Tüm alıştırma soruları

1526 soru

Soru 961Soru

A 250 cm3250\text{ cm}^3 sample of dry air is passed over excess heated phosphorus in a closed tube to remove all the oxygen gas present. Assuming oxygen constitutes 21%21\% by volume of dry air, what is the volume of the remaining gas mixture in cm3\text{cm}^3?

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Cevap: 197.5

Cevap

The volume of the remaining gas mixture is 197.5 cm3197.5\text{ cm}^3.
Because oxygen makes up 21%21\% by volume of dry air, a 250 cm3250\text{ cm}^3 sample contains 0.21×250 cm3=52.5 cm30.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3 of oxygen. Heated phosphorus reacts with all the oxygen to form solid phosphorus oxide, leaving behind 250 cm352.5 cm3=197.5 cm3250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3 of unreacted gases.

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1
Calculate the volume of oxygen gas in the initial sample
52.5 cm352.5\text{ cm}^3
Oxygen makes up 21%21\% by volume of dry air, so 0.21×250 cm3=52.5 cm30.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3.
2
Determine the remaining gas volume after complete removal of oxygen
197.5 cm3197.5\text{ cm}^3
Phosphorus reacts completely with oxygen, leaving the unreacted components of air: 250 cm352.5 cm3=197.5 cm3250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3.

Anahtar Kavram

Percentage composition of air by volume
Soru 962Soru

An electromagnetic microwave signal used in telecommunication has a wavelength of 0.02 m0.02\text{ m} in a vacuum. Given that the speed of light in a vacuum is 3.0×108 m/s3.0 \times 10^8\text{ m/s}, calculate the frequency of the signal in gigahertz (GHz\text{GHz}).

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Cevap: 15

Cevap

The frequency of the microwave signal is 15 GHz15\text{ GHz}.
Using the electromagnetic wave equation c=fλc = f \lambda, the frequency in Hz is calculated as f=cλ=3.0×108 m/s0.02 m=1.5×1010 Hzf = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.02\text{ m}} = 1.5 \times 10^{10}\text{ Hz}. Dividing by 10910^9 to convert into gigahertz gives 15 GHz15\text{ GHz}.

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1
Identify the given parameters and formula.
Speed of light c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}, wavelength λ=0.02 m\lambda = 0.02\text{ m}, and wave equation c=fλc = f \lambda.
Electromagnetic waves propagate at speed cc in a vacuum, relating frequency and wavelength.
2
Calculate the frequency in Hertz (Hz).
f=3.0×1080.02=1.5×1010 Hzf = \frac{3.0 \times 10^8}{0.02} = 1.5 \times 10^{10}\text{ Hz}.
Dividing the wave speed by the wavelength yields the frequency.
3
Convert the unit from Hz to GHz.
1.5×1010 Hz109 Hz/GHz=15 GHz\frac{1.5 \times 10^{10}\text{ Hz}}{10^9\text{ Hz/GHz}} = 15\text{ GHz}.
One gigahertz (1 GHz1\text{ GHz}) equals 109 Hz10^9\text{ Hz}.

Anahtar Kavram

Relationship between speed of light, frequency, and wavelength (c=fλc = f \lambda) for electromagnetic radiation.
Soru 963Soru

A thin converging lens forms a real image of an object on a screen placed 60 cm60\text{ cm} from the lens. If the object is located 30 cm30\text{ cm} in front of the lens, what is the focal length of the lens in centimeters (cm\text{cm})?

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Cevap: 20

Cevap

The focal length of the converging lens is 20 cm20\text{ cm}.
Using the thin lens formula 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} with an object distance u=30 cmu = 30\text{ cm} and a real image distance v=60 cmv = 60\text{ cm} yields 1f=130+160=120\frac{1}{f} = \frac{1}{30} + \frac{1}{60} = \frac{1}{20}, giving f=20 cmf = 20\text{ cm}.

