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13931 soru

Soru 6701Soru

A sound wave traveling through air at a speed of 340 m/s340\text{ m/s} has a wavelength of 0.85 m0.85\text{ m}. Upon entering a second gaseous medium, its speed decreases to 272 m/s272\text{ m/s}. What is the wavelength of the wave in the second medium?

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Cevap: 0.68 m0.68\text{ m}

Cevap

The wavelength of the wave in the second medium is 0.68 m0.68\text{ m}.
Because wave frequency is determined by the source, it stays constant at 400 Hz400\text{ Hz} across both media. Using λ2=v2f\lambda_2 = \frac{v_2}{f}, the wavelength in the second medium is 272 m/s400 Hz=0.68 m\frac{272\text{ m/s}}{400\text{ Hz}} = 0.68\text{ m}.

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1
Calculate the frequency of the wave in the first medium using the wave equation v=fλv = f \lambda.
f=v1λ1=340 m/s0.85 m=400 Hzf = \frac{v_1}{\lambda_1} = \frac{340\text{ m/s}}{0.85\text{ m}} = 400\text{ Hz}.
The source determines the wave frequency, which remains unchanged when passing into a new medium.
2
Apply the wave equation with the constant frequency to find the new wavelength in the second medium.
λ2=v2f=272 m/s400 Hz=0.68 m\lambda_2 = \frac{v_2}{f} = \frac{272\text{ m/s}}{400\text{ Hz}} = 0.68\text{ m}.
Wave speed changes in a new medium lead directly to proportional changes in wavelength since frequency is constant.

Anahtar Kavram

Invariance of wave frequency across media boundaries and application of the wave equation v=fλv = f \lambda.
Soru 6702Soru
Consider the thermal decomposition of potassium trioxochlorate(V) represented by the balanced equation:
2KClO3(s)2KCl(s)+3O2(g)2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)
What mass of KClO3KClO_3 is required to produce 6.72 dm36.72\text{ dm}^3 of oxygen gas measured at STP?
[K=39.0, Cl=35.5, O=16.0; Molar volume of gas at STP =22.4 dm3 mol1][K = 39.0,\text{ } Cl = 35.5,\text{ } O = 16.0;\text{ Molar volume of gas at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]
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Cevap: 24.5 g24.5\text{ g}

Cevap

24.5 g24.5\text{ g} of KClO3KClO_3 is required.
According to the balanced chemical equation, 2 moles2\text{ moles} of KClO3KClO_3 (245.0 g245.0\text{ g}) produce 3 moles3\text{ moles} of O2O_2 (67.2 dm367.2\text{ dm}^3 at STP). By direct proportion, 6.72 dm36.72\text{ dm}^3 of O2O_2 requires 6.7267.2×245.0=24.5 g\frac{6.72}{67.2} \times 245.0 = 24.5\text{ g} of KClO3KClO_3.

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1
Calculate the molar mass of KClO3KClO_3 and the total mass of 2 moles of KClO3KClO_3.
Molar mass of KClO3=39.0+35.5+3(16.0)=122.5 g mol1KClO_3 = 39.0 + 35.5 + 3(16.0) = 122.5\text{ g mol}^{-1}. Mass of 2 moles=2×122.5=245.0 g2\text{ moles} = 2 \times 122.5 = 245.0\text{ g}.
The balanced chemical equation shows 2 moles2\text{ moles} of KClO3KClO_3 undergo decomposition.
2
Calculate the volume of 3 moles of O2O_2 at STP.
Volume of 3 moles of O2=3×22.4 dm3=67.2 dm33\text{ moles of } O_2 = 3 \times 22.4\text{ dm}^3 = 67.2\text{ dm}^3.
At STP, 1 mole1\text{ mole} of any gas occupies 22.4 dm322.4\text{ dm}^3.
3
Set up a proportion to find the mass of KClO3KClO_3 needed to yield 6.72 dm36.72\text{ dm}^3 of O2O_2.
\text{Mass of } KClO_3 = \frac{6.72\text{ dm}^3}{67.2\text{ dm}^3} \times 245.0\text{ g} = 24.5\text{ g}.
Direct stoichiometric ratio relates mass of reactant to volume of gaseous product.

Anahtar Kavram

Mass-Volume Stoichiometry at STP
Soru 6703Soru

A merchant receives a written, unconditional promise signed by a debtor, undertaking to pay a specified sum of money on demand or at a fixed future date to a named payee. Which instrument of credit does this document represent?

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Cevap: Promissory Note

Cevap

Promissory Note
A promissory note is defined legally as an unconditional promise in writing, made and signed by the debtor (maker), promising to pay on demand or at a fixed or determinable future time a sum certain in money to a specified person or bearer.

