Tüm alıştırma soruları

58 soru

Soru 1Soru

In Khadija Abubakar Jalli's prescribed novel *The Life Changer*, what is the name of the introverted Lafayette resident whose compound is secretly used to detain a kidnapped child, illustrating the narrative function of how innocence can be exploited by criminals?

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Cevap: Talle; talle; The Silent One; the silent one

Cevap

Talle
Talle, popularly known as 'the silent one' in Lafayette, is the minor character whose introverted lifestyle and gullibility make him an unwitting accomplice to Zaki's kidnapping operation.

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1
Recall the key plot incident in the town of Lafayette from *The Life Changer*.
A young boy is abducted and hidden away in a quiet villager's compound.
This incident introduces the theme of crime creeping into peaceful rural communities.
2
Identify the specific minor character whose personality and home were involved in this event.
Talle, who earned the sobriquet 'the silent one' because of his reticent nature, was duped into harboring the victim by Zaki.
Talle's character functions as a structural catalyst to expose the dangers of gullibility and bad association.

Anahtar Kavram

Minor Character Identification and Narrative Function
Soru 2Soru

In Khadija Abubakar Jalli's prescribed prose text *The Life Changer*, which minor character serves as the traditional leader and village head of Lafayette?

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Cevap: Hakimi; The Hakimi; Hakimi the village head

Cevap

Hakimi is the minor character who functions as the traditional leader and village head of Lafayette in *The Life Changer*.
Hakimi is the traditional leader and village head of Lafayette in *The Life Changer*, responsible for maintaining order and handling community matters in the village.

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1
Recall the minor characters from the Lafayette community in *The Life Changer*.
Hakimi is identified as the titleholder governing Lafayette.
When strange occurrences happen in Lafayette (such as Talle's sudden spending and arrest), the community affairs and traditional leadership fall under Hakimi.

Anahtar Kavram

Minor Character Identification and Functions
Tahmini Süre:45s
Soru 3Soru

In Khadija Abubakar Jalli's prescribed prose text, The Life Changer, which minor character serves as the personal driver and trusted confidant of Honourable Habib?

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Cevap: Labaran; Mr. Labaran; Driver Labaran

Cevap

Labaran
In Khadija Abubakar Jalli's novella The Life Changer, Labaran is the minor character who functions as Honourable Habib's driver and confidant, assisting him in private matters and executing behind-the-scenes arrangements.

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1
Identify the character's functional role in relation to Honourable Habib in The Life Changer.
Honourable Habib relies on his personal driver to carry out confidential tasks, private errands, and discreet arrangements.
Understanding minor character identification and narrative functions.
2
Match the role of driver and confidant to the specific character name.
The minor character named Labaran fills this role throughout the narrative.
Direct text recall of character identity.

Anahtar Kavram

Minor Character Identification and Functions
Soru 4Soru

In Khadija Abubakar Jalli's prescribed prose text, *The Life Changer*, which minor character serves as Tomiwa's hostel roommate hailing from Anambra State, whose background and interactions contribute to the narrative's thematic representation of inter-ethnic unity?

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Cevap: Nkechi; nkechi

Cevap

Nkechi
In *The Life Changer*, Nkechi is the minor character who represents Anambra State (South-Eastern Nigeria) among Tomiwa's room members. Together with Ada from Benue and Clare from Ondo, her inclusion highlights the structural and thematic function of promoting ethnic harmony, accommodation, and peaceful coexistence among tertiary institution students from diverse backgrounds.

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1
Examine the narrative context surrounding Tomiwa's living arrangements at the university in *The Life Changer*.
Tomiwa shares a hostel room with three other female students from distinct geopolitical regions of Nigeria: Ada, Clare, and Nkechi.
Establishing the composition of Tomiwa's hostel room isolates the candidates for this minor character role.
2
Correlate each roommate with her state of origin and narrative function as described in the text.
Ada hails from Benue State, Clare from Ondo State, and Nkechi represents Anambra State (South-East).
Matching the regional origin specified in the question stem confirms Nkechi as the character who embodies this aspect of national diversity and integration.

Anahtar Kavram

Minor Character Identification and Cultural Functions in *The Life Changer*
Soru 5Soru

In Khadija Abubakar Jalli's prescribed novel *The Life Changer*, which minor character in Lafayette is commonly referred to as 'the silent one' prior to his involvement in a kidnapping incident?

