Tüm alıştırma soruları

13931 soru

Soru 3941Soru

While Fra Luca Pacioli is widely celebrated in accounting history, his 1494 work *Particularis de Computis et Scripturis* did not invent double-entry bookkeeping but rather codified existing mercantile practices. Which of the following historical conditions was the primary driver behind the original development of double-entry recording among Italian merchants prior to Pacioli's publication?

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Cevap: The expansion of long-distance commercial credit and merchant partnerships, which demanded a systematic method to track bilateral debt relationships and dual-aspect transactions

Cevap

The primary driver behind the development of double-entry bookkeeping in medieval Italy was the expansion of long-distance commercial credit and merchant partnerships, which required a systematic method to record bilateral debts and dual-aspect financial transactions.
Double-entry bookkeeping developed in Northern Italy during the 13th–14th centuries as commercial trade expanded rapidly. Merchants engaged in complex credit transactions, agency arrangements, and international trade routes, which necessitated a comprehensive method of tracking dual-aspect financial events (every transaction involves giving and receiving value). Pacioli's 1494 work documented these already established commercial practices.

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1
Analyze the historical context of pre-Pacioli accounting development
Pacioli did not invent double-entry bookkeeping; he documented practices developed by Italian merchants in trading hubs like Venice, Genoa, and Florence during the 13th and 14th centuries.
Understanding the distinction between inventing a system and codifying existing practice is key to accounting history.
2
Identify the socio-economic factors driving merchant bookkeeping needs
The Growth of international trade, credit transactions, and joint merchant ventures required tracking dynamic debtor/creditor positions and dual impact (debit and credit) for every transaction.
Primitive single-entry record-keeping could only track basic cash flow and owed amounts, making it insufficient for multi-party commercial credit networks.
3
Evaluate alternative distractors against historical timelines
Joint-stock acts and professional accounting bodies belong to the 19th-century Industrial Revolution era, while merging personal and business assets violates the business entity principle upon which accounting relies.
Distinguishing between medieval record-keeping origins and 19th-century industrial accounting evolution prevents historical misclassification.

Anahtar Kavram

Origins and Evolution of Double-Entry Bookkeeping
Tahmini Süre:2m 0s
Soru 3942Soru

In differentiating the quadratic function f(x)=2x2+5xf(x) = 2x^2 + 5x from first principles, which of the following expressions represents the fully simplified difference quotient f(x+h)f(x)h\frac{f(x+h) - f(x)}{h} prior to evaluating the limit as h0h \to 0?

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Cevap: 4x+2h+54x + 2h + 5

Cevap

The fully simplified difference quotient is 4x+2h+54x + 2h + 5.
Expanding f(x+h)=2(x+h)2+5(x+h)f(x+h) = 2(x+h)^2 + 5(x+h) gives 2x2+4xh+2h2+5x+5h2x^2 + 4xh + 2h^2 + 5x + 5h. Subtracting f(x)=2x2+5xf(x) = 2x^2 + 5x leaves 4xh+2h2+5h4xh + 2h^2 + 5h. Dividing this expression term-by-term by hh yields 4x+2h+54x + 2h + 5.

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1
Evaluate f(x+h)f(x+h) for the function f(x)=2x2+5xf(x) = 2x^2 + 5x.
f(x+h)=2(x+h)2+5(x+h)=2(x2+2xh+h2)+5x+5h=2x2+4xh+2h2+5x+5hf(x+h) = 2(x+h)^2 + 5(x+h) = 2(x^2 + 2xh + h^2) + 5x + 5h = 2x^2 + 4xh + 2h^2 + 5x + 5h
To apply first principles, we must replace every instance of xx with (x+h)(x+h) and expand completely.
2
Form the numerator of the difference quotient, f(x+h)f(x)f(x+h) - f(x).
f(x+h)f(x)=(2x2+4xh+2h2+5x+5h)(2x2+5x)=4xh+2h2+5hf(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 5x + 5h) - (2x^2 + 5x) = 4xh + 2h^2 + 5h
Subtracting f(x)f(x) cancels out the terms 2x22x^2 and 5x5x that do not contain hh.
3
Divide the result by hh to get the difference quotient.
\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 5h}{h} = 4x + 2h + 5
Factoring out hh from the numerator and canceling it with the denominator simplifies the quotient.

Anahtar Kavram

Differentiation from First Principles (Difference Quotient)
Tahmini Süre:1m 30s
Soru 3943Soru

In an X-ray tube, which of the following adjustments will increase the penetrating power (hardness) of the emitted X-rays?