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1
Analyze the core features described in the document.
The document is an unconditional promise written and signed directly by the debtor to pay a specified sum.
Identifying who initiates the instrument (debtor vs creditor) and the nature of the obligation (promise vs order) determines the specific credit instrument.
2
Distinguish between an order to pay and a promise to pay.
An unconditional promise to pay issued by the debtor is a Promissory Note, whereas an unconditional order to pay issued by a creditor is a Bill of Exchange.
Understanding key legal definitions under commercial credit instruments ensures accurate classification.

Anahtar Kavram

Forms and Instruments of Credit
Tahmini Süre:1m 0s
Soru 6704Soru

An industrial wastewater slurry contains suspended micro-fine latex particles, dissolved sodium chloride, and insoluble magnetite (Fe3O4Fe_3O_4) powder. To efficiently recover pure magnetite powder, isolate the micro-fine latex particles, and obtain pure water from the mixture without chemical additives, which sequence of separation techniques should be applied?

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Cevap: Application of a magnetic field to extract magnetite, high-speed centrifugation to sediment latex particles, followed by distillation to recover pure water

Cevap

Application of a magnetic field to extract magnetite, high-speed centrifugation to sediment latex particles, followed by distillation to recover pure water
Magnetite is ferromagnetic and can be selectively extracted using a magnetic field. Colloidal latex micro-particles do not settle under normal gravity due to thermal agitation (Brownian motion), requiring high-speed centrifugation to force sedimentation based on density. Distillation then vaporizes and condenses water to isolate it from dissolved non-volatile sodium chloride.

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1
Identify the physical properties of each component in the mixture
Magnetite (Fe3O4Fe_3O_4) is magnetic and insoluble; latex forms a micro-suspended colloid; sodium chloride forms a true aqueous solution; water is the liquid solvent.
Separation techniques must exploit differences in specific physical properties such as magnetism, particle mass/density, and boiling point.
2
Apply magnetization first
Insoluble magnetic Fe3O4Fe_3O_4 is attracted and cleanly separated from the liquid slurry.
Magnetic separation isolates the strongly magnetic solid without altering or contaminating liquid phase components.
3
Apply high-speed centrifugation to the remaining mixture
Suspended colloidal latex particles sediment rapidly out of the liquid phase.
Colloidal particles remain in suspension under normal gravity due to Brownian motion, but high centrifugal force forces them to settle based on density differences.
4
Perform distillation on the remaining true solution
Water evaporates and condenses as pure distillate while sodium chloride remains in the distillation flask.
Distillation separates volatile solvent liquid from non-volatile dissolved solid solutes based on boiling point differences.

Anahtar Kavram

Sequential physical separation of complex mixtures exploiting magnetism, centrifugal sedimentation of colloids, and differential boiling points.
Tahmini Süre:2m 0s
Soru 6705Soru

A saturated solution contains 12.0 g12.0\text{ g} of sodium hydroxide (NaOH\text{NaOH}) dissolved in 500 cm3500\text{ cm}^3 of solution at 25C25^\circ\text{C}. What is the solubility of sodium hydroxide in mol dm3\text{mol dm}^{-3} at this temperature?
[Molar mass of NaOH=40.0 g mol1\text{NaOH} = 40.0\text{ g mol}^{-1}]

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Cevap: 0.60 mol dm30.60\text{ mol dm}^{-3}

Cevap

0.60 mol dm30.60\text{ mol dm}^{-3}
To find the solubility in mol dm3\text{mol dm}^{-3}, first determine the number of moles of NaOH\text{NaOH} by dividing the mass (12.0 g12.0\text{ g}) by the molar mass (40.0 g mol140.0\text{ g mol}^{-1}), yielding 0.30 mol0.30\text{ mol}. Next, convert the volume to cubic decimeters (500 cm3=0.50 dm3500\text{ cm}^3 = 0.50\text{ dm}^3). Finally, divide the amount in moles by the volume in dm3\text{dm}^3 to get 0.60 mol dm30.60\text{ mol dm}^{-3}.

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1
Calculate the amount of NaOH\text{NaOH} in moles
Moles=12.0 g40.0 g mol1=0.30 mol\text{Moles} = \frac{12.0\text{ g}}{40.0\text{ g mol}^{-1}} = 0.30\text{ mol}
Solubility in mol dm3\text{mol dm}^{-3} requires expressing the amount of solute in moles.
2
Convert the solution volume from cm3\text{cm}^3 to dm3\text{dm}^3
Volume=500 cm31000 cm3 dm3=0.50 dm3\text{Volume} = \frac{500\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.50\text{ dm}^3
Standard concentration units require volume expressed in cubic decimeters.
3
Divide the number of moles by the volume in dm3\text{dm}^3
Solubility=0.30 mol0.50 dm3=0.60 mol dm3\text{Solubility} = \frac{0.30\text{ mol}}{0.50\text{ dm}^3} = 0.60\text{ mol dm}^{-3}
Solubility is the maximum amount of solute in moles that dissolves per dm3\text{dm}^3 of solution at a given temperature.

