Tüm alıştırma soruları

13931 soru

Soru 3921Soru

Which of the following best describes the fundamental attraction responsible for metallic bonding in solid metals?

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Cevap: Electrostatic attraction between positive metal cations and a sea of delocalized valence electrons

Cevap

The metallic bond is defined as the strong electrostatic attraction between positively charged metal cations fixed in a lattice and a surrounding sea of mobile, delocalized valence electrons.
The correct option accurately defines metallic bonding as the electrostatic force of attraction binding positive metal ions to a fluid sea of delocalized valence electrons, which accounts for characteristic metallic properties like electrical conductivity and malleability.

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1
Identify the valence electron behavior in metals
Metal atoms lose their outer valence electrons to form positive cations, producing a shared pool of delocalized electrons free to move throughout the giant structure.
Low ionization energies in metals allow valence electrons to become detached easily from individual atoms.
2
Determine the nature of the attractive force holding the lattice together
Strong electrostatic attraction acts non-directionally between the positive ions and the mobile electron sea.
Opposite charges attract each other, forming a stable metallic lattice.

Anahtar Kavram

Nature of Metallic Bonding
Tahmini Süre:45s
Soru 3922Soru

Three business partners incorporated a private limited company to process agricultural produce. One year after operations commenced, one partner decided to sell her equity stake to an outside investor without obtaining consent from the remaining members. Which regulation or characteristic of a private limited company prevents this transfer from occurring freely?

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Cevap: The statutory restriction on share transfer contained in the Articles of Association

Cevap

The statutory restriction on share transfer contained in the Articles of Association
Private limited companies are legally required to restrict the right to transfer shares in order to preserve their private character. The specific internal rules and restrictions governing share transfers are documented within the Articles of Association.

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1
Identify the legal characteristics governing share mobility in a private limited company.
Unlike public limited companies whose shares are freely transferable on a stock exchange, a private limited company restricts the transfer of its shares to protect the interest of existing members.
This restriction ensures that outside parties cannot acquire ownership control without the consent of existing shareholders.
2
Determine which company formation document regulates internal share transfer procedures.
The Articles of Association regulates internal management, director powers, and shareholder rights, including share transfer restrictions.
The Memorandum of Association covers external governance and object clauses, whereas the Articles of Association handles internal regulations.

Anahtar Kavram

Share Transfer Restrictions and Articles of Association in Private Limited Companies
Tahmini Süre:1m 15s
Soru 3923Soru

Match each vacuum flask component or surface feature on the left with its primary mechanism for controlling heat transfer on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Silvered inner surfaces of the double walls
Evacuated space (vacuum) between the walls
Cork stopper at the top opening
Dull black exterior casing

Eşleşmeler

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Cevap

Silvered inner surfaces match with radiation reflection; evacuated space matches with elimination of conduction and convection; cork stopper matches with prevention of convection and conduction at the opening; dull black casing matches with maximizing thermal radiation emission/absorption.
Each feature of the vacuum flask targets a specific heat transfer mode: silvering reflects infrared radiation; the vacuum eliminates particle-dependent transfer (conduction and convection); cork acts as an insulator preventing convection and conduction at the top; and black surfaces maximize thermal radiation emission/absorption.

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1
Analyze the silvered inner walls of a vacuum flask
Silver surfaces are good reflectors of heat rays (infrared waves).
Radiant heat travels via electromagnetic waves and is reflected by shiny metallic coatings, minimizing radiation heat loss.
2
Analyze the vacuum space between the glass walls
Conduction and convection cannot occur across a vacuum.
Both conduction (particle vibration/electron flow) and convection (fluid movement) strictly require a physical medium.
3
Analyze the cork/plastic stopper
Cork prevents hot air circulation and thermal conduction across the opening.
Cork is a poor conductor of heat and stops evaporative/convective air currents from leaving the container.
4
Analyze the dull black exterior
Dull black surfaces are efficient radiation emitters/absorbers.
According to radiation principles, black matte surfaces radiate energy much faster than polished surfaces.

Anahtar Kavram

Modes of Heat Transfer and Practical Applications in Thermal Insulation
Soru 3924Soru
A 4.00 g4.00\text{ g} sample of an impure copper(II) oxide ore is heated in a stream of dry hydrogen gas until reduction is complete according to the equation:
CuO(s)+H2(g)Cu(s)+H2O(g)\text{CuO}_{(s)} + \text{H}_{2(g)} \rightarrow \text{Cu}_{(s)} + \text{H}_2\text{O}_{(g)}
If 2.54 g2.54\text{ g} of pure copper metal is obtained, what is the percentage purity of the copper(II) oxide in the ore?
[Relative atomic masses: Cu=63.5,O=16.0][\text{Relative atomic masses: Cu} = 63.5, \text{O} = 16.0]
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Cevap: 79.5%79.5\%