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Cevap: Talle; Mallam Talle

Cevap

Talle
Talle is explicitly introduced in *The Life Changer* as 'the silent one' in Lafayette because he became introverted and withdrawn after losing his family.

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1
Identify the character described by the sobriquet 'the silent one' in the Lafayette community context of the novel.
In the text, Talle earns the epithet 'the silent one' due to his reclusive and quiet disposition following the death of his parents.
This specific characterization sets up his vulnerability to being manipulated into harboring a kidnapped child in his house.

Anahtar Kavram

Minor Character Identification
Soru 6Soru

In Khadija Abubakar Jalli's prescribed novel *The Life Changer*, which minor character is Salma's hostel roommate from Benue State, best known for her generosity in sharing home-cooked food with her roommates?

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Cevap: Ada; Ada.

Cevap

Ada
In *The Life Changer*, Ada is one of Salma's three hostel roommates at the university. Hailing from Benue State, she is characterized by her generosity and habit of bringing food items from home to share with her room members.

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1
Identify the character based on state of origin and behavioral details in the text.
Ada is identified as the roommate hailing from Benue State who frequently brings food to share in the hostel room.
In *The Life Changer*, the author highlights the harmonious relationship among Salma's diverse roommates, specifically noting Ada's food-sharing habit.

Anahtar Kavram

Minor Character Identification and Functions
Soru 7Soru

In the prescribed prose text *The Life Changer*, which minor character acts as Salim's close friend and confidant, listening to his narrative regarding an online dating encounter that led to an attempted carjacking?

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Cevap: Gimba; Gimba.

Cevap

Gimba
Gimba is the minor character who acts as Salim's confidant in *The Life Changer*, listening to Salim detail his encounter with Natasha on social media and his narrow escape from carjackers.

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1
Identify the subplot in *The Life Changer* involving Salim's social media acquaintance and attempted robbery.
Salim recounts his ordeal regarding his meeting with Natasha to his close friend.
Analyzing narrative subplots allows accurate identification of minor characters and their functional roles.
2
Determine the exact identity of the character listening to Salim's experience.
The friend is Gimba.
Gimba serves as a confidant, providing a sounding board that highlights the moral lesson against naive online interactions.

Anahtar Kavram

Minor Character Identification and Narrative Function
Soru 8Soru

In Khadija Abubakar Jalli's prescribed prose text, *The Life Changer*, what is the name of the exceptionally quiet Lafayette resident who is arrested by the police after unwittingly harboring a kidnapped child in his compound?

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Cevap: Talle; Tale; Talle (The Silent One); The Silent One

Cevap

Talle
In *The Life Changer*, Talle is the minor character known throughout Lafayette as 'The Silent One'. His quiet, secretive lifestyle makes him an easy target for Zaki, who uses Talle's house as a hideout for a kidnapped child, leading to Talle's unexpected arrest.

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1
Analyze the context of the Lafayette kidnapping subplot in the novel.
The narrative describes a well-known, introverted villager who lost his parents and rarely speaks to community members.
This background establishes the minor character's personality and vulnerability to manipulation.
2
Identify the character manipulated into storing the kidnapped victim.
The character Talle, nicknamed 'The Silent One', was coerced and paid by Zaki to hold the abducted boy in his house.
His naive participation leads directly to the police raid on his home and his subsequent arrest alongside Zaki.

Anahtar Kavram

Minor Character Identification and Narrative Functions
Soru 9Soru

In *The Life Changer*, what is the name of the university lecturer who makes uncharacteristic advances toward Salma during her registration, only to immediately regret his conduct?

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Cevap: Dr. Dabo; Dabo; Doctor Dabo; Dr Dabo

Cevap

Dr. Dabo
Dr. Dabo is the university lecturer who loses his composure and tries to solicit a personal relationship with Salma during her registration. Overcome with remorse immediately afterward, he asks God for forgiveness and returns to his strict ethical standards.

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1
Identify the character based on the described narrative event during registration at the university.
The scene involves a traditionally strict lecturer who briefly yields to temptation when meeting Salma.
This event serves as a narrative introduction to Salma's impact on others and tests the lecturer's moral character.
2
Recall the exact name of the lecturer from the prescribed text.
The lecturer's name is Dr. Dabo.
After Salma leaves his office, Dr. Dabo recognizes his lapse in judgment, prays for forgiveness, and vows never to repeat the mistake.