Anahtar Kavram

Solubility calculation in mol/dm³
Soru 6706Soru

A public limited company raises long-term corporate finance by issuing various financial instruments to investors. Which statement accurately describes the financial status and rights of debenture holders in a public limited company?

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Cevap: They are creditors of the company who receive fixed interest payments regardless of whether the company makes a profit.

Cevap

Debenture holders are creditors of the public limited company who receive fixed interest payments regardless of profit.
In a public limited company, debenture holders are lenders (creditors) rather than owners. Because debentures represent borrowed capital, the company is legally required to pay the agreed fixed rate of interest periodically as an expense, regardless of whether it registers a profit or a loss.

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1
Identify the nature of debentures as a corporate security instrument.
Debentures represent long-term borrowed capital (debt) rather than equity capital.
Understanding whether an instrument is debt or equity determines the legal rights and financial status of the holder.
2
Analyze the financial return and legal status associated with debentures.
Holders are creditors entitled to fixed interest charged against revenue, rather than owners receiving profit dividends.
Interest on debt is an obligation payable prior to profit distribution and does not depend on profit availability.

Anahtar Kavram

Debentures as Loan Capital in Public Limited Companies
Soru 6707Soru

A standard solution is prepared by dissolving 1.575 g1.575\text{ g} of hydrated ethanedioic acid (H2C2O4xH2O\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}) in distilled water to make 250.0 cm3250.0\text{ cm}^3 of solution. A 25.0 cm325.0\text{ cm}^3 sample of this acid solution requires 25.0 cm325.0\text{ cm}^3 of a 0.100 mol dm30.100\text{ mol dm}^{-3} sodium hydroxide (NaOH\text{NaOH}) solution for complete neutralization. What is the integer value of xx in the formula of the hydrated acid? [Relative atomic masses: H=1\text{H} = 1, C=12\text{C} = 12, O=16\text{O} = 16]

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Cevap: 2

Cevap

The integer value of xx is 2.
The value of xx is calculated as 22. Based on the reaction stoichiometry, 25.0 cm325.0\text{ cm}^3 of 0.100 mol dm30.100\text{ mol dm}^{-3} NaOH\text{NaOH} contains 0.0025 mol0.0025\text{ mol} of base, which neutralizes 0.00125 mol0.00125\text{ mol} of the diprotic acid in the 25.0 cm325.0\text{ cm}^3 sample. The entire 250.0 cm3250.0\text{ cm}^3 solution therefore contains 0.0125 mol0.0125\text{ mol} of acid. Dividing the mass (1.575 g1.575\text{ g}) by 0.0125 mol0.0125\text{ mol} gives a molar mass of 126 g mol1126\text{ g mol}^{-1} for H2C2O4xH2O\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}. Subtracting the molar mass of anhydrous H2C2O4\text{H}_2\text{C}_2\text{O}_4 (90 g mol190\text{ g mol}^{-1}) gives 36 g mol136\text{ g mol}^{-1} for water, which corresponds to x=36/18=2x = 36 / 18 = 2.

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1
Determine the mole ratio from the balanced chemical neutralization equation.
H2C2O4xH2O+2NaOHNa2C2O4+(x+2)H2O\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + (x+2)\text{H}_2\text{O}. The stoichiometric ratio of acid to base is 1:21 : 2.
Ethanedioic acid is a diprotic acid requiring two moles of hydroxide ions for complete neutralization per mole of acid.
2
Calculate the amount in moles of sodium hydroxide solution used in the titration.
Moles of NaOH=0.100 mol dm3×25.01000 dm3=0.0025 mol\text{Moles of NaOH} = 0.100\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.0025\text{ mol}.
Number of moles is equal to molar concentration multiplied by volume in cubic decimeters.
3
Calculate the total moles of hydrated acid present in the 250.0 cm3250.0\text{ cm}^3 volumetric flask.
Moles in 25.0 cm3 aliquot=0.00252=0.00125 mol\text{Moles in } 25.0\text{ cm}^3 \text{ aliquot} = \frac{0.0025}{2} = 0.00125\text{ mol}. Total moles in 250.0 cm3=0.00125×250.025.0=0.0125 mol250.0\text{ cm}^3 = 0.00125 \times \frac{250.0}{25.0} = 0.0125\text{ mol}.
Using the stoichiometric ratio (na/nb=1/2n_a/n_b = 1/2) and scaling up from the 25.0 cm325.0\text{ cm}^3 aliquot to the full 250.0 cm3250.0\text{ cm}^3 solution volume.
4
Compute the molar mass of the hydrated acid and solve for xx.
Molar mass=1.575 g0.0125 mol=126 g mol1\text{Molar mass} = \frac{1.575\text{ g}}{0.0125\text{ mol}} = 126\text{ g mol}^{-1}. Molar mass of anhydrous H2C2O4=2(1)+2(12)+4(16)=90 g mol1\text{H}_2\text{C}_2\text{O}_4 = 2(1) + 2(12) + 4(16) = 90\text{ g mol}^{-1}. Mass of xH2O=12690=36 g mol1x\text{H}_2\text{O} = 126 - 90 = 36\text{ g mol}^{-1}. Thus, x=3618=2x = \frac{36}{18} = 2.
Subtracting the molar mass of the anhydrous acid from the total molar mass gives the mass of the water of crystallization, which is divided by the molar mass of water (18 g mol118\text{ g mol}^{-1}) to find xx.