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Cevap: Increasing the accelerating potential difference between the anode and cathode

Cevap

Increasing the accelerating potential difference between the anode and cathode increases the penetrating power of X-rays.
Increasing the accelerating potential difference between the anode and cathode increases the energy transferred to the colliding electrons. According to the Duane-Hunt law (eV=hfmaxe V = h f_{\text{max}}), higher potential difference leads to higher frequency X-ray photons, which increases their penetrating power (hardness).

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1
Identify the factor controlling electron kinetic energy
Maximum kinetic energy of electrons is Emax=eVE_{\text{max}} = e V, directly proportional to the tube voltage VV.
Electrons gain energy from the electric field established by the potential difference.
2
Relate electron energy to X-ray photon energy and penetrating power
Higher kinetic energy produces X-ray photons of higher maximum frequency (fmax=eVhf_{\text{max}} = \frac{e V}{h}) and shorter minimum wavelength (hard X-rays).
Hard X-rays with higher frequencies have greater energy and higher penetrating power.

Anahtar Kavram

X-ray Hardness vs. Intensity Control
Tahmini Süre:45s
Soru 3944Soru

A flat circular coil consisting of 100100 turns and having a cross-sectional area of 4.0×103 m24.0 \times 10^{-3}\text{ m}^2 is positioned perpendicular to a uniform magnetic field of 0.50 T0.50\text{ T}. If the direction of the magnetic field is completely reversed in a time interval of 0.02 s0.02\text{ s}, what is the magnitude of the average electromotive force (e.m.f.) induced in the coil?

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Cevap: 20 V20\text{ V}

Cevap

20 V20\text{ V}
According to Faraday's law of electromagnetic induction, the magnitude of induced electromotive force equals the rate of change of magnetic flux linkage. Because the magnetic field direction is completely reversed, the change in field magnitude is ΔB=0.50(0.50)=1.0 T\Delta B = 0.50 - (-0.50) = 1.0\text{ T}. Multiplying by the area (4.0×103 m24.0 \times 10^{-3}\text{ m}^2) and turn count (100100), then dividing by the duration (0.02 s0.02\text{ s}) gives 20 V20\text{ V}.

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1
Determine the change in magnetic flux density (ΔB\Delta B)
ΔB=BfBi=0.50 T0.50 T=1.0 T\Delta B = B_f - B_i = -0.50\text{ T} - 0.50\text{ T} = -1.0\text{ T}, so ΔB=1.0 T|\Delta B| = 1.0\text{ T}
Reversing the magnetic field direction changes the sign of the flux density relative to the coil area.
2
Calculate the total change in magnetic flux linkage (ΔΦtotal\Delta \Phi_{total})
ΔΦtotal=NAΔB=100×(4.0×103 m2)×1.0 T=0.40 Wb-turns\Delta \Phi_{total} = N \cdot A \cdot |\Delta B| = 100 \times (4.0 \times 10^{-3}\text{ m}^2) \times 1.0\text{ T} = 0.40\text{ Wb-turns}
The total magnetic flux linkage is proportional to the number of turns, the surface area, and the net field change.
3
Apply Faraday's law of electromagnetic induction to find the induced e.m.f. (E\mathcal{E})
E=ΔΦtotalΔt=0.40 Wb-turns0.02 s=20 V\mathcal{E} = \frac{\Delta \Phi_{total}}{\Delta t} = \frac{0.40\text{ Wb-turns}}{0.02\text{ s}} = 20\text{ V}
The induced e.m.f. magnitude is equal to the rate of change of magnetic flux linkage.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction
Soru 3945Soru

A progressive transverse wave with a frequency of 50 Hz50\text{ Hz} travels through a medium with a wave speed of 300 m/s300\text{ m/s}. If the wave crosses a boundary into a second medium where its speed decreases to 150 m/s150\text{ m/s}, what is the frequency of the wave in the second medium?

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Cevap: 50 Hz50\text{ Hz}

Cevap

The frequency of the wave in the second medium is 50 Hz50\text{ Hz}.
The frequency of a wave is determined entirely by the vibrating source that creates it. When a wave passes from one medium to another, its speed and wavelength change, but its frequency remains constant. Therefore, the frequency remains 50 Hz50\text{ Hz}.