Anahtar Kavram

Minor Character Identification and Functions
Soru 10Soru

Tayo Enterprises earned a gross profit of N150,000\text{N}150,000 for the year ended 31st December 2025. The books of the business revealed the following additional items at the end of the accounting period:

- Rent paid: N18,000\text{N}18,000 (includes N3,000\text{N}3,000 prepaid for the next accounting period)
- Discount received: N2,500\text{N}2,500
- Commission received: N4,000\text{N}4,000 (N1,000\text{N}1,000 is accrued and not yet received)
- Salaries paid: N35,000\text{N}35,000
- Reduction in provision for doubtful debts: N1,500\text{N}1,500

What is the net profit of Tayo Enterprises for the year?

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Cevap: N109,000; ��109,000; 109,000; N109000; ₦109000; 109000; N 109,000; ₦ 109,000

Cevap

The net profit of Tayo Enterprises for the year is N109,000\text{N}109,000.
To determine the net profit, add all income earned during the period (Gross Profit N150,000\text{N}150,000 + Discount received N2,500\text{N}2,500 + Total Commission earned N5,000\text{N}5,000 + Decrease in provision for doubtful debts N1,500=N159,000\text{N}1,500 = \text{N}159,000) and subtract all operating expenses incurred (Adjusted Rent N15,000\text{N}15,000 + Salaries N35,000=N50,000\text{N}35,000 = \text{N}50,000). This gives N159,000N50,000=N109,000\text{N}159,000 - \text{N}50,000 = \text{N}109,000.

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1
Calculate total income to be added to Gross Profit.
Discount received: N2,500\text{N}2,500; Commission earned: N4,000+N1,000=N5,000\text{N}4,000 + \text{N}1,000 = \text{N}5,000; Reduction in provision for doubtful debts: N1,500\text{N}1,500. Total additional income = N2,500+N5,000+N1,500=N9,000\text{N}2,500 + \text{N}5,000 + \text{N}1,500 = \text{N}9,000.
Accrued income is added to income received, and reductions in provision for doubtful debts represent gains credited to the Profit and Loss Account.
2
Calculate total adjusted operating expenses.
Adjusted Rent expense: N18,000N3,000=N15,000\text{N}18,000 - \text{N}3,000 = \text{N}15,000; Salaries: N35,000\text{N}35,000. Total expenses = N15,000+N35,000=N50,000\text{N}15,000 + \text{N}35,000 = \text{N}50,000.
Prepaid expenses must be deducted from the total paid to reflect only the expense incurred for the current accounting period.
3
Calculate Net Profit.
Net Profit=Gross Profit(N150,000)+Total Additional Income(N9,000)Total Expenses(N50,000)=N109,000\text{Net Profit} = \text{Gross Profit} (\text{N}150,000) + \text{Total Additional Income} (\text{N}9,000) - \text{Total Expenses} (\text{N}50,000) = \text{N}109,000.
Net profit is computed by adding non-operating/other revenue to gross profit and deducting total operating expenses for the period.

Anahtar Kavram

Determination of Net Profit in Profit and Loss Account with year-end adjustments
Tahmini Süre:2m 0s
Soru 11Soru

Consider the unsaturated hydrocarbon 2-methylbut-1-en-3-yne, which has the condensed structural formula CH2=C(CH3)CCHCH_2=C(CH_3)C\equiv CH. What is the total number of sigma (σ\sigma) bonds present in one molecule of this compound?

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Cevap: 10; 10 sigma bonds; 10 bonds

Cevap

The total number of sigma (σ\sigma) bonds present in one molecule of 2-methylbut-1-en-3-yne is 10.
In 2-methylbut-1-en-3-yne (CH2=C(CH3)CCHCH_2=C(CH_3)C\equiv CH), there are 6 carbon-hydrogen single bonds (2 from C1, 3 from the methyl group, and 1 from C4) and 4 carbon-carbon sigma bonds (1 from the C1=C2 double bond, 1 connecting C2 to the methyl carbon, 1 connecting C2 to C3, and 1 from the C3\equiv C4 triple bond). Summing these gives 10 sigma bonds in total.