Anahtar Kavram

Volumetric Analysis and Water of Crystallization Determination
Soru 6708Soru

A light rigid bar PQPQ of length 5.0 m5.0\text{ m} is hinged at end PP and held horizontally. An upward force of 40 N40\text{ N} is applied at end QQ at an angle of 3030^\circ to the bar. To keep the bar in horizontal equilibrium, a vertical downward force FF is applied at a distance of 2.0 m2.0\text{ m} from PP. What is the magnitude of the force FF?

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Cevap: 50 N50\text{ N}

Cevap

The magnitude of the force FF required to maintain equilibrium is 50 N50\text{ N}.
The force of 40 N40\text{ N} applied at an angle of 3030^\circ to the bar has a perpendicular component of 40sin30=20 N40\sin 30^\circ = 20\text{ N}. The counterclockwise moment about pivot PP is 20 N×5.0 m=100 Nm20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}. For the bar to remain in equilibrium, the clockwise moment created by FF must equal 100 Nm100\text{ N}\cdot\text{m}, giving F×2.0 m=100 NmF \times 2.0\text{ m} = 100\text{ N}\cdot\text{m}, which yields F=50 NF = 50\text{ N}.

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1
Calculate the perpendicular component of the force applied at end QQ.
F=40 N×sin(30)=40×0.5=20 NF_{\perp} = 40\text{ N} \times \sin(30^\circ) = 40 \times 0.5 = 20\text{ N}
Only the force component perpendicular to the bar produces a moment about the pivot PP.
2
Calculate the counterclockwise moment produced by the force at QQ about pivot PP.
\text{Moment}_{Q} = 20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}
Moment is defined as perpendicular force multiplied by distance from the pivot.
3
Apply the Principle of Moments about pivot PP to find force FF.
F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m} \implies F = \frac{100}{2.0} = 50\text{ N}
For rotational equilibrium, total clockwise moment about PP must equal total counterclockwise moment about PP.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium with Forces at an Angle
Soru 6709Soru

A uniform cylindrical wire of length 2.0m2.0\,\text{m} and radius 1.0mm1.0\,\text{mm} is connected to a direct-current source. When a potential difference of 12V12\,\text{V} is applied across its ends, a steady current of 4.0A4.0\,\text{A} flows through it. Taking π3.142\pi \approx 3.142, what is the resistivity of the material of the wire?

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Cevap: 4.71×106Ωm4.71 \times 10^{-6}\,\Omega\cdot\text{m}

Cevap

The resistivity of the material of the wire is 4.71×106Ωm4.71 \times 10^{-6}\,\Omega\cdot\text{m}.
Using Ohm's law (R=V/IR = V/I), the wire's resistance is 3.0Ω3.0\,\Omega. Substituting this resistance, the wire length (2.0m2.0\,\text{m}), and the circular area (A=πr2=3.142×106m2A = \pi r^2 = 3.142 \times 10^{-6}\,\text{m}^2) into ρ=RA/L\rho = R A / L yields 4.71×106Ωm4.71 \times 10^{-6}\,\Omega\cdot\text{m}.

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1
Calculate the resistance of the wire using Ohm's Law.
R=VI=12V4.0A=3.0ΩR = \frac{V}{I} = \frac{12\,\text{V}}{4.0\,\text{A}} = 3.0\,\Omega
Ohm's Law relates potential difference, current, and resistance.
2
Calculate the cross-sectional area of the wire from its radius.
A=πr2=3.142×(1.0×103m)2=3.142×106m2A = \pi r^2 = 3.142 \times (1.0 \times 10^{-3}\,\text{m})^2 = 3.142 \times 10^{-6}\,\text{m}^2
The wire has a circular cross-section.
3
Rearrange the resistance formula R=ρLAR = \frac{\rho L}{A} to solve for resistivity ρ\rho.
ρ=RAL=3.0Ω×3.142×106m22.0m=4.713×106Ωm4.71×106Ωm\rho = \frac{R A}{L} = \frac{3.0\,\Omega \times 3.142 \times 10^{-6}\,\text{m}^2}{2.0\,\text{m}} = 4.713 \times 10^{-6}\,\Omega\cdot\text{m} \approx 4.71 \times 10^{-6}\,\Omega\cdot\text{m}
Resistivity is an intrinsic property derived from resistance, length, and cross-sectional area.