Cevap

The percentage purity of the copper(II) oxide in the ore is 79.5%79.5\%.
The option stating 79.5%79.5\% is correct because 2.54 g2.54\text{ g} of copper corresponds to 0.04 mol0.04\text{ mol} of Cu\text{Cu}. According to the chemical equation, 0.04 mol0.04\text{ mol} of Cu\text{Cu} requires 0.04 mol0.04\text{ mol} of pure CuO\text{CuO}, which weighs 0.04×79.5 g mol1=3.18 g0.04 \times 79.5\text{ g mol}^{-1} = 3.18\text{ g}. Dividing 3.18 g3.18\text{ g} of pure CuO\text{CuO} by the total sample mass of 4.00 g4.00\text{ g} yields 79.5%79.5\%.

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1
Calculate the molar mass of copper(II) oxide (CuO) and the moles of copper metal produced.
Molar mass of CuO=63.5+16.0=79.5 g mol1\text{Molar mass of CuO} = 63.5 + 16.0 = 79.5\text{ g mol}^{-1}. Moles of Cu=2.54 g63.5 g mol1=0.04 mol\text{Moles of Cu} = \frac{2.54\text{ g}}{63.5\text{ g mol}^{-1}} = 0.04\text{ mol}.
Converting the given mass of product into moles allows stoichiometric ratio calculations.
2
Determine the moles and mass of pure copper(II) oxide in the sample.
From the equation, 1 mol CuO1 mol Cu1\text{ mol CuO} \rightarrow 1\text{ mol Cu}. Moles of pure CuO=0.04 mol\text{CuO} = 0.04\text{ mol}. Mass of pure CuO=0.04 mol×79.5 g mol1=3.18 g\text{CuO} = 0.04\text{ mol} \times 79.5\text{ g mol}^{-1} = 3.18\text{ g}.
Stoichiometry dictates that 1 mole of CuO produces 1 mole of Cu upon complete reduction.
3
Calculate the percentage purity of the copper(II) oxide sample.
Percentage purity=Mass of pure CuOTotal mass of sample×100%=3.18 g4.00 g×100%=79.5%\text{Percentage purity} = \frac{\text{Mass of pure CuO}}{\text{Total mass of sample}} \times 100\% = \frac{3.18\text{ g}}{4.00\text{ g}} \times 100\% = 79.5\%.
Percentage purity is the ratio of pure reactive compound mass to total impure sample mass expressed as a percentage.

Anahtar Kavram

Determining percentage purity using stoichiometric reduction yields
Soru 3925Soru

What is the pH of an aqueous solution prepared by dissolving 0.49 g0.49\text{ g} of tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4) in distilled water to make 500 cm3500\text{ cm}^3 of solution? [Molar mass of H2SO4=98 g mol1\text{H}_2\text{SO}_4 = 98\text{ g mol}^{-1}, log102=0.301\log_{10} 2 = 0.301]

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Cevap: 1.701.70

Cevap

The pH of the solution is 1.701.70.
Dissolving 0.49 g0.49\text{ g} of H2SO4\text{H}_2\text{SO}_4 (molar mass 98 g mol198\text{ g mol}^{-1}) in 0.5 dm30.5\text{ dm}^3 yields a 0.01 mol dm30.01\text{ mol dm}^{-3} solution. Because tetraoxosulfate(VI) acid fully ionizes into 2H+2\text{H}^+ and SO42\text{SO}_4^{2-}, the hydrogen ion concentration [H+][\text{H}^+] is 0.02 mol dm30.02\text{ mol dm}^{-3}. Taking the negative logarithm base 10 gives pH=log10(0.02)=1.70\text{pH} = -\log_{10}(0.02) = 1.70.

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1
Calculate the amount of H2SO4\text{H}_2\text{SO}_4 in moles
Moles of H2SO4=0.49 g98 g mol1=0.005 mol\text{Moles of } \text{H}_2\text{SO}_4 = \frac{0.49\text{ g}}{98\text{ g mol}^{-1}} = 0.005\text{ mol}
Converting mass of solute to moles using molar mass.
2
Determine the molar concentration of the acid solution
Volume=500 cm3=0.5 dm3\text{Volume} = 500\text{ cm}^3 = 0.5\text{ dm}^3; Molarity=0.005 mol0.5 dm3=0.01 mol dm3\text{Molarity} = \frac{0.005\text{ mol}}{0.5\text{ dm}^3} = 0.01\text{ mol dm}^{-3}
Molarity is defined as moles of solute per cubic decimetre of solution.
3
Calculate the hydrogen ion concentration [H+][\text{H}^+]
[H+]=2×0.01 mol dm3=0.02 mol dm3=2.0×102 mol dm3[\text{H}^+] = 2 \times 0.01\text{ mol dm}^{-3} = 0.02\text{ mol dm}^{-3} = 2.0 \times 10^{-2}\text{ mol dm}^{-3}
H2SO4\text{H}_2\text{SO}_4 is a strong dibasic acid that ionizes completely to produce two H+\text{H}^+ ions per molecule.
4
Calculate the pH of the solution
pH=log10[H+]=log10(2.0×102)=2log102=20.301=1.6991.70\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(2.0 \times 10^{-2}) = 2 - \log_{10} 2 = 2 - 0.301 = 1.699 \approx 1.70
Applying the logarithmic definition of pH.