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1
Identify the given physical parameters.
Initial frequency f1=50 Hzf_1 = 50\text{ Hz}, initial speed v1=300 m/sv_1 = 300\text{ m/s}, new speed v2=150 m/sv_2 = 150\text{ m/s}.
Recognize which wave properties are given and which property is being asked for.
2
Apply the fundamental principle of wave propagation across different media boundaries.
Frequency ff depends only on the source of vibration and remains constant during refraction/transmission across media boundaries.
While wave speed vv and wavelength λ\lambda change when entering a new medium, frequency ff remains unaffected.
3
Determine the frequency in the second medium.
f2=f1=50 Hzf_2 = f_1 = 50\text{ Hz}.
Since frequency is invariant across media boundaries, f2f_2 must equal 50 Hz50\text{ Hz}.

Anahtar Kavram

Invariance of wave frequency across media boundaries
Tahmini Süre:45s
Soru 3946Soru

A body is projected vertically upwards from the ground with an initial velocity uu. It passes a point at a height of 40 m40\text{ m} above the ground at t=2 st = 2\text{ s} while ascending and again at t=4 st = 4\text{ s} while descending. Taking acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2, what is the initial speed of projection uu of the body?

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Cevap: 30 m/s30\text{ m/s}

Cevap

The initial speed of projection of the body is 30 m/s30\text{ m/s}.
The position of a body thrown vertically upwards is given by h=ut12gt2h = ut - \frac{1}{2}gt^2. Rearranging this equation into standard quadratic form gives t2(2ug)t+2hg=0t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0. The roots t1t_1 and t2t_2 correspond to the times the body reaches height hh. By Vieta's formulas, the sum of the times is t1+t2=2ugt_1 + t_2 = \frac{2u}{g}. Substituting t1=2 st_1 = 2\text{ s}, t2=4 st_2 = 4\text{ s}, and g=10 m/s2g = 10\text{ m/s}^2 gives 6=2u106 = \frac{2u}{10}, which solves to u=30 m/su = 30\text{ m/s}. Alternatively, the time to reach maximum height is the midpoint ttop=t1+t22=3 st_{\text{top}} = \frac{t_1 + t_2}{2} = 3\text{ s}, and at maximum height v=0=ugttopv = 0 = u - gt_{\text{top}}, yielding u=10×3=30 m/su = 10 \times 3 = 30\text{ m/s}.

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1
Set up the vertical motion displacement equation
h=ut12gt2h = ut - \frac{1}{2}gt^2
The equation describes the vertical position hh at any time tt for a projectile launched from ground level with initial speed uu.
2
Rearrange the equation into standard quadratic form for tt
12gt2ut+h=0    t2(2ug)t+2hg=0\frac{1}{2}gt^2 - ut + h = 0 \implies t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0
The two solutions t1t_1 and t2t_2 represent the times at which the body reaches the specific height hh.
3
Apply Vieta's formulas for the sum of roots of the quadratic equation
t1+t2=2ugt_1 + t_2 = \frac{2u}{g}
The sum of the roots of a quadratic equation t2Bt+C=0t^2 - Bt + C = 0 is equal to the coefficient BB.
4
Substitute given values to solve for uu
2+4=2u10    6=u5    u=30 m/s2 + 4 = \frac{2u}{10} \implies 6 = \frac{u}{5} \implies u = 30\text{ m/s}
Given t1=2 st_1 = 2\text{ s}, t2=4 st_2 = 4\text{ s}, and g=10 m/s2g = 10\text{ m/s}^2, direct substitution yields the initial velocity.

Anahtar Kavram

Vertical Motion under Gravity and Time Symmetry
Tahmini Süre:1m 0s
Soru 3947Soru

The presence of a soluble non-volatile impurity in a pure liquid sample depresses its boiling point and causes it to boil over a narrow temperature range.

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Cevap: False

Cevap

False. Soluble non-volatile impurities elevate the boiling point of liquids and broaden the temperature range over which boiling occurs.
Soluble non-volatile impurities elevate the boiling point of liquids rather than depressing it, and cause the liquid to boil over a broad range of temperatures instead of a narrow, sharp temperature point.

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1
Analyze how non-volatile impurities affect the vapor pressure of a liquid.
Solute molecules occupy surface area, reducing the rate of solvent evaporation and lowering the vapor pressure.
Lower vapor pressure means the liquid cannot boil at its normal boiling point.
2
Determine the effect on boiling point and boiling range.
The liquid must be heated to a higher temperature to reach atmospheric pressure, causing boiling point elevation and a non-sharp boiling range.
A sharp, fixed boiling point is a criterion of purity; impure liquids boil over a range of elevated temperatures.