Anahtar Kavram

Relationship between resistance, potential difference, current, and material resistivity.
Soru 6710Soru

A sample of well water forms a white precipitate when boiled and requires a large amount of soap to produce lather. Which soluble compound present in the water is responsible for this temporary hardness?

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Cevap: Ca(HCO3)2\text{Ca(HCO}_3)_2

Cevap

Calcium hydrogentrioxocarbonate(IV), Ca(HCO3)2\text{Ca(HCO}_3)_2, is responsible for temporary hardness.
Temporary water hardness is caused by dissolved hydrogentrioxocarbonate(IV) salts such as calcium hydrogentrioxocarbonate(IV), Ca(HCO3)2\text{Ca(HCO}_3)_2. Heating the water decomposes this soluble compound into insoluble calcium trioxocarbonate(IV), carbon dioxide gas, and water, thereby removing the calcium ions responsible for hardness.

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1
Identify the type of water hardness based on the behavior upon heating.
Because boiling removes the hardness by forming a precipitate, the water sample exhibits temporary hardness.
Temporary hardness is caused by hydrogentrioxocarbonate(IV) salts of calcium or magnesium, which undergo thermal decomposition.
2
Determine the chemical formula and reaction involved.
Ca(HCO3)2(aq)ΔCaCO3(s)+H2O(l)+CO2(g)\text{Ca(HCO}_3)_2(aq) \xrightarrow{\Delta} \text{CaCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)
Soluble calcium hydrogentrioxocarbonate(IV) breaks down on heating to yield insoluble calcium trioxocarbonate(IV), effectively precipitating out calcium ions.

Anahtar Kavram

Temporary hardness in water is caused by dissolved calcium and magnesium hydrogentrioxocarbonates(IV) and can be removed by boiling.
Soru 6711Soru

Calculate the mass, in grams, of nitrogen contained in a 16.4 g16.4\text{ g} sample of pure calcium trioxonitrate(V), Ca(NO3)2\text{Ca(NO}_3)_2. [Ca=40,N=14,O=16][\text{Ca} = 40, \text{N} = 14, \text{O} = 16]

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Cevap: 2.8

Cevap

The mass of nitrogen in the sample is 2.8 g2.8\text{ g}.
The molar mass of Ca(NO3)2\text{Ca(NO}_3)_2 is calculated as 40+2(14+3×16)=164 g/mol40 + 2(14 + 3 \times 16) = 164\text{ g/mol}. A 16.4 g16.4\text{ g} sample corresponds to 16.4164=0.1 mol\frac{16.4}{164} = 0.1\text{ mol} of Ca(NO3)2\text{Ca(NO}_3)_2. Because each formula unit contains 2 nitrogen atoms, 0.1 mol0.1\text{ mol} of compound yields 0.2 mol0.2\text{ mol} of nitrogen. Multiplying 0.2 mol0.2\text{ mol} by the molar mass of atomic nitrogen (14 g/mol14\text{ g/mol}) gives 2.8 g2.8\text{ g}.

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1
Calculate the molar mass of calcium trioxonitrate(V), Ca(NO3)2\text{Ca(NO}_3)_2
164 g/mol164\text{ g/mol}
Sum the relative atomic masses of all atoms present: 40+2(14+3×16)=164 g/mol40 + 2(14 + 3 \times 16) = 164\text{ g/mol}.
2
Calculate the number of moles of Ca(NO3)2\text{Ca(NO}_3)_2 present in 16.4 g16.4\text{ g}
0.1 mol0.1\text{ mol}
Use the formula moles=massmolar mass=16.4 g164 g/mol=0.1 mol\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{16.4\text{ g}}{164\text{ g/mol}} = 0.1\text{ mol}.
3
Determine the number of moles of nitrogen atoms in 0.1 mol0.1\text{ mol} of Ca(NO3)2\text{Ca(NO}_3)_2
0.2 mol0.2\text{ mol} of N atoms
Each formula unit of Ca(NO3)2\text{Ca(NO}_3)_2 contains 2 nitrogen atoms.
4
Calculate the mass of the nitrogen atoms
2.8 g2.8\text{ g}
Multiply the moles of nitrogen by its atomic mass: 0.2 mol×14 g/mol=2.8 g0.2\text{ mol} \times 14\text{ g/mol} = 2.8\text{ g}.

Anahtar Kavram

Mole Concept and Mass Composition of Compounds
Tahmini Süre:1m 30s
Soru 6712Soru

A progressive wave traveling along a stretched string has a frequency of 250 Hz250\text{ Hz} and a speed of 300 m/s300\text{ m/s}. What is the minimum distance, in meters, between two points on the string that differ in phase by π3 rad\frac{\pi}{3}\text{ rad}?