Anahtar Kavram

pH Calculation of Dibasic Strong Acids
Soru 3926Soru

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 3927Soru

In the Kano Closed-Settlement Zone of Northern Nigeria, agricultural land is cropped continuously every year with intensive manure application, whereas in surrounding lower-density rural districts, farmers rely on bush fallowing with multi-year land rotation. Which of the following statements best explains this spatial variation in agricultural land-use intensity?

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Cevap: High population pressure creates severe land scarcity, compelling farmers to adopt labor-intensive soil conservation and continuous cultivation to maximize yield per unit area.

Cevap

High population pressure creates severe land scarcity, compelling farmers to adopt labor-intensive soil conservation and continuous cultivation to maximize yield per unit area.
In agricultural geography, high rural population density leads to land fragmentation and scarcity. Farmers can no longer afford to leave land idle under traditional bush fallowing, forcing an intensification of land use through continuous cropping, heavy application of animal manure, and labor-intensive soil management as seen in the Kano Closed-Settlement Zone.

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1
Analyze the spatial contrast in land use intensity between the Kano Closed-Settlement Zone and surrounding rural areas.
Identified that the Kano Closed-Settlement Zone is characterized by extreme population density and fragmented land holdings compared to outlying districts.
Agricultural land-use intensity is strongly governed by man-land ratios and population pressure on land resources (Boserup's agricultural intensification theory).
2
Evaluate the relationship between land scarcity and fallow duration.
As population density increases, land scarcity prevents long fallow periods, forcing a transition from bush fallowing to permanent/continuous cultivation supported by organic fertilization (manuring).
Farmers must substitute land expansion with intensive labor and land management practices to sustain yields per unit of land.
3
Distinguish correct demographic and land-use principles from distractors involving crop belts, location economics, and migration dynamics.
Confirmed that land scarcity driven by high rural population density is the primary driver of continuous intensive cropping in semi-arid West African agricultural systems.
Other options misattribute cash crop climatic belts, industrial location principles, or migration push/pull dynamics.

Anahtar Kavram

Agricultural Land-Use Intensification and Population Density
Tahmini Süre:2m 0s
Soru 3928Soru

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 3929Soru

A naturally occurring element XX consists of two stable isotopes, 63X^{63}X and 65X^{65}X. If the relative atomic mass of element XX is 63.663.6, what is the percentage abundance of the heavier isotope 65X^{65}X?

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Cevap: 30%30\%

Cevap

The percentage abundance of the heavier isotope 65X^{65}X is 30%30\%.
The relative atomic mass of an element with isotopes is calculated using the weighted average formula: RAM=(mass×fractional abundance)\text{RAM} = \sum (\text{mass} \times \text{fractional abundance}). Setting pp as the fraction of 65X^{65}X gives 63.6=63(1p)+65p=63+2p63.6 = 63(1 - p) + 65p = 63 + 2p. Solving yields 2p=0.62p = 0.6, so p=0.30p = 0.30, which corresponds to 30%30\%.

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1
Set up the relative atomic mass weighted average equation
63.6=(m1×x1)+(m2×x2)10063.6 = \frac{(m_1 \times x_1) + (m_2 \times x_2)}{100}
Relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes based on their fractional abundances.
2
Express isotopic abundances in terms of a single variable pp
Let pp be the fraction of 65X^{65}X, so the fraction of 63X^{63}X is 1p1 - p.
The sum of the fractional abundances of all isotopes of an element must equal 1 (or 100%100\%).
3
Substitute known values and solve for pp
63.6=63(1p)+65p    63.6=63+2p    2p=0.6    p=0.3063.6 = 63(1 - p) + 65p \implies 63.6 = 63 + 2p \implies 2p = 0.6 \implies p = 0.30
Algebraic expansion isolates the variable corresponding to the heavier isotope abundance.
4
Convert the fractional abundance to percentage
0.30×100%=30%0.30 \times 100\% = 30\%
Multiplying the decimal fraction by 100 yields the percentage abundance.

Anahtar Kavram

Calculation of Relative Atomic Mass from Isotopic Abundance
Tahmini Süre:1m 30s
Soru 3930Soru

An expanding commercial enterprise structured as a public limited company intends to raise long-term capital by inviting members of the general public to subscribe to its newly issued shares. Which official document is the company legally required to prepare, register, and issue to the public for this purpose?

