydrogen atoms and oxygen atoms in one formula unit of (NH4)2CO3(\text{NH}_4)_2\text{CO}_3.
One formula unit contains 2×4=82 \times 4 = 8 hydrogen atoms and 33 oxygen atoms.
The subscript 2 outside the ammonium group (NH4)(\text{NH}_4) multiplies both N and H inside.
3
Calculate the moles of oxygen atoms present in the sample.
Moles of O atoms=6.0 moles of H×3 mol O8 mol H=2.25 moles of O\text{Moles of O atoms} = 6.0\text{ moles of H} \times \frac{3\text{ mol O}}{8\text{ mol H}} = 2.25\text{ moles of O}.
The mole ratio of O to H in the chemical formula is 3:83 : 8.
4
Convert the moles of oxygen atoms to mass.
Mass of O=2.25 mol×16 g/mol=36 g\text{Mass of O} = 2.25\text{ mol} \times 16\text{ g/mol} = 36\text{ g}.
Mass is obtained by multiplying the number of moles by the molar mass of oxygen.

Key Concept

Stoichiometric mole relationships between constituent elements in a chemical compound using Avogadro's constant and molar mass.
Question 6671Question

Match each dramatic character type on the left with its corresponding defining feature or dramatic function on the right.

Click a left item, then click its matching right item

Items

Dynamic character
Flat character
Foil character
Stock character

Matches

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Answer

Dynamic character pairs with the description of internal evolution; Flat character pairs with the description of a single static trait; Foil character pairs with the description of contrasting qualities; Stock character pairs with the description of a conventional dramatic stereotype.
Each character type matches its precise function: dynamic characters undergo psychological transformation; flat characters remain fixed around a single trait; foil characters highlight specific traits through strategic contrast; and stock characters represent familiar theatrical archetypes.

Step-by-Step Solution

1
Examine the degree of internal change and complexity for the first two terms.
Dynamic characters evolve psychologically across the narrative, whereas flat characters are defined by a single static dimension.
Character classification in drama relies primarily on the capacity for internal development and personal transformation.
2
Analyze the structural role and dramatic convention of the remaining two terms.
Foils function relationally to highlight traits of another character through contrast, while stock characters embody recurring societal archetypes.
Foils create thematic and psychological contrast, whereas stock characters rely on audience recognition of theatrical tropes.

Key Concept

Classification and dramatic functions of character types in drama
Question 6672Question

Match each prose narrative scenario to the central theme it predominantly illustrates in literary prose analysis.

Click a left item, then click its matching right item

Items

A dedicated scholar rejects personal relationships and emotional intimacy to pursue absolute knowledge, ending life in total solitary isolation.
A displaced protagonist attempts to navigate two conflicting judicial systems in a post-colonial urban center, feeling alienated from both indigenous traditions and foreign legal codes.
A modest rural artisan refuses to adopt modern mechanized machinery, sustaining hand-weaving practices despite growing financial hardship.
An ambitious political candidate turns against lifelong childhood companions to secure a lucrative endorsement from a corrupt political cabal.

Matches

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Answer

The scholar's pursuit of absolute knowledge aligns with intellectual hubris; the protagonist caught between legal systems aligns with cultural alienation; the artisan preserving hand-weaving aligns with defense of heritage against industrialization; and the candidate betraying companions aligns with the cost of unbridled ambition.
Each narrative scenario reflects a specific abstract theme in prose literature: the isolated pursuit of absolute knowledge demonstrates the danger of intellectual hubris; navigating conflicting post-colonial structures expresses cultural alienation; preserving artisan traditions against modern machinery depicts defense of heritage; and betraying loyalties for political gain portrays moral corruption driven by ambition.

Step-by-Step Solution

1
Analyze each narrative scenario on the left to identify the core conflict and character motivation.
Identified four distinct prose motifs: intellectual isolation, post-colonial cultural conflict, traditional craftsmanship vs. industry, and unethical political ascension.
Themes in prose are derived from analyzing character choices, central conflicts, and plot resolutions.
2
Map each identified conflict to its corresponding overarching thematic concept on the right.
Matched intellectual isolation with intellectual hubris; dual legal systems with cultural alienation; hand-weaving resistance with defense of heritage; and political betrayal with the cost of ambition.
Direct alignment between narrative action and abstract thematic interpretation is required.

Key Concept

Themes and Thematic Interpretation in Prose Literature
Question 6673Question

Consider the structural arrangement of the eight-line passage below:

Upon the ancient shore the billows break,
And scatter foam across the darkened sand;
The lonely traveller steps upon the land,
Uncertain of the journey he must take.
No distant beacon shines for safety's sake,
Nor guidance comes from any friendly hand;
Yet firm in purpose does he proudly stand,
Determined that his courage will not shake.