Anahtar Kavram

Effect of Soluble Impurities on Boiling Point and Purity Criteria
Soru 3948Soru

Monochromatic light of frequency 9.0×1014 Hz9.0 \times 10^{14}\text{ Hz} is incident on a clean potassium emitter surface having a threshold frequency of 5.0×1014 Hz5.0 \times 10^{14}\text{ Hz}. Given that Planck's constant h=6.6×1034 Jsh = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} and the elementary charge e=1.6×1019 Ce = 1.6 \times 10^{-19}\text{ C}, what is the stopping potential in volts required to completely arrest the photoelectric current?

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Cevap: 1.65

Cevap

The stopping potential required to arrest the emitted photoelectrons is 1.65 V1.65\text{ V}.
Applying Einstein's photoelectric equation Kmax=hfhf0=h(ff0)K_{\max} = h f - h f_0 = h(f - f_0) gives a maximum kinetic energy of 2.64×1019 J2.64 \times 10^{-19}\text{ J}. Dividing this by the electronic charge e=1.6×1019 Ce = 1.6 \times 10^{-19}\text{ C} yields the stopping potential Vs=1.65 VV_s = 1.65\text{ V}.

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1
Determine the net energy available for photoelectron kinetic energy using Einstein's photoelectric equation
Kmax=h(ff0)=6.6×1034 Js×(9.0×10145.0×1014) Hz=2.64×1019 JK_{\max} = h(f - f_0) = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \times (9.0 \times 10^{14} - 5.0 \times 10^{14})\text{ Hz} = 2.64 \times 10^{-19}\text{ J}
Photoelectron emission occurs only when photon energy hfhf exceeds the work function W0=hf0W_0 = h f_0, with the excess energy appearing as maximum kinetic energy.
2
Express stopping potential in terms of electron charge and maximum kinetic energy
Vs=Kmaxe=2.64×1019 J1.6×1019 C=1.65 VV_s = \frac{K_{\max}}{e} = \frac{2.64 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ C}} = 1.65\text{ V}
The stopping potential VsV_s does work eVse V_s equal to the maximum kinetic energy KmaxK_{\max} of the fastest photoelectrons to bring them to rest.

Anahtar Kavram

Einstein's Photoelectric Equation and Stopping Potential Relationship
Soru 3949Soru

A steady electric current of 2.5A2.5\,\text{A} flows through a conductor for 4.0minutes4.0\,\text{minutes}. What is the total electric charge that passes through any cross-section of the conductor during this period?

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Cevap: 600C600\,\text{C}

Cevap

600C600\,\text{C}
Electric current (II) is defined as the rate of charge flow (QQ) per unit time (tt), expressed as Q=I×tQ = I \times t. Converting 4.0minutes4.0\,\text{minutes} into seconds gives 4.0×60=240s4.0 \times 60 = 240\,\text{s}. Multiplying the current of 2.5A2.5\,\text{A} by 240s240\,\text{s} yields 600C600\,\text{C}.

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1
Convert the time duration from minutes to seconds
t=4.0minutes=4.0×60s=240secondst = 4.0\,\text{minutes} = 4.0 \times 60\,\text{s} = 240\,\text{seconds}
The standard SI unit for time in electromagnetism formulas is seconds.
2
Apply the formula relating electric charge, current, and time (Q=I×tQ = I \times t)
Q=2.5A×240s=600CQ = 2.5\,\text{A} \times 240\,\text{s} = 600\,\text{C}
Electric current is defined as the rate of flow of electric charge (I=QtI = \frac{Q}{t}).

Anahtar Kavram

Relationship between Electric Current, Charge, and Time
Soru 3950Soru

A resistance thermometer has a resistance of 4.0Ω4.0\,\Omega at the ice point (0C0^\circ\text{C}) and 6.0Ω6.0\,\Omega at the steam point (100C100^\circ\text{C}). When immersed in a liquid bath, its resistance is measured to be 5.2Ω5.2\,\Omega. What is the temperature of the bath?

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Cevap: 60.0C60.0^\circ\text{C}

Cevap

60.0C60.0^\circ\text{C}
The temperature of the bath is found by calculating the fraction of resistance change relative to the total change between the ice point and steam point: θ=5.24.06.04.0×100C=60.0C\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}.