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Cevap: Prospectus

Cevap

The Prospectus is the statutory document issued by a public limited company to invite the public to subscribe to its shares or debentures.
A Prospectus is specifically designed and legally mandated to invite members of the general public to purchase or subscribe for shares or debentures in a public limited company.

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1
Identify the primary operational goal described in the question stem.
The company needs to invite the general public to purchase its newly issued shares.
Only public limited companies are permitted to invite public subscriptions for capital securities.
2
Evaluate the statutory function of company documents.
The Prospectus contains financial performance, director information, and share offering details designed to inform prospective public investors.
Company law mandates the publication and registration of a Prospectus before public offer of securities.

Anahtar Kavram

Public Share Offerings and the Role of a Prospectus
Soru 3931Soru

A 6.95 g6.95\text{ g} sample of hydrated iron(II) tetraoxosulfate(VI), FeSO4xH2O\text{FeSO}_4 \cdot x\text{H}_2\text{O}, is dissolved in dilute tetraoxosulfate(VI) acid and made up to 250 cm3250\text{ cm}^3 in a volumetric flask. A 25.0 cm325.0\text{ cm}^3 portion of this solution requires 25.0 cm325.0\text{ cm}^3 of 0.020 mol dm30.020\text{ mol dm}^{-3} acidified potassium tetraoxomanganate(VII), KMnO4\text{KMnO}_4, for complete titration. Given the relative atomic masses Fe=56\text{Fe} = 56, S=32\text{S} = 32, O=16\text{O} = 16, and H=1\text{H} = 1, what is the integer value of xx, the number of molecules of water of crystallization per formula unit?

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Cevap: 7

Cevap

The integer value of x is 7.
By using the titration volume and concentration of acidified potassium tetraoxomanganate(VII), the amount of iron(II) ions in the sample is calculated. Knowing the total mole amount of anhydrous iron(II) tetraoxosulfate(VI) allows determination of the mass of the anhydrous salt component (3.80 g). Subtracting this from the initial hydrated sample mass (6.95 g) gives the mass of water of crystallization (3.15 g). Dividing the moles of water (0.175 mol) by the moles of anhydrous salt (0.025 mol) yields exactly 7 water molecules of crystallization per formula unit.

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1
Calculate the moles of KMnO4\text{KMnO}_4 consumed in the titration
n(KMnO4)=0.020 mol dm3×0.0250 dm3=0.00050 moln(\text{KMnO}_4) = 0.020 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 = 0.00050 \text{ mol}
Concentration and volume of titrant are provided.
2
Determine moles of Fe2+\text{Fe}^{2+} present in the 25.0 cm325.0\text{ cm}^3 aliquot using the redox reaction stoichiometry
n(Fe2+)25cm3=5×0.00050 mol=0.0025 moln(\text{Fe}^{2+})_{25\text{cm}^3} = 5 \times 0.00050 \text{ mol} = 0.0025 \text{ mol}
The mole ratio of MnO4\text{MnO}_4^- to Fe2+\text{Fe}^{2+} in acidic redox titration is 1:51:5 according to MnO4+5Fe2++8H+Mn2++5Fe3++4H2O\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}.
3
Calculate the total moles of FeSO4\text{FeSO}_4 in the original 250 cm3250\text{ cm}^3 volumetric flask
n(FeSO4)total=0.0025 mol×(250 cm325.0 cm3)=0.025 moln(\text{FeSO}_4)_{\text{total}} = 0.0025 \text{ mol} \times \left(\frac{250\text{ cm}^3}{25.0\text{ cm}^3}\right) = 0.025 \text{ mol}
The aliquot represents one-tenth of the total solution volume.
4
Calculate the mass of anhydrous FeSO4\text{FeSO}_4 in the sample
Molar mass of FeSO4=56+32+(4×16)=152 g mol1\text{FeSO}_4 = 56 + 32 + (4 \times 16) = 152 \text{ g mol}^{-1}. Mass =0.025 mol×152 g mol1=3.80 g= 0.025 \text{ mol} \times 152 \text{ g mol}^{-1} = 3.80 \text{ g}.
Converting moles of anhydrous salt to mass using molar mass.
5
Determine the mass and moles of water of crystallization
Mass of H2O=6.95 g3.80 g=3.15 g\text{H}_2\text{O} = 6.95 \text{ g} - 3.80 \text{ g} = 3.15 \text{ g}. Moles of H2O=3.15 g18 g mol1=0.175 mol\text{H}_2\text{O} = \frac{3.15 \text{ g}}{18 \text{ g mol}^{-1}} = 0.175 \text{ mol}.
Subtracting anhydrous mass from initial mass gives water of crystallization mass, converted to moles using molar mass of H2O=18 g mol1\text{H}_2\text{O} = 18 \text{ g mol}^{-1}.
6
Compute the hydration coefficient x=n(H2O)n(FeSO4)x = \frac{n(\text{H}_2\text{O})}{n(\text{FeSO}_4)}
x=0.175 mol0.025 mol=7x = \frac{0.175 \text{ mol}}{0.025 \text{ mol}} = 7
The coefficient xx represents the mole ratio of water of crystallization to anhydrous salt.