Which of the following best describes the stanzaic structure and rhyme scheme of this passage?

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Answer: An Italian (Petrarchan) octave with an enclosed rhyme scheme of abba abba

Answer

An Italian (Petrarchan) octave with an enclosed rhyme scheme of abba abba
The correct response accurately identifies the eight-line stanza as an Italian (Petrarchan) octave. By examining the end words—break (a), sand (b), land (b), take (a), sake (a), hand (b), stand (b), shake (a)—we observe an enclosed rhyme pattern of abba abba, which defines the classical opening octave of a Petrarchan sonnet.

Step-by-Step Solution

1
Analyze the line count and stanza arrangement of the excerpt
The excerpt consists of eight lines of verse (an octave).
Identifying line count establishes the primary stanzaic form.
2
Assign lowercase letters to each line based on matching end-rhyme sounds
Line 1 (break = a), Line 2 (sand = b), Line 3 (land = b), Line 4 (take = a), Line 5 (sake = a), Line 6 (hand = b), Line 7 (stand = b), Line 8 (shake = a).
Mapping end rhymes determines the exact rhyme scheme.
3
Correlate the line count and rhyme scheme with standard poetic forms
An eight-line stanza rhyming abba abba is classified as a Petrarchan (Italian) octave.
Matching the observed structure to established literary classification yields the correct term.

Key Concept

Petrarchan Octave Structure and Enclosed Rhyme Scheme
Question 6674Question

A capacitor of capacitance 5 μF5\text{ }\mu\text{F} is charged to a potential difference of 200 V200\text{ V} and then disconnected from the power supply. If it is subsequently connected in parallel across an uncharged capacitor of capacitance 15 μF15\text{ }\mu\text{F}, what is the final common potential difference across the combination in volts?

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Answer: 50

Answer

The final common potential difference across the combination is 50 V.
By charge conservation, the total charge Q=C1V1=5 μF×200 V=1000 μCQ = C_1 V_1 = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C} is shared across the two parallel capacitors. The total equivalent capacitance is Ctotal=C1+C2=20 μFC_{\text{total}} = C_1 + C_2 = 20\text{ }\mu\text{F}. Therefore, the final potential difference is V=QCtotal=1000 μC20 μF=50 VV = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}.

Step-by-Step Solution

1
Calculate the initial electric charge (QQ) stored on the charged capacitor
Q=5 μF×200 V=1000 μCQ = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}
Before connection, all charge is stored solely on the 5 μF5\text{ }\mu\text{F} capacitor.
2
Calculate the total equivalent capacitance (CtotalC_{\text{total}}) of the parallel network
Ctotal=5 μF+15 μF=20 μFC_{\text{total}} = 5\text{ }\mu\text{F} + 15\text{ }\mu\text{F} = 20\text{ }\mu\text{F}
Capacitors in parallel add directly (Ctotal=C1+C2C_{\text{total}} = C_1 + C_2).
3
Apply the law of conservation of charge to find the final common voltage (VV)
V=QCtotal=1000 μC20 μF=50 VV = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}
The total charge remains conserved and redistributes across the total combined capacitance.

Key Concept

Charge Redistribution and Conservation in Parallel Capacitors
Question 6675Question

An enterprise presents the following financial balances at the end of its trading year:

- Premises and Machinery: 1,850,000\text{₦}1,850,000
- Stock (Inventory): 410,000\text{₦}410,000
- Trade Debtors: 240,000\text{₦}240,000
- Cash at Bank: 150,000\text{₦}150,000
- Trade Creditors: 270,000\text{₦}270,000
- Accrued Expenses: 130,000\text{₦}130,000
- Long-term Loan: 600,000\text{₦}600,000

What is the Capital Employed of the enterprise in Naira (\text{₦})?

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Answer: 2250000

Answer

The Capital Employed of the enterprise is 2,250,000\text{₦}2,250,000.
Capital Employed represents the total resources employed in the business operations. It is computed as Fixed Assets plus Working Capital. First, calculate Current Assets (Stock 410,000\text{₦}410,000 + Debtors 240,000\text{₦}240,000 + Bank 150,000\text{₦}150,000 = 800,000\text{₦}800,000) and Current Liabilities (Creditors 270,000\text{₦}270,000 + Accrued Expenses 130,000\text{₦}130,000 = 400,000\text{₦}400,000). Working Capital is 800,000400,000=400,000\text{₦}800,000 - \text{₦}400,000 = \text{₦}400,000. Adding Working Capital to Fixed Assets (Premises and Machinery 1,850,000\text{₦}1,850,000) gives a Capital Employed of 2,250,000\text{₦}2,250,000.