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1
Identify the given thermometric parameters
R0=4.0ΩR_0 = 4.0\,\Omega, R100=6.0ΩR_{100} = 6.0\,\Omega, and Rθ=5.2ΩR_\theta = 5.2\,\Omega
These are the measured resistance values corresponding to the lower fixed point, upper fixed point, and unknown temperature.
2
Apply the linear temperature interpolation formula for a resistance thermometer
θ=RθR0R100R0×100C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}
Temperature on the Celsius scale is proportional to the relative change in the thermometric property between fixed points.
3
Substitute the values and compute the temperature
θ=5.24.06.04.0×100C=1.22.0×100C=60.0C\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = \frac{1.2}{2.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}
Evaluating the expression yields the exact temperature of the liquid bath.

Anahtar Kavram

Linear interpolation on temperature scales using thermometric properties
Soru 3951Soru

Which of the following gases is evolved when dilute hydrochloric acid (HClHCl) reacts with calcium trioxocarbonate(IV) (CaCO3CaCO_3)?

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Cevap: Carbon(IV) oxide (CO2CO_2)

Cevap

Carbon(IV) oxide (CO2CO_2)
A characteristic chemical property of acids is that they react with trioxocarbonate(IV) salts (such as CaCO3CaCO_3) to produce a salt, water, and carbon(IV) oxide gas (CO2CO_2).

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1
Identify the reactants and the class of reaction
Dilute hydrochloric acid (HClHCl) reacts with calcium trioxocarbonate(IV) (CaCO3CaCO_3), which is an acid-carbonate reaction.
Acids characteristically react with trioxocarbonates to form salt, water, and carbon dioxide.
2
Write the balanced chemical equation
2HCl(aq)+CaCO3(s)CaCl2(aq)+H2O(l)+CO2(g)2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)
The equation shows that carbon(IV) oxide gas (CO2CO_2) is liberated during the chemical process.

Anahtar Kavram

Action of acids on trioxocarbonate(IV) salts
Tahmini Süre:45s
Soru 3952Soru

Two 12 μF12\text{ }\mu\text{F} capacitors are connected in series, and this combination is placed in parallel with a third capacitor of capacitance 6 μF6\text{ }\mu\text{F}. What is the total equivalent capacitance of the arrangement?

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Cevap: 12.0 μF12.0\text{ }\mu\text{F}

Cevap

The total equivalent capacitance of the network is 12.0 μF12.0\text{ }\mu\text{F}.
Combining the two identical 12 μF12\text{ }\mu\text{F} capacitors in series gives an equivalent capacitance of 6.0 μF6.0\text{ }\mu\text{F}. Adding this in parallel with the 6.0 μF6.0\text{ }\mu\text{F} capacitor yields 6.0 μF+6.0 μF=12.0 μF6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}.

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1
Calculate the equivalent capacitance (CsC_s) of the two 12 μF12\text{ }\mu\text{F} capacitors connected in series.
Cs=12×1212+12=6.0 μFC_s = \frac{12 \times 12}{12 + 12} = 6.0\text{ }\mu\text{F}
Capacitors in series combine reciprocally: 1Cs=1C1+1C2\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}.
2
Calculate the total equivalent capacitance (CeqC_{eq}) by adding CsC_s to the parallel 6.0 μF6.0\text{ }\mu\text{F} capacitor.
Ceq=Cs+C3=6.0 μF+6.0 μF=12.0 μFC_{eq} = C_s + C_3 = 6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}
Capacitors in parallel add directly: Ceq=Cseries+CparallelC_{eq} = C_{series} + C_{parallel}.

Anahtar Kavram

Equivalent capacitance of series-parallel capacitor networks
Soru 3953Soru

Under fixed laboratory conditions of temperature and pressure, Gas A has a density of 0.09 g/dm30.09\text{ g/dm}^3 while Gas B has a density of 1.44 g/dm31.44\text{ g/dm}^3. What is the ratio of the rate of diffusion of Gas A to the rate of diffusion of Gas B?

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Cevap: 4.04.0

Cevap

The ratio of the rate of diffusion of Gas A to Gas B is 4.04.0.
Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its density or molar mass. Calculating 1.440.09=16=4.0\sqrt{\frac{1.44}{0.09}} = \sqrt{16} = 4.0 correctly gives the relative rate ratio, showing that the lighter gas diffuses 4 times faster than the denser gas.