Anahtar Kavram

Quantitative determination of water of crystallization in hydrated salts via redox volumetric analysis
Soru 3932Soru

A 0.04 mol dm30.04\text{ mol dm}^{-3} aqueous solution of a weak monobasic acid, HA\text{HA}, is 2.0%2.0\% ionized at 25C25^\circ\text{C}. If distilled water is added to this solution until its total volume is quadrupled, what is the degree of ionization of the acid in the diluted solution?

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Cevap: 4.0%4.0\%

Cevap

The degree of ionization of the acid in the diluted solution is 4.0%4.0\%.
By Ostwald's dilution law for weak monobasic acids, Ka=α2CK_a = \alpha^2 C, which rearranges to α=KaC\alpha = \sqrt{\frac{K_a}{C}}. Since KaK_a is constant at a fixed temperature, the degree of ionization α\alpha is inversely proportional to the square root of the concentration (α1C\alpha \propto \frac{1}{\sqrt{C}}). Quadrupling the volume decreases the concentration by a factor of 4 (C2=C14C_2 = \frac{C_1}{4}), which increases the degree of ionization by a factor of 4=2\sqrt{4} = 2. Therefore, the new degree of ionization is 2.0%×2=4.0%2.0\% \times 2 = 4.0\%.

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1
Calculate the acid dissociation constant (KaK_a) using the initial concentration and degree of ionization.
Initial concentration C1=0.04 mol dm3C_1 = 0.04\text{ mol dm}^{-3}, initial degree of ionization α1=2.0%=0.02\alpha_1 = 2.0\% = 0.02. Using Kaα12C1K_a \approx \alpha_1^2 C_1, we get Ka=(0.02)2×0.04=4.0×104×0.04=1.6×105 mol dm3K_a = (0.02)^2 \times 0.04 = 4.0 \times 10^{-4} \times 0.04 = 1.6 \times 10^{-5}\text{ mol dm}^{-3}.
The value of KaK_a depends only on temperature and remains constant upon dilution.
2
Determine the new concentration (C2C_2) after quadrupling the solution volume.
C2=C14=0.04 mol dm34=0.01 mol dm3C_2 = \frac{C_1}{4} = \frac{0.04\text{ mol dm}^{-3}}{4} = 0.01\text{ mol dm}^{-3}.
Diluting a solution to 4 times its original volume reduces its molar concentration by a factor of 4.
3
Calculate the new degree of ionization (α2\alpha_2) in the diluted solution.
\alpha_2 = \sqrt{\frac{K_a}{C_2}} = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = \sqrt{1.6 \times 10^{-3}} = \sqrt{16 \times 10^{-4}} = 4.0 \times 10^{-2} = 0.04 = 4.0\%$.
Applying Ostwald's dilution law to find the updated degree of dissociation at the lower concentration.

Anahtar Kavram

Ostwald's Dilution Law and the relationship between dilution, concentration, and degree of ionization of weak electrolytes.
Tahmini Süre:2m 30s
Soru 3933Soru

Match each atmospheric pollutant or component in Column I with its corresponding chemical mode of action or environmental transformation mechanism in Column II.

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Öğeler

Carbon monoxide (CO\text{CO})
Sulfur dioxide (SO2\text{SO}_2)
Chlorofluorocarbons (CFCs\text{CFCs})
Nitrogen dioxide (NO2\text{NO}_2)

Eşleşmeler

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Cevap

Carbon monoxide matches with irreversible hemoglobin binding; Sulfur dioxide matches with oxidation to an acid anhydride causing limestone corrosion; Chlorofluorocarbons match with UV-induced photolysis producing chlorine radicals; Nitrogen dioxide matches with solar photolysis initiating photochemical smog.
Each atmospheric pollutant is accurately matched to its chemical mechanism: carbon monoxide forms carboxyhemoglobin in blood, sulfur dioxide forms acid anhydrides causing marble/limestone corrosion, chlorofluorocarbons generate ozone-depleting chlorine radicals under UV radiation, and nitrogen dioxide photolyzes in sunlight to drive photochemical smog formation.