Step-by-Step Solution

1
Determine total Current Assets
Current Assets = 410,000+240,000+150,000=800,000\text{₦}410,000 + \text{₦}240,000 + \text{₦}150,000 = \text{₦}800,000
Current assets consist of short-term liquid assets including stock, debtors, and cash at bank.
2
Determine total Current Liabilities
Current Liabilities = 270,000+130,000=400,000\text{₦}270,000 + \text{₦}130,000 = \text{₦}400,000
Current liabilities consist of short-term obligations payable within a year, including trade creditors and accrued expenses.
3
Calculate Working Capital
Working Capital = 800,000400,000=400,000\text{₦}800,000 - \text{₦}400,000 = \text{₦}400,000
Working capital is the net operational buffer calculated as Current Assets minus Current Liabilities.
4
Compute Capital Employed
Capital Employed = 1,850,000+400,000=2,250,000\text{₦}1,850,000 + \text{₦}400,000 = \text{₦}2,250,000
Capital Employed represents the total long-term assets and funds financing the business, calculated as Fixed Assets plus Working Capital (or Total Assets minus Current Liabilities).

Key Concept

Capital Employed represents the total funds actively utilized in running a business. It can be computed either as Fixed Assets + Working Capital or Total Assets - Current Liabilities.
Question 6676Question

A laboratory technician needs to recover pure hydrated copper(II) tetraoxosulfate(VI) crystals (CuSO45H2O\text{CuSO}_4\cdot5\text{H}_2\text{O}) from an aqueous suspension containing insoluble copper(II) oxide residue. Which of the following procedures correctly describes the optimal method to obtain the dry, hydrated crystals?

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Answer: Filter the mixture to remove insoluble copper(II) oxide, concentrate the filtrate by gentle heating until saturation, allow the saturated solution to cool for crystallization, and dry the crystals between filter papers.

Answer

Filter the mixture to remove insoluble copper(II) oxide, concentrate the filtrate by gentle heating until saturation, allow the saturated solution to cool for crystallization, and dry the crystals between filter papers.
Filtration removes insoluble solid residue from the solution. Heating the filtrate only until a saturated solution is obtained, followed by controlled cooling, enables the formation of hydrated copper(II) tetraoxosulfate(VI) pentahydrate crystals (CuSO45H2O\text{CuSO}_4\cdot5\text{H}_2\text{O}) without destroying the water of crystallization.

Step-by-Step Solution

1
Filter the aqueous suspension
Insoluble copper(II) oxide is collected as residue, while dissolved copper(II) tetraoxosulfate(VI) passes through into the filtrate.
Filtration effectively separates insoluble solids from dissolved solutes in liquid solutions.
2
Concentrate the filtrate to saturation point
Water evaporates until a saturated solution is formed at elevated temperature.
Evaporating to dryness must be avoided because strong heating destroys the hydration shell (H2O\text{H}_2\text{O} of crystallization).
3
Cool the saturated solution slowly
Hydrated copper(II) tetraoxosulfate(VI) pentahydrate crystals (CuSO45H2O\text{CuSO}_4\cdot5\text{H}_2\text{O}) precipitate out.
Solubility of salts generally decreases upon cooling, causing excess dissolved solute to crystallize.
4
Isolate and dry the crystals
Pure, intact hydrated crystals are retrieved.
Pressing crystals between filter papers removes surface moisture without causing thermal dehydration.

Key Concept

Distinction between evaporation to dryness and crystallization for thermally sensitive or hydrated salts
Question 6677Question

Match each chemical substance with its characteristic atmospheric behavior or hydration property when exposed to ambient laboratory air or heat:

Click a left item, then click its matching right item

Items

Na2CO310H2O\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} (Washing soda crystals)
Anhydrous CaCl2\text{CaCl}_2 (Calcium chloride pellets)
Concentrated H2SO4\text{H}_2\text{SO}_4 (Sulfuric acid)
CuSO45H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O} (Blue vitriol)

Matches

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Answer

Washing soda crystals undergo efflorescence; anhydrous calcium chloride undergoes deliquescence; concentrated sulfuric acid is hygroscopic; copper(II) sulfate pentahydrate loses its water of crystallization upon heating.
Washing soda crystals (Na2CO310H2O\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}) undergo efflorescence when exposed to dry air. Anhydrous calcium chloride (CaCl2\text{CaCl}_2) absorbs moisture until it forms a solution, which is deliquescence. Concentrated sulfuric acid (H2SO4\text{H}_2\text{SO}_4) absorbs moisture without dissolving a solid matrix, illustrating hygroscopic behavior. Copper(II) sulfate pentahydrate (CuSO45H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}) loses its water of crystallization thermally to yield a white anhydrous salt.