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1
Expand the condensed structural formula to identify all individual carbon-hydrogen (C-H) bonds.
The =CH2=CH_2 group contains 2 C-H σ\sigma bonds, the methyl group (CH3-CH_3) contains 3 C-H σ\sigma bonds, and the terminal alkynyl group (CH\equiv CH) contains 1 C-H σ\sigma bond, giving a total of 6 C-H σ\sigma bonds.
Every single bond between carbon and hydrogen is a single sigma bond.
2
Identify all carbon-carbon (C-C) sigma bonds in the backbone and side chain.
The C=CC=C double bond contributes 1 C-C σ\sigma bond, the single bond to the methyl branch contributes 1 C-C σ\sigma bond, the C2-C3 single bond contributes 1 C-C σ\sigma bond, and the CCC\equiv C triple bond contributes 1 C-C σ\sigma bond, giving a total of 4 C-C σ\sigma bonds.
Multiple bonds (double or triple) contain exactly one sigma bond each, with the remaining bonds being pi (π\pi) bonds.
3
Sum the total number of C-H and C-C sigma bonds.
6 (C-H σ bonds)+4 (C-C σ bonds)=10 total σ bonds6 \text{ (C-H } \sigma\text{ bonds)} + 4 \text{ (C-C } \sigma\text{ bonds)} = 10 \text{ total } \sigma \text{ bonds}.
Adding all localized σ\sigma bonds yields the total count for the molecule.

Anahtar Kavram

Determination of sigma (\sigma) and pi (\pi) bond counts in complex open-chain hydrocarbons
Soru 12Soru

Given the universal set U={xZ:1x15}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 15\}, with subsets P={xU:x is a prime number}P = \{x \in \mathcal{U} : x \text{ is a prime number}\} and Q={xU:x is an odd number}Q = \{x \in \mathcal{U} : x \text{ is an odd number}\}, what is the number of elements in the set (PQ)(P \cup Q)'?

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Cevap: 6; six; 6 elements

Cevap

The number of elements in (PQ)(P \cup Q)' is 6.
The set PQP \cup Q consists of all numbers from 1 to 15 that are either prime or odd: {1,2,3,5,7,9,11,13,15}\{1, 2, 3, 5, 7, 9, 11, 13, 15\}. The complement (PQ)(P \cup Q)' relative to U\mathcal{U} contains all elements of U\mathcal{U} that are neither prime nor odd, which are the even composite numbers: {4,6,8,10,12,14}\{4, 6, 8, 10, 12, 14\}. Counting these elements gives 6.

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1
List all elements of the universal set U\mathcal{U}, subset PP, and subset QQ.
U={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}, P={2,3,5,7,11,13}P = \{2, 3, 5, 7, 11, 13\}, and Q={1,3,5,7,9,11,13,15}Q = \{1, 3, 5, 7, 9, 11, 13, 15\}.
Listing the explicit elements allows precise execution of set union and complement operations.
2
Find the union PQP \cup Q.
PQ={1,2,3,5,7,9,11,13,15}P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15\}.
The union combines all unique elements that belong to either set PP, set QQ, or both.
3
Determine the complement set (PQ)(P \cup Q)' relative to U\mathcal{U} and count its elements.
(PQ)={4,6,8,10,12,14}(P \cup Q)' = \{4, 6, 8, 10, 12, 14\}, which contains 6 elements.
The complement set consists of all elements in U\mathcal{U} that are not present in PQP \cup Q.

Anahtar Kavram

Complement of Set Union
Soru 13Soru

Using the first principles of differentiation for the reciprocal function f(x)=4xf(x) = \frac{4}{x}, evaluate the limit of the difference quotient as h0h \to 0: limh0f(x+h)f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. What is the simplified expression for the derivative dydx\frac{dy}{dx}?

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Cevap: -\frac{4}{x^2}; -4/x^2; -4 / x^2; -\frac{4}{x^{2}}

Cevap

The derivative dydx\frac{dy}{dx} is 4x2-\frac{4}{x^2}.
Substituting f(x)=4xf(x) = \frac{4}{x} into the first principles limit formula yields limh04x4(x+h)hx(x+h)=limh04hhx(x+h)\lim_{h \to 0} \frac{4x - 4(x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-4h}{h \cdot x(x+h)}. Canceling hh gives limh04x(x+h)=4x2\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x^2}.