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Cevap: 0.2

Cevap

The minimum distance between the two points is 0.2 m0.2\text{ m}.
The wavelength is found using λ=vf=300250=1.2 m\lambda = \frac{v}{f} = \frac{300}{250} = 1.2\text{ m}. Substituting λ=1.2 m\lambda = 1.2\text{ m} and Δϕ=π3 rad\Delta \phi = \frac{\pi}{3}\text{ rad} into the phase difference formula Δϕ=2πΔxλ\Delta \phi = \frac{2\pi \Delta x}{\lambda} gives Δx=(π/3)×1.22π=0.2 m\Delta x = \frac{(\pi / 3) \times 1.2}{2\pi} = 0.2\text{ m}.

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1
Calculate the wavelength (\(\lambda\)) from wave speed (\(v\)) and frequency (\(f\))
\(\lambda = \frac{300}{250} = 1.2\text{ m}\)
The fundamental wave equation relates wave speed, frequency, and wavelength as \(v = f\lambda\).
2
Apply the phase difference formula to solve for spatial separation (\(\Delta x\))
\(\Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} = \frac{(\pi / 3) \cdot 1.2}{2\pi} = 0.2\text{ m}\)
A full cycle of \(2\pi\text{ radians}\) corresponds to a spatial displacement of one wavelength (\(\lambda\)).

Anahtar Kavram

Relationship between phase difference and spatial displacement
Soru 6713Soru

Read the poetic excerpt below and answer the question that follows:

"Count not the gold that fills the merchant’s chest,
Nor praise the marble where the tyrants rest;
For iron rusts, and gilded crowns decay,
And silent dust consumes their proud display."

What is the prevailing tone of the speaker in the lines above?

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Cevap: Didactic and contemptuous

Cevap

Didactic and contemptuous
The correct option is 'Didactic and contemptuous' because the speaker uses instructive language ('Count not', 'Nor praise') to convey a moral lesson while expressing scorn toward material wealth and political power by emphasizing their inevitable decay.

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1
Analyze the speaker's stance and diction toward the subject matter.
The imperatives 'Count not' and 'Nor praise' show an instructive (didactic) stance aimed at teaching a moral lesson.
Tone is defined as the speaker's attitude toward the subject.
2
Evaluate the imagery and choice of words describing wealth and power.
Phrases like 'iron rusts', 'crowns decay', and 'silent dust consumes' dismiss earthly power as temporary and meaningless, establishing a contemptuous tone.
Diction and imagery objectively justify the specific emotional stance of the speaker.

Anahtar Kavram

Tone, Mood, and Persona in Poetry
Soru 6714Soru

Read the prose extract below:

"Throughout the harsh Harmattan season, the dried riverbed became a winding highway of scorched earth and white pebbles."

Which literary device is employed in the phrase "the dried riverbed became a winding highway"?

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Cevap: Metaphor

Cevap

Metaphor
The correct answer is Metaphor because the author asserts directly that the dried riverbed was a winding highway, making an implicit comparison without using explicit connective words such as 'like' or 'as'.

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1
Analyze the targeted phrase in the prose extract
The phrase compares 'the dried riverbed' directly to 'a winding highway'.
Identifying the subject and what it is being compared to establishes the figurative framework.
2
Check for comparative connective markers
No comparative words such as 'like' or 'as' are present in the sentence.
The absence of 'like' or 'as' distinguishes a metaphor from a simile.
3
Determine the dominant literary device
Directly asserting that one thing is another creates a metaphor.
A metaphor makes an implicit, direct comparison between two unlike entities.

Anahtar Kavram

Literary Devices in Prose: Metaphor vs. Simile
Soru 6715Soru

A manufacturing firm distributes its consumer goods directly to buyers by issuing illustrated catalogs and accepting orders primarily through postal or electronic communication, completely dispensing with physical storefronts. Which large-scale retail trading format is this enterprise employing?

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Cevap: Mail order business

Cevap

Mail order business
The mail order business is a form of large-scale retailing where trading is conducted entirely without physical shop premises, relying instead on descriptive catalogs, advertisements, and postal or courier delivery services.

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1
Identify the key operational characteristics in the question scenario
The business uses catalog distribution, remote ordering, and lacks physical shop premises.
Retail formats are classified based on whether they operate via physical storefronts or direct non-store communication channels.
2
Match the features to the appropriate large-scale retail format
Mail order retailing is specifically designed to eliminate physical stores by relying on catalogs and mail/postal delivery.
Department stores, tied shops, and supermarkets all depend on physical retail locations for customer visits.

Anahtar Kavram

Mail Order Retailing
Soru 6716Soru

Match each physical property of metals on the left with the microscopic structural explanation on the right that best accounts for it.