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1
Identify given parameters and apply correct sign conventions
u=+30 cmu = +30\text{ cm} (real object) and v=+60 cmv = +60\text{ cm} (real image on screen)
In thin lens calculations for real objects and images formed on screens, both distances are positive.
2
Substitute parameters into the thin lens formula
1f=130+160\frac{1}{f} = \frac{1}{30} + \frac{1}{60}
The thin lens equation relates focal length ff, object distance uu, and image distance vv.
3
Perform fraction addition and solve for focal length
1f=360=120    f=20 cm\frac{1}{f} = \frac{3}{60} = \frac{1}{20} \implies f = 20\text{ cm}
Taking the common denominator gives 120 cm1\frac{1}{20}\text{ cm}^{-1}, which yields f=20 cmf = 20\text{ cm}.

Anahtar Kavram

Thin Lens Formula for Real Image Formation
Soru 964Soru

An alternating voltage source of root-mean-square (RMS) voltage 200 V200\ \text{V} is connected across a series combination of a resistor with resistance 60 Ω60\ \Omega and a capacitor with capacitive reactance 80 Ω80\ \Omega. Calculate the root-mean-square current in the circuit.

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Cevap: 2

Cevap

The root-mean-square current flowing through the circuit is 2 A.
The total impedance of the series RC circuit is obtained via quadrature sum Z=R2+XC2=602+802=100 ΩZ = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = 100\ \Omega. Dividing the RMS supply voltage (200 V200\ \text{V}) by this impedance gives an RMS current of 2 A2\ \text{A}.

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1
Calculate the total impedance of the series RC circuit.
Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = 100\ \Omega
In an AC circuit with resistance and capacitive reactance in series, the total Opposition (impedance) is found by phasor addition.
2
Apply Ohm's law for alternating current circuits to find RMS current.
I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\ \text{V}}{100\ \Omega} = 2\ \text{A}
The RMS current is equal to the RMS voltage divided by the total circuit impedance.

Anahtar Kavram

Impedance and RMS Current in AC Series Circuits
Soru 965Soru

When 10.0 g10.0\text{ g} of pure calcium carbonate (CaCO3\text{CaCO}_3) is strongly heated, it completely decomposes into solid calcium oxide (CaO\text{CaO}) and carbon(IV) oxide gas (CO2\text{CO}_2). According to the Law of Conservation of Mass, if 5.6 g5.6\text{ g} of calcium oxide remains in the container, what is the mass of carbon(IV) oxide gas released in grams?

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Cevap: 4.4

Cevap

The mass of carbon(IV) oxide gas released is 4.4 g4.4\text{ g}.
According to the Law of Conservation of Mass, the total mass of reactants must equal the total mass of products in a chemical change. For the reaction CaCO3(s)CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g), the initial mass of 10.0 g10.0\text{ g} of CaCO3\text{CaCO}_3 must equal the combined mass of CaO\text{CaO} (5.6 g5.6\text{ g}) and CO2\text{CO}_2. Subtracting 5.6 g5.6\text{ g} from 10.0 g10.0\text{ g} yields 4.4 g4.4\text{ g} for the gas produced.

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1
Apply the Law of Conservation of Mass
Total mass of reactants (10.0 g10.0\text{ g}) = Total mass of products (solid residue + gas)
Mass cannot be created or destroyed in a chemical reaction.
2
Calculate the missing mass of carbon(IV) oxide gas
Mass of CO2=10.0 g5.6 g=4.4 g\text{Mass of CO}_2 = 10.0\text{ g} - 5.6\text{ g} = 4.4\text{ g}
Subtracting the mass of the solid product from the initial mass of reactant gives the mass of the gaseous product evolved.

Anahtar Kavram

Law of Conservation of Mass
Tahmini Süre:45s
Soru 966Soru

A satellite revolves around a planet in a circular orbit of radius 1.0×104 km1.0 \times 10^4 \text{ km} with an orbital period of 12 hours12 \text{ hours}. Calculate the orbital period, in hours, of a second satellite orbiting the same planet in a circular path of radius 4.0×104 km4.0 \times 10^4 \text{ km}.