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1
Write the first principles formula and substitute f(x)=4xf(x) = \frac{4}{x}.
\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{4}{x+h} - \frac{4}{x}}{h}
Apply the definition of differentiation from first principles.
2
Combine the fractions in the numerator using a common denominator x(x+h)x(x+h).
\frac{4}{x+h} - \frac{4}{x} = \frac{4x - 4(x+h)}{x(x+h)} = \frac{4x - 4x - 4h}{x(x+h)} = \frac{-4h}{x(x+h)}
Simplify the numerator into a single fractional expression.
3
Divide the simplified numerator by hh and cancel the common factor of hh.
\frac{\frac{-4h}{x(x+h)}}{h} = \frac{-4h}{h \cdot x(x+h)} = -\frac{4}{x(x+h)}
Eliminate the indeterminate factor of hh from the denominator.
4
Evaluate the limit as h0h \to 0.
\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x(x+0)} = -\frac{4}{x^2}
Substitute h=0h = 0 into the simplified algebraic expression.

Anahtar Kavram

Differentiation from First Principles
Soru 14Soru

An experimental effusion cell measures gas diffusion rates through a micro-porous membrane at fixed temperature and pressure. In a calibration run, 120 cm3120\text{ cm}^3 of neon gas (Ne\text{Ne}, atomic mass =20 g/mol= 20\text{ g/mol}) effuses through the membrane in 30 seconds30\text{ seconds}. Calculate the volume (in cm3\text{cm}^3) of sulfur trioxide gas (SO3\text{SO}_3, atomic masses: S=32 g/mol\text{S} = 32\text{ g/mol}, O=16 g/mol\text{O} = 16\text{ g/mol}) that will effuse through the exact same membrane in 50 seconds50\text{ seconds}.

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Cevap: 100; 100 cm3; 100cm3; 100 cm^3; 100 cm³

Cevap

100 cm³
The effusion rate of neon is rNe=120 cm330 s=4 cm3/sr_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}. Given MNe=20 g/molM_{\text{Ne}} = 20\text{ g/mol} and MSO3=80 g/molM_{\text{SO}_3} = 80\text{ g/mol}, Graham's law dictates rNerSO3=8020=2\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = 2. Thus, rSO3=42=2 cm3/sr_{\text{SO}_3} = \frac{4}{2} = 2\text{ cm}^3/\text{s}. Over a period of 50 seconds50\text{ seconds}, the volume of SO3\text{SO}_3 effused is 2 cm3/s×50 s=100 cm32\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3.

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1
Calculate the rate of effusion of neon gas (rNer_{\text{Ne}})
rNe=120 cm330 s=4 cm3/sr_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}
Effusion rate is defined as the volume of gas effusing per unit time.
2
Determine the molar mass of sulfur trioxide (SO3\text{SO}_3)
MSO3=32+3(16)=80 g/molM_{\text{SO}_3} = 32 + 3(16) = 80\text{ g/mol}
The molar mass is calculated from the constituent relative atomic masses of sulfur and oxygen.
3
Apply Graham's Law of Effusion to determine the effusion rate of sulfur trioxide (rSO3r_{\text{SO}_3})
rNerSO3=MSO3MNe    4rSO3=8020=4=2    rSO3=2 cm3/s\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{M_{\text{SO}_3}}{M_{\text{Ne}}}} \implies \frac{4}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = \sqrt{4} = 2 \implies r_{\text{SO}_3} = 2\text{ cm}^3/\text{s}
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
4
Compute the total volume of sulfur trioxide effused in 50 seconds50\text{ seconds}
VSO3=rSO3×t=2 cm3/s×50 s=100 cm3V_{\text{SO}_3} = r_{\text{SO}_3} \times t = 2\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3
Multiplying the calculated rate of effusion by the given time yields the total volume.

Anahtar Kavram

Graham's Law of Diffusion and Effusion
Soru 15Soru

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.

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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 16Soru

What is the equation of the locus of a point P(x,y)P(x, y) that moves in a plane such that its distance from the origin (0,0)(0,0) is always 5 units?

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Cevap: x^2 + y^2 = 25; x^2+y^2=25; x²+y²=25

Cevap

x2+y2=25x^2 + y^2 = 25
By definition, the locus of a point moving at a fixed distance of 5 units from the origin (0,0)(0,0) is a circle centered at (0,0)(0,0) with radius 5. Substituting into the standard circle equation x2+y2=r2x^2 + y^2 = r^2 yields x2+y2=52=25x^2 + y^2 = 5^2 = 25.

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1
Apply the distance formula between a general point P(x,y)P(x, y) and the origin (0,0)(0,0).
d=(x0)2+(y0)2=x2+y2d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}
The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is (x2x1)2+(y2y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.
2
Equate the distance formula to the given fixed distance of 5 units and square both sides.
x2+y2=5    x2+y2=25\sqrt{x^2 + y^2} = 5 \implies x^2 + y^2 = 25
Squaring both sides eliminates the square root to give the algebraic equation of the locus.