Adım Adım Çözüm

1
Analyze Carbon monoxide (CO\text{CO})
It readily binds to blood hemoglobin to form carboxyhemoglobin.
CO\text{CO} toxicological impact relies on inhibiting cellular respiration by reducing oxygen transport.
2
Analyze Sulfur dioxide (SO2\text{SO}_2)
It oxidizes to SO3\text{SO}_3 (sulfur trioxide, an acid anhydride), forming sulfuric acid rain.
Acid rain dissolves building materials containing calcium carbonate (CaCO3\text{CaCO}_3).
3
Analyze Chlorofluorocarbons (CFCs\text{CFCs})
High-energy solar ultraviolet radiation breaks C-Cl\text{C-Cl} bonds in CFCs\text{CFCs}, producing free chlorine radicals.
Chlorine radicals act as catalysts in the breakdown of stratospheric ozone molecules into oxygen gas.
4
Analyze Nitrogen dioxide (NO2\text{NO}_2)
In the troposphere, NO2\text{NO}_2 absorbs near-UV sunlight, splitting into NO\text{NO} and atomic oxygen (O\text{O}).
Atomic oxygen combines with O2\text{O}_2 to form tropospheric ozone, a key ingredient in photochemical smog.

Anahtar Kavram

Chemical transformations and atmospheric impacts of primary air pollutants
Soru 3934Soru

A 200 cm3200\text{ cm}^3 sample of a saturated potassium trioxonitrate(V) solution, KNO3\text{KNO}_3, at 60C60^\circ\text{C} has a concentration of 1.5 mol/dm31.5\text{ mol/dm}^3. The solution is cooled to 25C25^\circ\text{C} to form a supersaturated solution. When a seed crystal is added, excess solute crystallizes until a new saturated concentration of 0.5 mol/dm30.5\text{ mol/dm}^3 is reached at 25C25^\circ\text{C}. What mass of KNO3\text{KNO}_3 crystallizes out of the solution? [Molar mass of KNO3=101 g/mol\text{KNO}_3 = 101\text{ g/mol}]

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Cevap: 20.2 g20.2\text{ g}

Cevap

20.2 g20.2\text{ g} of KNO3\text{KNO}_3 crystallizes out of the solution.
Cooling a saturated solution creates a unstable supersaturated state. Adding a seed crystal induces rapid crystallization of the excess solute until saturation equilibrium is re-established at the lower temperature. The difference in concentration is 1.0 mol/dm31.0\text{ mol/dm}^3. For 0.200 dm30.200\text{ dm}^3, this equals 0.20 mol0.20\text{ mol} of KNO3\text{KNO}_3, which has a mass of 0.20 mol×101 g/mol=20.2 g0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}.

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1
Convert the volume of the solution from cm3\text{cm}^3 to dm3\text{dm}^3.
Volume=200 cm31000=0.200 dm3\text{Volume} = \frac{200\text{ cm}^3}{1000} = 0.200\text{ dm}^3.
Concentrations are given in mol/dm3\text{mol/dm}^3, so volume must be in dm3\text{dm}^3.
2
Calculate the moles of KNO3\text{KNO}_3 dissolved initially at 60C60^\circ\text{C} and remaining at 25C25^\circ\text{C}.
Initial moles=1.5 mol/dm3×0.200 dm3=0.30 mol\text{Initial moles} = 1.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.30\text{ mol}. Final moles=0.5 mol/dm3×0.200 dm3=0.10 mol\text{Final moles} = 0.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.10\text{ mol}.
Determining the mole difference shows how much solute leaves the supersaturated state upon seeding.
3
Determine the amount of moles precipitated and convert to mass.
Moles precipitated=0.30 mol0.10 mol=0.20 mol\text{Moles precipitated} = 0.30\text{ mol} - 0.10\text{ mol} = 0.20\text{ mol}. Mass precipitated=0.20 mol×101 g/mol=20.2 g\text{Mass precipitated} = 0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}.
Multiplying the precipitated moles by the molar mass gives the required mass in grams.

Anahtar Kavram

Mass of solute crystallized from a supersaturated solution upon reaching saturation equilibrium
Soru 3935Soru

In ammonium trioxonitrate(V), NH4NO3NH_4NO_3, nitrogen exists in two distinct ionic environments. What are the respective oxidation numbers of the nitrogen atom in the cation and the nitrogen atom in the anion, and what is the systematic IUPAC name of the nitrogen-containing oxoanion formed when the anion is reduced by gaining two electrons?

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Cevap: 3-3 and +5+5; dioxonitrate(III) ion

Cevap

The respective oxidation numbers are 3-3 for nitrogen in the ammonium cation and +5+5 for nitrogen in the trioxonitrate(V) anion. The reduced species formed upon gaining two electrons is the dioxonitrate(III) ion.
In the ionic compound NH4NO3NH_4NO_3, the nitrogen atom in the cation NH4+NH_4^+ has an oxidation state of 3-3 because x+4(+1)=+1x + 4(+1) = +1. The nitrogen atom in the anion NO3NO_3^- has an oxidation state of +5+5 because y+3(2)=1y + 3(-2) = -1. When NO3NO_3^- is reduced by two electrons, the oxidation number of nitrogen decreases from +5+5 to +3+3, converting NO3NO_3^- to NO2NO_2^-. In IUPAC nomenclature, NO2NO_2^- is named the dioxonitrate(III) ion due to its two oxo ligands and nitrogen oxidation state of +3+3.