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1
State Graham's Law of Diffusion in terms of gas densities.
rArB=dBdA\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}}
Graham's law dictates that the rate of diffusion (rr) of a gas is inversely proportional to the square root of its density (dd) at constant temperature and pressure.
2
Substitute the given density values into the equation.
rArB=1.44 g/dm30.09 g/dm3\frac{r_A}{r_B} = \sqrt{\frac{1.44\text{ g/dm}^3}{0.09\text{ g/dm}^3}}
Gas B's density (dB=1.44d_B = 1.44) goes into the numerator and Gas A's density (dA=0.09d_A = 0.09) into the denominator due to the inverse relationship.
3
Simplify the fraction inside the square root and calculate the final square root.
rArB=16=4.0\frac{r_A}{r_B} = \sqrt{16} = 4.0
Dividing 1.441.44 by 0.090.09 gives 1616, and taking the square root of 1616 yields 4.04.0.

Anahtar Kavram

Graham's Law relating gas diffusion rates to density
Tahmini Süre:1m 15s
Soru 3954Soru

In dramatic structure, when a playwright juxtaposes a secondary character with the main character to accentuate specific moral qualities, behavioral traits, or flaws of the protagonist—such as Banquo's moral restraint highlighting Macbeth's unchecked ambition—which character type does this contrasting character represent?

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Cevap: Foil

Cevap

The contrasting character represents a dramatic foil.
A dramatic foil is a character whose values, decisions, or traits directly contrast with those of another character (usually the protagonist), thereby illuminating and sharpening the audience's understanding of the primary character's nature.

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1
Analyze the functional relationship between the characters described in the stem.
Banquo's moral choices serve to illuminate Macbeth's moral corruption by comparison.
Dramatic function relies on literary technique, not just character opposition.
2
Match the comparative function to standard dramatic character definitions.
The literary device where one character underscores the distinctive traits of another through contrast is a foil.
Distinguishes structural contrast (foil) from active plot opposition (antagonist).

Anahtar Kavram

Dramatic Foil and Characterization Functions
Soru 3955Soru

If the solubility of sodium hydroxide (NaOH\text{NaOH}) in water at 25C25^\circ\text{C} is 0.50 mol/dm30.50\text{ mol/dm}^3, what is its concentration in g/dm3\text{g/dm}^3? [Na=23,O=16,H=1][\text{Na} = 23, \text{O} = 16, \text{H} = 1]

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Cevap: 20.0 g/dm320.0\text{ g/dm}^3

Cevap

The concentration of sodium hydroxide in g/dm3\text{g/dm}^3 is 20.0 g/dm320.0\text{ g/dm}^3.
To convert concentration from mol/dm3\text{mol/dm}^3 to g/dm3\text{g/dm}^3, multiply the molar concentration by the molar mass of the solute. For NaOH\text{NaOH}, the molar mass is 23+16+1=40 g/mol23 + 16 + 1 = 40\text{ g/mol}. Multiplying 0.50 mol/dm30.50\text{ mol/dm}^3 by 40 g/mol40\text{ g/mol} yields 20.0 g/dm320.0\text{ g/dm}^3.

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1
Calculate the molar mass of sodium hydroxide (NaOH\text{NaOH}).
Molar mass of NaOH=23+16+1=40 g/mol\text{Molar mass of NaOH} = 23 + 16 + 1 = 40\text{ g/mol}.
Molar mass is required to convert concentration from moles per cubic decimetre to grams per cubic decimetre.
2
Convert molar solubility to mass concentration using the formula: Concentration in g/dm3=Molar concentration in mol/dm3×Molar Mass\text{Concentration in g/dm}^3 = \text{Molar concentration in mol/dm}^3 \times \text{Molar Mass}.
Concentration in g/dm3=0.50 mol/dm3×40 g/mol=20.0 g/dm3\text{Concentration in g/dm}^3 = 0.50\text{ mol/dm}^3 \times 40\text{ g/mol} = 20.0\text{ g/dm}^3.
Multiplying the amount of substance per volume by mass per mole yields mass per unit volume.

Anahtar Kavram

Conversion between molarity (mol/dm³) and mass concentration (g/dm³)
Soru 3956Soru

Match each trade classification in Column A with its correct operational description in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Wholesale Trade
Retail Trade
Entrepôt Trade
Export Trade

Eşleşmeler

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Cevap

Wholesale Trade matches with buying merchandise in bulk from manufacturers and selling in smaller quantities to distributors; Retail Trade matches with selling items in small units directly to final consumers; Entrepôt Trade matches with importing commodities from one foreign country to re-export them to another nation; Export Trade matches with selling goods produced locally to buyers in foreign markets.
Each trade classification correctly pairs with its fundamental commercial role: Wholesale trade deals in bulk distribution to retailers; Retail trade supplies the final consumer directly; Entrepôt trade handles re-exported transit goods; and Export trade sells domestic products abroad.