Step-by-Step Solution

1
Analyze the behavior of Na2CO310H2O\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}.
It loses 9 molecules of water of crystallization to dry air, forming a powdery monohydrate Na2CO3H2O\text{Na}_2\text{CO}_3 \cdot \text{H}_2\text{O} (efflorescence).
Efflorescence occurs when the vapor pressure of the hydrated salt is greater than the partial pressure of water vapor in the atmosphere.
2
Analyze anhydrous CaCl2\text{CaCl}_2.
It absorbs moisture from air and eventually dissolves in the absorbed water to form a saturated solution (deliquescence).
Deliquescence occurs when a solid compound's saturated solution has a lower vapor pressure than the atmospheric water vapor pressure.
3
Analyze concentrated H2SO4\text{H}_2\text{SO}_4.
It absorbs moisture from the atmosphere without changing state to form a new dissolved phase, acting as a liquid hygroscopic substance.
Hygroscopic substances absorb water vapor from the surrounding environment without dissolving into a liquid solution from a solid state.
4
Analyze CuSO45H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}.
Heating drives off the 5 water molecules of crystallization, converting blue crystalline pentahydrate to white anhydrous copper(II) sulfate.
Water of crystallization is chemically bound in the crystal lattice and can be expelled thermally, altering optical/crystalline properties.

Key Concept

Water of Crystallization, Deliquescence, Efflorescence, and Hygroscopy
Question 6678Question

A machine with a velocity ratio of 66 is used to raise a load of 540 N540\text{ N} by applying an effort of 120 N120\text{ N}. What is the efficiency of the machine?

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Answer: 75%75\%

Answer

The efficiency of the machine is 75%75\%.
The correct option is 75%75\%. Mechanical Advantage is found by dividing the load (540 N540\text{ N}) by the effort (120 N120\text{ N}), giving 4.54.5. Dividing this mechanical advantage by the given velocity ratio (66) and expressing as a percentage yields 4.56×100%=75%\frac{4.5}{6} \times 100\% = 75\%.

Step-by-Step Solution

1
Calculate Mechanical Advantage (MA)
MA=LoadEffort=540 N120 N=4.5\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{540\text{ N}}{120\text{ N}} = 4.5
Mechanical advantage measures the force magnification produced by the machine.
2
CalculateEfficiency(η)Calculate Efficiency (\eta)
η=(MAVR)×100%=(4.56)×100%=75%\eta = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4.5}{6}\right) \times 100\% = 75\%
Efficiency is the ratio of Mechanical Advantage to Velocity Ratio multiplied by 100%.

Key Concept

Efficiency of a Simple Machine
Estimated Time:1m 0s
Question 6679Question

What is the percentage by mass of water of crystallization in copper(II) tetraoxosulfate(VI) pentahydrate (CuSO45H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O})? [Cu=64,S=32,O=16,H=1][\text{Cu} = 64, \text{S} = 32, \text{O} = 16, \text{H} = 1]

Show answer & explanation

Answer: 36.0%36.0\%

Answer

The percentage by mass of water of crystallization in CuSO45H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O} is 36.0%36.0\%.
The total molar mass of CuSO45H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O} is 250 g/mol250\text{ g/mol} and the mass contributed by the five water molecules is 90 g/mol90\text{ g/mol}. Dividing 9090 by 250250 and multiplying by 100%100\% yields 36.0%36.0\%.

Step-by-Step Solution

1
Calculate the molar mass of water (H2O\text{H}_2\text{O}) and the total mass of five moles of water.
Molar mass of H2O=(2×1)+16=18 g/mol\text{H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}. For 5H2O5\text{H}_2\text{O}, mass =5×18=90 g/mol= 5 \times 18 = 90\text{ g/mol}.
Water of crystallization in the formula consists of five water molecules per formula unit.
2
Calculate the total molar mass of hydrated copper(II) tetraoxosulfate(VI) (CuSO45H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}).
Molar mass =64+32+(4×16)+90=64+32+64+90=250 g/mol= 64 + 32 + (4 \times 16) + 90 = 64 + 32 + 64 + 90 = 250\text{ g/mol}.
The percentage composition must be based on the complete formula weight of the hydrated compound.
3
Calculate the percentage by mass of water of crystallization.
Percentage of H2O=(90250)×100%=36.0%\text{Percentage of } \text{H}_2\text{O} = \left(\frac{90}{250}\right) \times 100\% = 36.0\%.
Percentage composition by mass is the mass of the component divided by the total molar mass of the compound multiplied by 100.