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Öğeler

Thermal conductivity
High metallic lustre
Ductility

Eşleşmeler

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Cevap

Thermal conductivity matches with rapid transfer of thermal energy by mobile delocalized electrons through the lattice. High metallic lustre matches with re-emission of absorbed light by oscillating free electrons on the metal surface. Ductility matches with sliding of positive cation layers past each other into wires without breaking metallic bonds.
Thermal conductivity is caused by mobile delocalized electrons rapidly transferring heat throughout the metal lattice. Metallic lustre occurs because surface free electrons absorb and re-emit light photons. Ductility is enabled by positive cation layers sliding past each other within the sea of delocalized electrons without disrupting the non-directional metallic bonding.

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1
Identify the cause of heat transport in metals
Thermal conductivity relies on delocalized valence electrons carrying thermal energy rapidly through the crystal lattice.
Free electrons gain kinetic energy when heated and collide with surrounding ions and electrons to distribute heat.
2
Identify the cause of optical reflection (lustre)
Metallic lustre is produced by surface delocalized electrons oscillating in response to incoming light waves and reflecting them.
The un-bound nature of free valence electrons allows immediate absorption and re-radiation of visible light photons.
3
Identify the mechanical feature enabling deformation into wires
Ductility relies on layers of cations sliding past one another while held together by non-directional metallic attraction.
Because metallic bonding is non-directional, moving cation layers do not cause repulsive strain that breaks the crystal structure.

Anahtar Kavram

Properties of metals in terms of the delocalized electron sea model
Soru 6717Soru

Match each type of cooperative society with the primary operational focus or economic need it addresses for its members.

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Öğeler

Consumers' Cooperative Society
Producers' Cooperative Society
Credit and Thrift Cooperative Society
Farmers' Cooperative Society

Eşleşmeler

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Cevap

Consumers' Cooperative Society matches with protection against artificial scarcity and inflated retail prices; Producers' Cooperative Society matches with elimination of middlemen markup on raw materials and collective marketing of outputs; Credit and Thrift Cooperative Society matches with encouragement of regular savings and micro-loans; Farmers' Cooperative Society matches with provision of affordable agricultural inputs and equipment hiring.
Each cooperative type specializes in solving specific financial or operational hurdles: consumer cooperatives eliminate retail middleman profits, producer cooperatives support manufacturing production and joint sales, credit and thrift societies provide micro-finance, and farmers' cooperatives provide agricultural inputs and machinery access.

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1
Identify the primary mandate of Consumers' Cooperative Societies.
Connect Consumers' Cooperative Society to retail price protection and direct bulk purchasing for domestic items.
Consumer cooperatives focus on making essential consumer goods affordable by eliminating retail distribution intermediaries.
2
Analyze the core functions of Producers' Cooperative Societies.
Match Producers' Cooperative Society to raw material sourcing and joint output sales.
Producer cooperatives assist small-scale manufacturing units with affordable raw materials and joint market access.
3
Examine the role of Credit and Thrift Cooperative Societies.
Link Credit and Thrift Cooperative Society to savings mobilization and accessible credit.
Credit and thrift societies encourage financial discipline and supply short-term loans without conventional banking hurdles.
4
Determine the objective of Farmers' Cooperative Societies.
Link Farmers' Cooperative Society to farm input supply and agricultural mechanization support.
Farmers' cooperatives specifically address agricultural productivity by reducing input procurement costs for growers.

Anahtar Kavram

Classification and functional roles of different types of cooperative societies.
Tahmini Süre:1m 30s
Soru 6718Soru

Complete the statement below regarding the volumetric analysis of a mixture containing sodium carbonate and sodium hydrogen carbonate using double indicators.

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During the titration of a mixture of sodium carbonate and sodium hydrogen carbonate against standard hydrochloric acid, the first equivalence point marking the half-neutralization of sodium carbonate to sodium hydrogen carbonate is indicated by , which turns from pink to at a pH range of approximately 8.3 to 10.0.
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Cevap

The first equivalence point is detected using phenolphthalein indicator, which transitions from pink to colorless (or colourless).
Phenolphthalein is the correct indicator for the first equivalence point because its transition interval (pH 8.3–10.0) matches the pH of the sodium hydrogen carbonate solution formed when all carbonate ions are neutralized. Its color changes from pink in alkaline medium to colorless at the endpoint.

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1
Identify the chemical reaction occurring at the first equivalence point of the carbonate-bicarbonate titration.
At the first endpoint, all CO32\text{CO}_3^{2-} ions are protonated to form HCO3\text{HCO}_3^- according to CO32+H+HCO3\text{CO}_3^{2-} + \text{H}^+ \rightarrow \text{HCO}_3^-. The pH at this stage is weakly alkaline (around pH 8.3).
Sodium hydrogen carbonate is an amphiprotic salt that forms a solution with a pH near 8.3, requiring an indicator with a pH transition range in the alkaline region.
2
Determine the appropriate indicator and its observed color change at this pH transition.
Phenolphthalein has a pH transition range of 8.3–10.0. In basic solution containing carbonate ions, it displays a pink color, and turns colorless once all carbonate ions are converted to hydrogen carbonate ions.
Methyl orange changes color at pH 3.1–4.4, which corresponds to the second equivalence point where hydrogen carbonate is completely converted to carbon dioxide and water.