Adım Adım Çözüm

1
Analyze Home Trade classifications
Home trade consists of wholesale trade (bulk purchasing from manufacturers to break bulk) and retail trade (selling small units directly to ultimate consumers).
Home trade occurs within the geographical boundaries of a single country and is divided based on scale and customer type.
2
Analyze Foreign Trade classifications
Foreign trade consists of import, export (selling home-made goods abroad), and entrepôt trade (re-exporting imported goods to another country).
Foreign trade crosses national borders and is categorized by the direction of physical trade flows.

Anahtar Kavram

Classification of Trade into Home Trade (Wholesale, Retail) and Foreign Trade (Import, Export, Entrepôt)
Soru 3957Soru

A 13.90 g13.90\text{ g} sample of hydrated iron(II) tetraoxosulfate(VI), FeSO4xH2O\text{FeSO}_4 \cdot x\text{H}_2\text{O}, was heated strongly until all water of crystallization was driven off. The mass of the remaining anhydrous salt was 7.60 g7.60\text{ g}. What is the value of xx? [Fe=56,S=32,O=16,H=1][\text{Fe} = 56, \text{S} = 32, \text{O} = 16, \text{H} = 1]

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Cevap: 7

Cevap

The value of xx in the hydrated salt formula FeSO4xH2O\text{FeSO}_4 \cdot x\text{H}_2\text{O} is 7.
Heating the sample drives off all water of crystallization, leaving only anhydrous FeSO4\text{FeSO}_4. The mass of water is 13.90 g7.60 g=6.30 g13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}. Converting both components to moles gives 0.05 mol0.05\text{ mol} of FeSO4\text{FeSO}_4 and 0.35 mol0.35\text{ mol} of H2O\text{H}_2\text{O}. The ratio 0.350.05=7\frac{0.35}{0.05} = 7, yielding x=7x = 7.

Adım Adım Çözüm

1
Find the mass of water lost
Mass of H2O=13.90 g7.60 g=6.30 g\text{H}_2\text{O} = 13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}
The loss in mass upon heating represents the driven-off water of crystallization.
2
Calculate molar masses of anhydrous salt and water
Molar mass of FeSO4=152 g/mol\text{FeSO}_4 = 152\text{ g/mol}, Molar mass of H2O=18 g/mol\text{H}_2\text{O} = 18\text{ g/mol}
Molar masses are required to convert the measured masses into mole quantities.
3
Calculate the amount in moles of both components
Moles of FeSO4=7.60152=0.05 mol\text{FeSO}_4 = \frac{7.60}{152} = 0.05\text{ mol}; Moles of H2O=6.3018=0.35 mol\text{H}_2\text{O} = \frac{6.30}{18} = 0.35\text{ mol}
Stoichiometric coefficient xx represents the mole ratio between water and anhydrous salt.
4
Determine the mole ratio
x=0.35 mol0.05 mol=7x = \frac{0.35\text{ mol}}{0.05\text{ mol}} = 7
Dividing the moles of water of crystallization by the moles of anhydrous salt yields the integer coefficient xx.

Anahtar Kavram

Determining the formula of a hydrated salt from gravimetric data
Tahmini Süre:1m 30s
Soru 3958Soru

A retail merchant in Lagos intends to stock goods from a manufacturing firm in Ibadan. Arrange the following commercial documents in the correct chronological sequence of their issuance, from the initiation of the transaction to its final settlement.

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The correct chronological order of commercial documents in home trade is: Letter of Inquiry, Quotation, Order, Invoice, and Receipt.
A home trade transaction begins when the prospective buyer sends a Letter of Inquiry to ask about prices and terms. The seller replies with a Quotation containing unit prices, discounts, and delivery terms. After accepting the terms, the buyer sends an Order for the specified goods. Upon dispatching the goods, the seller issues an Invoice outlining the total debt incurred. Finally, when the buyer pays the bill, the seller provides a Receipt as conclusive evidence of payment.

Adım Adım Çözüm

1
Identify the document that initiates trade contact.
The Letter of Inquiry is placed first.
Trade transactions begin when a buyer requests information regarding goods, prices, and terms.
2
Identify the document sent by the seller in response to the inquiry.
The Quotation is placed second.
The seller replies to the inquiry by providing specific prices, discounts, and delivery conditions.
3
Identify the buyer's formal purchase commitment.
The Order is placed third.
After evaluating the terms stated in the quotation, the buyer formally requests the goods.
4
Identify the document billing the buyer upon or following dispatch.
The Invoice is placed fourth.
The seller issues an invoice to summarize the goods supplied and state the total debt due for settlement.
5
Identify the final document confirming payment.
The Receipt is placed fifth.
Once the buyer settles the invoice, the seller issues a receipt as proof of payment.