Key Concept

Percentage Water of Crystallization in Hydrated Salts
Question 6680Question

Two weak monobasic acids, HX\text{HX} (Ka=1.0×105 mol dm3K_a = 1.0 \times 10^{-5}\text{ mol dm}^{-3}) and HY\text{HY} (Ka=4.0×105 mol dm3K_a = 4.0 \times 10^{-5}\text{ mol dm}^{-3}), are prepared as aqueous solutions. Solution X contains 0.40 mol dm30.40\text{ mol dm}^{-3} of HX\text{HX}, whereas Solution Y contains 0.10 mol dm30.10\text{ mol dm}^{-3} of HY\text{HY}. Based on Ostwald's dilution law and the principles of acid ionization, which of the following statements correctly compares the two solutions?

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Answer: Both solutions have the same hydrogen ion concentration of 2.0×103 mol dm32.0 \times 10^{-3}\text{ mol dm}^{-3}, but HY\text{HY} has a four-fold greater degree of ionization than HX\text{HX}.

Answer

Both solutions have the exact same hydrogen ion concentration of 2.0×103 mol dm32.0 \times 10^{-3}\text{ mol dm}^{-3}, but HY\text{HY} has a four-fold greater degree of ionization than HX\text{HX}.
Evaluating [H+]=KaC[H^+] = \sqrt{K_a \cdot C} for both solutions gives [H+]X=1.0×105×0.40=2.0×103 mol dm3[H^+]_X = \sqrt{1.0 \times 10^{-5} \times 0.40} = 2.0 \times 10^{-3}\text{ mol dm}^{-3} and [H+]Y=4.0×105×0.10=2.0×103 mol dm3[H^+]_Y = \sqrt{4.0 \times 10^{-5} \times 0.10} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}, showing both solutions have identical hydrogen ion concentrations. Calculating the degree of ionization α=Ka/C\alpha = \sqrt{K_a / C} yields αX=0.005\alpha_X = 0.005 (0.5%0.5\%) and αY=0.020\alpha_Y = 0.020 (2.0%2.0\%), demonstrating that HY\text{HY} is four times more ionized than HX\text{HX} in these conditions.

Step-by-Step Solution

1
Calculate [H+][H^+] for Solution X
[H+]X=Ka,X×CX=(1.0×105)×0.40=4.0×106=2.0×103 mol dm3[H^+]_X = \sqrt{K_{a,X} \times C_X} = \sqrt{(1.0 \times 10^{-5}) \times 0.40} = \sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}
For a weak monobasic acid, [H+]=KaC[H^+] = \sqrt{K_a \cdot C} derived from Ostwald's dilution law.
2
Calculate [H+][H^+] for Solution Y
[H+]Y=Ka,Y×CY=(4.0×105)×0.10=4.0×106=2.0×103 mol dm3[H^+]_Y = \sqrt{K_{a,Y} \times C_Y} = \sqrt{(4.0 \times 10^{-5}) \times 0.10} = \sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}
Applying the weak acid ionization expression to Solution Y.
3
Calculate the degree of ionization (\alpha) for both acids
αX=Ka,XCX=1.0×1050.40=5.0×103=0.5%\alpha_X = \sqrt{\frac{K_{a,X}}{C_X}} = \sqrt{\frac{1.0 \times 10^{-5}}{0.40}} = 5.0 \times 10^{-3} = 0.5\%; αY=Ka,YCY=4.0×1050.10=2.0×102=2.0%\alpha_Y = \sqrt{\frac{K_{a,Y}}{C_Y}} = \sqrt{\frac{4.0 \times 10^{-5}}{0.10}} = 2.0 \times 10^{-2} = 2.0\%
The degree of ionization is given by α=KaC\alpha = \sqrt{\frac{K_a}{C}}.
4
Compare the calculated parameters
[H+]X=[H+]Y=2.0×103 mol dm3[H^+]_X = [H^+]_Y = 2.0 \times 10^{-3}\text{ mol dm}^{-3} and αYαX=2.0%0.5%=4\frac{\alpha_Y}{\alpha_X} = \frac{2.0\%}{0.5\%} = 4
Comparing [H+][H^+] and the ratio of degrees of ionization.

Key Concept

Relative Strength and Ionization of Acids and Bases
Estimated Time:2m 0s
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