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1
Determine the oxidation state of nitrogen in the ammonium cation (NH4+NH_4^+).
Let the oxidation state of N be xx. Since hydrogen is +1+1 and the net charge is +1+1: x+4(+1)=+1    x=3x + 4(+1) = +1 \implies x = -3.
Ammonium is a polyatomic cation where the sum of oxidation numbers equals the net ionic charge.
2
Determine the oxidation state of nitrogen in the trioxonitrate(V) anion (NO3NO_3^-).
Let the oxidation state of N be yy. Since oxygen is 2-2 and the net charge is 1-1: y+3(2)=1    y=+5y + 3(-2) = -1 \implies y = +5.
Nitrate is a polyatomic anion where oxygen exhibits an oxidation number of 2-2.
3
Calculate the oxidation state of nitrogen after two-electron reduction of the anion and determine its IUPAC name.
Reduction by gaining 2 electrons decreases the oxidation number of nitrogen: +52=+3+5 - 2 = +3. The resulting species containing nitrogen in the +3+3 state with two oxygen atoms (NO2NO_2^-) is systematically named the dioxonitrate(III) ion.
Reduction corresponds to a gain of electrons (decrease in oxidation number), and IUPAC rules for oxoanions specify the prefix for oxygen count ('dioxo-') followed by the central element and its oxidation state in Roman numerals.

Anahtar Kavram

Polyatomic Ion Oxidation Numbers and IUPAC Nomenclature of Oxoanions
Soru 3936Soru

Calculate the mass number and atomic number of the final daughter nuclide in the given radioactive decay series.

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When a nucleus of Thorium-232 (90232Th^{232}_{90}\text{Th}) undergoes a decay series emitting 66 alpha particles (α\alpha) and 44 beta particles (β\beta^-), the resulting stable daughter nuclide has a mass number of and an atomic number of .
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Cevap

The final stable daughter nuclide has a mass number of 208 and an atomic number of 82.
Emission of 6 alpha particles reduces the mass number by 6×4=246 \times 4 = 24 (from 232 to 208) and the atomic number by 6×2=126 \times 2 = 12. The subsequent emission of 4 beta particles increases the atomic number by 4×1=44 \times 1 = 4, resulting in a final atomic number of 9012+4=8290 - 12 + 4 = 82.

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1
Calculate the total change in mass number (AA) caused by alpha and beta particle emissions.
Each alpha particle (24He^{4}_{2}\text{He}) reduces AA by 4, while beta particles (10e^{0}_{-1}\text{e}) do not change AA. Total decrease in A=6×4=24A = 6 \times 4 = 24.
Alpha particles carry 4 atomic mass units, whereas beta particles have virtually zero mass number.
2
Determine the final mass number.
Final mass number = 23224=208232 - 24 = 208.
Subtracting the lost nucleon mass from the initial mass number gives the mass number of the resulting nucleus.
3
Calculate the total change in atomic number (ZZ) caused by alpha and beta particle emissions.
Each alpha particle reduces ZZ by 2, while each beta particle increases ZZ by 1. Total change in Z=(6×2)+(4×+1)=12+4=8Z = (6 \times -2) + (4 \times +1) = -12 + 4 = -8.
Alpha decay removes 2 protons, whereas beta minus decay converts a neutron into a proton, increasing nuclear charge by 1.
4
Determine the final atomic number.
Final atomic number = 908=8290 - 8 = 82.
Applying the net charge change to the initial atomic number of Thorium (90) yields 82, which corresponds to Lead (Pb).

Anahtar Kavram

Conservation of mass number and atomic number in decay series
Soru 3937Soru

Complete the sentence below by filling in the blank with the appropriate idiomatic expression that best fits the context.

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Despite explicit warnings from the regulatory agency, the investment directors continued to with clients' capital, ultimately precipitating the collapse of the fund.
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Cevap

play fast and loose
The expression 'play fast and loose' is an established English idiom meaning to act irresponsibly, dishonestly, or recklessly with someone or something, which accurately matches the scenario of reckless handling of investment capital.

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1
Analyze the context and tone of the sentence
The context involves investment directors engaging in reckless, deceitful, and irresponsible financial actions with clients' funds despite regulatory warnings.
Identifying the context helps determine the precise meaning required for the missing idiomatic expression.
2
Identify the idiomatic phrase that matches this figurative meaning
The idiom 'play fast and loose' means to behave dishonestly, deceitfully, or irresponsibly, especially by taking improper liberties with rules or trust.
This exact phrase fits both the syntax of the infinitive 'to [idiom]' and the semantic requirement of reckless financial handling.

Anahtar Kavram

Idiomatic Expressions - Expressing Deceit and Recklessness
Soru 3938Soru

During the public inquiry, the panel meticulously cross-examined the key witness regarding the submitted documentation. Which word nearest in meaning to 'truthfulness' appropriately completes the sentence below?