Anahtar Kavram

The locus of points at a constant distance rr from a fixed point (h,k)(h, k) forms a circle with equation (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2.
Tahmini Süre:45s
Soru 17Soru

If set A={a,b,c,d,e}A = \{a, b, c, d, e\} and set B={c,d,e,f,g}B = \{c, d, e, f, g\}, what is the number of elements in the set (AB)(BA)(A \setminus B) \cup (B \setminus A)?

Cevabı ve açıklamayı göster

Cevap: 4; four

Cevap

The number of elements in (AB)(BA)(A \setminus B) \cup (B \setminus A) is 4.
The set difference ABA \setminus B consists of elements in AA that are not in BB, which gives {a,b}\{a, b\}. Similarly, BAB \setminus A consists of elements in BB that are not in AA, giving {f,g}\{f, g\}. The union (AB)(BA)(A \setminus B) \cup (B \setminus A) is {a,b,f,g}\{a, b, f, g\}, which has a cardinality of 4.

Adım Adım Çözüm

1
Find the relative difference ABA \setminus B
AB={a,b}A \setminus B = \{a, b\}
Remove elements of BB present in AA.
2
Find the relative difference BAB \setminus A
BA={f,g}B \setminus A = \{f, g\}
Remove elements of AA present in BB.
3
Take the union of the two set differences
(AB)(BA)={a,b,f,g}(A \setminus B) \cup (B \setminus A) = \{a, b, f, g\}
Combine elements from both set differences.
4
Count the number of elements in the resulting set
4 elements
The set {a,b,f,g}\{a, b, f, g\} contains 4 distinct elements.

Anahtar Kavram

Symmetric Difference of Two Sets
Tahmini Süre:45s
Soru 18Soru

In a group of 100 candidates preparing for an entrance examination, 48 registered for Mathematics, 45 for Physics, and 40 for Chemistry. If 18 candidates registered for both Mathematics and Physics, 15 for both Physics and Chemistry, 20 for both Mathematics and Chemistry, and 8 registered for none of these three subjects, how many candidates registered for all three subjects?

Cevabı ve açıklamayı göster

Cevap: 12; 12 candidates; 12 students

Cevap

12 candidates registered for all three subjects.
By the Principle of Inclusion-Exclusion, the total number of candidates taking at least one subject is 1008=92100 - 8 = 92. Summing the individual totals gives 48+45+40=13348 + 45 + 40 = 133. Subtracting the pairwise intersections gives 133(18+15+20)=80133 - (18 + 15 + 20) = 80. Adding the intersection of all three sets must equal 92, yielding 9280=1292 - 80 = 12.

Adım Adım Çözüm

1
Determine the total number of candidates who registered for at least one of the three subjects.
n(MPC)=1008=92n(M \cup P \cup C) = 100 - 8 = 92
Subtracting the number of candidates who registered for none of the subjects from the universal set gives the cardinality of the union.
2
Apply the Principle of Inclusion-Exclusion for three sets.
n(MPC)=n(M)+n(P)+n(C)n(MP)n(PC)n(MC)+n(MPC)n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)
This formula relates the individual set sizes, pair intersections, and triple intersection to the union.
3
Substitute the known values into the inclusion-exclusion equation.
92=48+45+40181520+n(MPC)92 = 48 + 45 + 40 - 18 - 15 - 20 + n(M \cap P \cap C)
Insert the given cardinalities into the formula.
4
Simplify and solve for n(MPC)n(M \cap P \cap C).
92=80+n(MPC)    n(MPC)=9280=1292 = 80 + n(M \cap P \cap C) \implies n(M \cap P \cap C) = 92 - 80 = 12
Isolate the unknown triple intersection term.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets
Soru 19Soru

Find the equation of the locus of a point P(x,y)P(x, y) that moves such that it is equidistant from the fixed points A(1,3)A(-1, 3) and B(5,1)B(5, -1).