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Cevap: 96

Cevap

96 hours
According to Kepler's Third Law (T2r3T^2 \propto r^3), the orbital period TT scales with radius rr as Tr3/2T \propto r^{3/2}. Increasing the orbital radius by a factor of 4 increases the period by a factor of 43/2=84^{3/2} = 8. Multiplying the original period of 12 hours by 8 yields 96 hours.

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1
Set up Kepler's Third Law equation relating orbital period and orbital radius.
T22T12=r23r13\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}
Kepler's Third Law states that the square of the orbital period of a body in circular orbit is directly proportional to the cube of the radius of its orbit.
2
Substitute the given orbital radii and evaluate the scaling factor.
\frac{r_2}{r_1} = \frac{4.0 \times 10^4 \text{ km}}{1.0 \times 10^4 \text{ km}} = 4
Simplifying the ratio of the two orbital radii gives a factor of 4 increase in radius.
3
Calculate the period multiplier by taking the ratio to the power of 3/2.
4^{3/2} = (\sqrt{4})^3 = 2^3 = 8
Taking T2=T1×(r2r1)3/2T_2 = T_1 \times \left(\frac{r_2}{r_1}\right)^{3/2} shows the period scales by a factor of 8.
4
Multiply the initial orbital period by the scaling factor to find the final answer.
T_2 = 12 \text{ hours} \times 8 = 96 \text{ hours}
Multiplying the baseline period of 12 hours by 8 yields the new orbital period.

Anahtar Kavram

Kepler's Third Law of Planetary Motion
Tahmini Süre:1m 30s
Soru 967Soru

A satellite of mass 500 kg500\text{ kg} orbits a spherical planet of radius R=6.0×106 mR = 6.0 \times 10^6\text{ m} with surface gravitational acceleration g=10 m/s2g = 10\text{ m/s}^2. The satellite is transferred from an initial circular orbit of radius 2R2R to a higher circular orbit of radius 3R3R. What is the minimum energy required, in megajoules (MJ\text{MJ}), to perform this transfer?

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Cevap: 2500

Cevap

The minimum energy required to perform the orbital transfer is 2500 MJ2500\text{ MJ}.
The minimum energy needed to move a satellite between circular orbits is equal to the change in its total mechanical energy (E=GMm2rE = -\frac{GMm}{2r}). Expressing GMGM as gR2gR^2, the energy difference between radii 2R2R and 3R3R simplifies to ΔE=gRm12\Delta E = \frac{gRm}{12}, which evaluates to 2500 MJ2500\text{ MJ}.

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1
Relate surface acceleration due to gravity to planet mass and radius.
GM=gR2GM = gR^2
At the planet's surface (r=Rr = R), gravitational acceleration is g=GMR2g = \frac{GM}{R^2}.
2
Formulate the total mechanical energy equation for a circular orbit.
E=GMm2r=gR2m2rE = -\frac{GMm}{2r} = -\frac{gR^2 m}{2r}
Total energy is kinetic energy GMm2r\frac{GMm}{2r} plus gravitational potential energy GMmr-\frac{GMm}{r}.
3
Calculate initial and final total energies.
E1=gRm4E_1 = -\frac{gRm}{4} and E2=gRm6E_2 = -\frac{gRm}{6}
Substitute the orbit radii r1=2Rr_1 = 2R and r2=3Rr_2 = 3R into the total energy equation.
4
Determine the net work required for the transfer.
ΔE=E2E1=gRm12\Delta E = E_2 - E_1 = \frac{gRm}{12}
The energy required equals the difference in total mechanical energy between the final and initial orbits.
5
Substitute given numerical values and convert joules to megajoules.
ΔE=10×(6.0×106)×50012=2.5×109 J=2500 MJ\Delta E = \frac{10 \times (6.0 \times 10^6) \times 500}{12} = 2.5 \times 10^9\text{ J} = 2500\text{ MJ}
Dividing 2.5×109 J2.5 \times 10^9\text{ J} by 10610^6 converts the value to megajoules.