Anahtar Kavram

Double Indicator Titration of Carbonate and Hydrogen Carbonate Mixtures
Soru 6719Soru

Match each prose extract below to the dominant literary device or figurative element it demonstrates.

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Öğeler

"The solitary lantern wept yellow tears onto the damp cobblestones of the narrow alleyway."
"The village council hall became a buzzing hive of discontent as soon as the decree was announced."
"His voice during the heated debate was as smooth and reassuring as cool morning oil."
"Standing alone in the ruins of his ancestral home, he was wrapped in a deafening silence."

Eşleşmeler

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Cevap

Extract 1 ("The solitary lantern wept...") matches Personification. Extract 2 ("The village council hall became a buzzing hive...") matches Metaphor. Extract 3 ("His voice... was as smooth... as cool morning oil") matches Simile. Extract 4 ("...wrapped in a deafening silence") matches Oxymoron.
Each extract contains distinct stylistic markers: giving human actions like weeping to a lamp is personification; directly stating a hall is a beehive is a metaphor; comparing voice to oil using 'as' is a simile; and placing 'deafening' next to 'silence' creates an oxymoron.

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1
Analyze Extract 1 for human attributes given to non-human entities.
The lantern is described as weeping tears, which is a human action, identifying the device as Personification.
Attributing human qualities to non-living objects defines personification.
2
Analyze Extract 2 for direct figurative comparison.
The hall is directly equated to a buzzing hive, establishing a Metaphor.
Directly asserting that one thing is another without comparative signal words constitutes a metaphor.
3
Analyze Extract 3 for explicit comparison markers.
The comparison uses the marker "as... as", identifying it as a Simile.
Comparisons using "like" or "as" are similes.
4
Analyze Extract 4 for contradictory word combinations.
"Deafening silence" combines opposing concepts of loud sound and complete quiet, identifying it as an Oxymoron.
Two adjacent contradictory terms form an oxymoron.

Anahtar Kavram

Literary Devices and Figurative Language in Prose
Soru 6720Soru

An organic compound contains 40.0%40.0\% carbon, 6.7%6.7\% hydrogen, and 53.3%53.3\% oxygen by mass. If the vapour density of the compound is 3030, what is its molecular formula? [C=12,H=1,O=16][C = 12, H = 1, O = 16]

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Cevap: C2H4O2C_2H_4O_2

Cevap

The molecular formula of the compound is C2H4O2C_2H_4O_2.
Dividing mass percentage by relative atomic mass yields a 1:2:11:2:1 mole ratio for C:H:O\text{C}:\text{H}:\text{O}, establishing an empirical formula of CH2OCH_2O (mass =30 g/mol= 30\text{ g/mol}). Multiplying the vapour density (3030) by 22 gives a molar mass of 60 g/mol60\text{ g/mol}. The molecular formula multiplier n=60/30=2n = 60 / 30 = 2, yielding C2H4O2C_2H_4O_2.

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1
Determine the mole ratio of each constituent element
Moles of C=40.012=3.33C = \frac{40.0}{12} = 3.33, Moles of H=6.71=6.70H = \frac{6.7}{1} = 6.70, Moles of O=53.316=3.33O = \frac{53.3}{16} = 3.33
Mass percentages are divided by their respective relative atomic masses to yield mole quantities.
2
Calculate the simplest whole-number ratio to obtain the empirical formula
Ratio C:H:O=3.333.33:6.703.33:3.333.33=1:2:1C : H : O = \frac{3.33}{3.33} : \frac{6.70}{3.33} : \frac{3.33}{3.33} = 1 : 2 : 1. Empirical formula is CH2OCH_2O.
Dividing all mole values by the smallest value (3.333.33) converts the mole ratio into simple integers.
3
Compute the molar mass from the given vapour density
Molar Mass=2×Vapour Density=2×30=60 g/mol\text{Molar Mass} = 2 \times \text{Vapour Density} = 2 \times 30 = 60\text{ g/mol}
The molar mass of a volatile substance is equal to twice its vapour density.
4
Find the molecular formula multiplier nn and determine the molecular formula
Empirical formula mass of CH2O=12+(2×1)+16=30 g/molCH_2O = 12 + (2 \times 1) + 16 = 30\text{ g/mol}. n=6030=2n = \frac{60}{30} = 2. Molecular formula = (CH2O)2=C2H4O2(CH_2O)_2 = C_2H_4O_2.
The molecular formula is obtained by multiplying the subscripts of the empirical formula by nn, where n=Molar MassEmpirical Massn = \frac{\text{Molar Mass}}{\text{Empirical Mass}}.

Anahtar Kavram

Deriving Empirical and Molecular Formulae using Vapour Density
ÖncekiSayfa 336 / 697Sonraki
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