Cevabı ve açıklamayı göster

Cevap: 3x - 2y - 4 = 0; 3x-2y-4=0; 3x - 2y = 4; 3x-2y=4; 12x - 8y - 16 = 0; y = (3/2)x - 2; y = 1.5x - 2

Cevap

The equation of the locus is 3x2y4=03x - 2y - 4 = 0 (or 3x2y=43x - 2y = 4).
The locus of a point equidistant from two fixed points A(1,3)A(-1, 3) and B(5,1)B(5, -1) is the perpendicular bisector of the line segment joining them. Equating the squared distances (x+1)2+(y3)2=(x5)2+(y+1)2(x+1)^2 + (y-3)^2 = (x-5)^2 + (y+1)^2 and simplifying yields the linear equation 3x2y4=03x - 2y - 4 = 0.

Adım Adım Çözüm

1
Set up the distance equality condition using the distance formula.
sqrt(x(1))2+(y3)2=sqrt(x5)2+(y(1))2\\sqrt{(x - (-1))^2 + (y - 3)^2} = \\sqrt{(x - 5)^2 + (y - (-1))^2}
Since point P(x,y)P(x, y) is equidistant from AA and BB, PA=PBPA = PB.
2
Square both sides to remove the radical signs.
(x+1)2+(y3)2=(x5)2+(y+1)2(x + 1)^2 + (y - 3)^2 = (x - 5)^2 + (y + 1)^2
Squaring both sides eliminates square roots and simplifies polynomial expansion.
3
Expand all squared terms on both sides.
x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 + 2y + 1
Expanding allows gathering like terms.
4
Cancel x2x^2 and y2y^2 from both sides and collect all terms on one side.
(2x + 10x) + (-6y - 2y) + (10 - 26) = 0 \\Rightarrow 12x - 8y - 16 = 0
Combining like terms simplifies the locus equation into standard linear form.
5
Divide the entire equation by the common factor of 4.
3x - 2y - 4 = 0
Expressing the linear equation in its simplest form gives the perpendicular bisector of line segment ABAB.

Anahtar Kavram

Locus equidistant from two fixed points (Perpendicular Bisector)
Tahmini Süre:2m 0s
Soru 20Soru

An acyclic hydrocarbon XX with the molecular formula C5H8C_5H_8 rapidly decolourises bromine water. However, when XX is treated with ammoniacal silver nitrate solution, no precipitate is observed. Upon complete catalytic hydrogenation in the presence of a nickel catalyst, XX is converted into pentane. What is the IUPAC name of hydrocarbon XX?

Cevabı ve açıklamayı göster

Cevap: pent-2-yne; 2-pentyne; Pent-2-yne; 2-Pentyne

Cevap

Pent-2-yne (or 2-pentyne)
Hydrocarbon XX has the molecular formula C5H8C_5H_8, corresponding to two degrees of unsaturation. Complete hydrogenation to pentane confirms an unbranched five-carbon chain. Decolourisation of bromine water verifies unsaturation. Because XX yields no precipitate with ammoniacal silver nitrate solution, it lacks acidic terminal acetylenic hydrogens (RCCHR-C \equiv C-H). Therefore, the triple bond must be located internally between carbon-2 and carbon-3, making the compound pent-2-yne.

Adım Adım Çözüm

1
Determine the degree of unsaturation and carbon skeleton of hydrocarbon XX.
Degree of unsaturation is 2, and the carbon skeleton is a straight 5-carbon chain.
The molecular formula C5H8C_5H_8 corresponds to CnH2n2C_nH_{2n-2}, indicating two degrees of unsaturation (an alkyne or alkadiene). Complete catalytic hydrogenation yields pentane (C5H12C_5H_{12}), proving an unbranched five-carbon chain.
2
Analyze the reaction with bromine water.
Hydrocarbon XX contains carbon-carbon multiple bonds.
Decolourisation of bromine water confirms the presence of unsaturation.
3
Evaluate the test with ammoniacal silver nitrate solution.
Hydrocarbon XX is an internal (non-terminal) alkyne.
Terminal alkynes possess acidic acetylenic hydrogen atoms (RCCHR-C \equiv C-H) that react with ammoniacal silver nitrate to form a characteristic white silver acetylide precipitate. The absence of a precipitate rules out pent-1-yne and confirms that the triple bond is located internally at C-2.
4
Deduce the final IUPAC name of hydrocarbon XX.
pent-2-yne
Combining a straight 5-carbon chain with an internal triple bond between carbon-2 and carbon-3 gives pent-2-yne.

Anahtar Kavram

Distinction between terminal and non-terminal alkynes using ammoniacal silver nitrate test and carbon skeleton determination via hydrogenation
Tahmini Süre:2m 0s
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