Anahtar Kavram

Chronological Sequence of Commercial Documents in Home Trade
Soru 3959Soru

A weak monobasic acid has a concentration of 0.20 mol dm30.20\text{ mol dm}^{-3} in aqueous solution. At equilibrium, the hydrogen ion concentration, [H+][H^+], is measured to be 1.0×103 mol dm31.0 \times 10^{-3}\text{ mol dm}^{-3}. What is the percentage degree of ionization of the acid in this solution?

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Cevap: 0.5

Cevap

The percentage degree of ionization of the weak acid is 0.5%.
The percentage degree of ionization measures the fraction of acid molecules that ionize in water, expressed as a percentage. By substituting [H+]=1.0×103 mol dm3[H^+] = 1.0 \times 10^{-3}\text{ mol dm}^{-3} and initial concentration C=0.20 mol dm3C = 0.20\text{ mol dm}^{-3} into Percentage ionization=([H+]C)×100%\text{Percentage ionization} = \left(\frac{[H^+]}{C}\right) \times 100\%, we obtain (0.0010.20)×100%=0.5%\left(\frac{0.001}{0.20}\right) \times 100\% = 0.5\%.

Adım Adım Çözüm

1
Identify the relationship between degree of ionization (\alpha), hydrogen ion concentration ([H+][H^+]), and initial acid concentration (CC).
α=[H+]C\alpha = \frac{[H^+]}{C}
For a weak monobasic acid ionizing according to HAH++AHA \rightleftharpoons H^+ + A^-, the concentration of ionized hydrogen ions equals αC\alpha C.
2
Calculate the fractional degree of ionization (\alpha).
\alpha = \frac{1.0 \times 10^{-3}\text{ mol dm}^{-3}}{0.20\text{ mol dm}^{-3}} = 0.005
Dividing the equilibrium hydrogen ion concentration by the initial acid concentration gives the fraction of acid molecules that ionized.
3
Convert the fractional degree of ionization into a percentage.
\text{Percentage ionization} = 0.005 \times 100\% = 0.5\%
Multiplying the decimal fraction by 100 converts the degree of ionization into percentage form.

Anahtar Kavram

Degree of Ionization of Weak Acids
Tahmini Süre:1m 30s
Soru 3960Soru

When potassium ethanoate, CH3COOKCH_3COOK, is dissolved in distilled water, the resulting solution has a pH greater than 7. Which ion is responsible for this alkalinity by undergoing hydrolysis?

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Cevap: The ethanoate ion, CH3COOCH_3COO^-

Cevap

The ethanoate ion, CH3COOCH_3COO^-
The ethanoate ion (CH3COOCH_3COO^-) is the conjugate base of the weak acid ethanoic acid (CH3COOHCH_3COOH). When potassium ethanoate dissolves in water, CH3COOCH_3COO^- hydrolyzes by accepting a proton from water to form CH3COOHCH_3COOH and releasing OHOH^- ions into the solution: CH3COO(aq)+H2O(l)CH3COOH(aq)+OH(aq)CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq). The accumulation of free hydroxide ions makes the solution alkaline (pH>7pH > 7).

Adım Adım Çözüm

1
Identify the ions formed when potassium ethanoate dissolves in water.
Potassium ethanoate dissociates completely into potassium ions (K+K^+) and ethanoate ions (CH3COOCH_3COO^-).
Soluble ionic salts completely dissociate into their constituent cations and anions in aqueous solution.
2
Determine which ion reacts with water (hydrolyzes).
Potassium (K+K^+) is derived from a strong base (KOHKOH) and does not hydrolyze. Ethanoate (CH3COOCH_3COO^-) is derived from a weak acid (CH3COOHCH_3COOH) and reacts with water.
Anions derived from weak acids act as weak conjugate bases and undergo anion hydrolysis.
3
Write the hydrolysis equation to identify the species giving rise to alkalinity.
CH3COO(aq)+H2O(l)CH3COOH(aq)+OH(aq)CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)
Hydrolysis of ethanoate generates free OHOH^- ions, making the solution basic (pH>7pH > 7).

Anahtar Kavram

Anion hydrolysis of salts formed from a weak acid and a strong base
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