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The members of the commission expressed serious reservations concerning the of the statements contained in the report.
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Cevap

veracity (or truthfulness)
The word 'veracity' means conformity to facts, accuracy, or truthfulness. In the context of investigating official statements, questioning their veracity means doubting whether they are true.

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1
Analyze the context of the sentence to identify the required meaning of the blank.
The sentence discusses doubts surrounding whether the statements in a report are factual and honest.
The phrase 'reservations concerning the...' requires a noun denoting factual accuracy or truthfulness.
2
Identify a precise formal synonym for 'truthfulness' suitable for an administrative or academic register.
The noun 'veracity' signifies adherence to truth, accuracy, or correctness.
'Veracity' fits the formal register of the sentence and directly substitutes for 'truthfulness'.

Anahtar Kavram

Synonyms and Contextual Word Meaning
Tahmini Süre:1m 0s
Soru 3939Soru

A mountain climber of mass 60 kg60\text{ kg} climbs a vertical height of 15 m15\text{ m} in a time of 30 s30\text{ s}. Taking the acceleration due to gravity g=10 m s2g = 10\text{ m s}^{-2}, calculate the average power expended by the climber in Watts.

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Cevap: 300

Cevap

The average power expended by the climber is 300 W300\text{ W}.
The work done in lifting a mass mm through vertical height hh is given by W=mgh=60×10×15=9000 JW = mgh = 60 \times 10 \times 15 = 9000\text{ J}. The average power is the rate of doing work, P=Wt=9000 J30 s=300 WP = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}.

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1
Identify the given values and formula for work done against gravity.
Mass m=60 kgm = 60\text{ kg}, vertical displacement h=15 mh = 15\text{ m}, acceleration due to gravity g=10 m s2g = 10\text{ m s}^{-2}, and time t=30 st = 30\text{ s}. Work done formula: W=mghW = mgh.
The climber works against gravity to increase their potential energy by an amount equal to mghmgh.
2
Calculate the total work done.
W=60 kg×10 m s2×15 m=9000 JW = 60\text{ kg} \times 10\text{ m s}^{-2} \times 15\text{ m} = 9000\text{ J}.
Multiplying force (mgmg) by vertical distance (hh) yields work done in Joules.
3
Calculate the average power.
P=Wt=9000 J30 s=300 WP = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}.
Power is defined as work done divided by the time interval (P=WtP = \frac{W}{t}).

Anahtar Kavram

Power as the rate of doing work against gravity
Tahmini Süre:45s
Soru 3940Soru

Match each commercial contract scenario described in Column I with its corresponding legal concept, mode of discharge, or legal remedy in Column II.

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Öğeler

An agreement to supply agricultural produce becomes impossible to perform after contract formation due to a sudden statutory export prohibition passed by the government.
A seller genuinely believes and states that a commercial vehicle's engine is newly overhauled, inducing a buyer to purchase it, though it later turns out to be false without intent to deceive.
A structural engineer completes 70% of a design project before the client wrongfully repudiates the contract, prompting the engineer to seek recovery for the work already rendered.
A property vendor refuses to convey title to a unique parcel of commercial land after receiving full payment, where financial compensation is demonstrably inadequate.

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Cevap

The scenarios correctly match as follows: the statutory export prohibition constitutes discharge by frustration due to supervening illegality; the honest but false representation of vehicle condition is innocent misrepresentation rendering the contract voidable; the partial project claim following wrongful repudiation is a remedy of quantum meruit; and the refusal to transfer unique real estate where damages are inadequate warrants specific performance.
Each legal situation aligns with established common law contract principles: supervening statutory prohibition causes discharge by frustration; false statements made without fraud constitute innocent misrepresentation; recovery for partial work executed before wrongful repudiation is sought via quantum meruit; and specific performance is decreed for land contracts due to the unique nature of real property.

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1
Analyze Scenario 1 (Statutory export prohibition)
Performance becomes legally impossible after contract formation due to external legal changes without fault of either party.
This satisfies the legal criteria for discharge by frustration (supervening illegality).
2
Analyze Scenario 2 (Believed true statement inducing contract)
An untrue statement of material fact was made without fraudulent intent or deceit.
This is an innocent misrepresentation, acting as a vitiating element that makes the contract voidable.
3
Analyze Scenario 3 (Partial performance interrupted by repudiation)
Work was performed before the contract was wrongfully terminated by the counterparty.
The aggrieved party can claim quantum meruit ('as much as he has earned') rather than suing purely for unliquidated breach damages.
4
Analyze Scenario 4 (Refusal to transfer unique land parcel)
Breach occurs over unique property where monetary damages cannot restore the injured buyer.
Courts grant the equitable remedy of specific performance to compel performance of the express contract.

Anahtar Kavram

Law of Contract: Discharge, Vitiating Elements, and Remedies for Breach
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