Anahtar Kavram

Total Mechanical Energy of a Satellite in Circular Orbit and Orbital Transfer Energy
Tahmini Süre:3m 0s
Soru 968Soru

An agricultural economist recorded the prices per bag of fertilizer (in thousands of Naira) across five regional markets as 12₦12, 15₦15, 18₦18, 21₦21, and 24₦24. What is the mean deviation of the fertilizer prices (in thousands of Naira)?

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Cevap: 3.6

Cevap

The mean deviation of the fertilizer prices is 3.63.6 thousand Naira.
The mean deviation is 3.63.6 because the arithmetic mean of the prices is 1818. The absolute differences of the data values from 1818 are 66, 33, 00, 33, and 66. The sum of these absolute deviations is 1818, and dividing by 55 observations yields 3.63.6.

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1
Calculate the arithmetic mean (xˉ\bar{x}) of the data set
xˉ=12+15+18+21+245=905=18\bar{x} = \frac{12 + 15 + 18 + 21 + 24}{5} = \frac{90}{5} = 18
The arithmetic mean provides the central reference point required to evaluate individual deviations.
2
Determine the absolute deviation xxˉ|x - \bar{x}| for each value
1218=6|12 - 18| = 6, 1518=3|15 - 18| = 3, 1818=0|18 - 18| = 0, 2118=3|21 - 18| = 3, 2418=6|24 - 18| = 6
Mean deviation measures dispersion using absolute differences to prevent positive and negative deviations from canceling out.
3
Compute the average of the absolute deviations
\text{Mean Deviation} = \frac{6 + 3 + 0 + 3 + 6}{5} = \frac{18}{5} = 3.6
Dividing the sum of absolute deviations by the total number of observations gives the mean deviation.

Anahtar Kavram

Mean Deviation
Soru 969Soru

A gaseous mixture containing carbon monoxide (CO\text{CO}) and carbon dioxide (CO2\text{CO}_2) has a total mass of 10.0 g10.0\text{ g}. If the mixture contains a total of 2.408×10232.408 \times 10^{23} oxygen atoms, calculate the mass, in grams, of carbon dioxide (CO2\text{CO}_2) present in the mixture. [C=12.0,O=16.0,NA=6.02×1023 mol1][\text{C} = 12.0, \text{O} = 16.0, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]

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Cevap: 4.4

Cevap

The mass of carbon dioxide (CO2\text{CO}_2) present in the mixture is 4.4 g4.4\text{ g}.
The correct calculation yields 4.4 g by converting the oxygen atom count to 0.40 moles of O atoms, formulating the system of equations for total mass (28x + 44y = 10.0) and total oxygen moles (x + 2y = 0.40), and solving for the mass of CO₂.

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1
Determine the molar masses of carbon monoxide and carbon dioxide.
Molar mass of CO=12.0+16.0=28.0 g mol1\text{Molar mass of CO} = 12.0 + 16.0 = 28.0\text{ g mol}^{-1}; Molar mass of CO2=12.0+2(16.0)=44.0 g mol1\text{Molar mass of CO}_2 = 12.0 + 2(16.0) = 44.0\text{ g mol}^{-1}.
Molar masses are required to relate the mass of each component to its molar quantity.
2
Calculate the total number of moles of oxygen atoms in the mixture using Avogadro's constant.
nO=2.408×10236.02×1023 mol1=0.40 mol of O atomsn_{\text{O}} = \frac{2.408 \times 10^{23}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 0.40\text{ mol of O atoms}.
Avogadro's constant converts particle count to mole quantity.
3
Set up a system of linear equations representing the total mass and total moles of oxygen atoms.
Let x=moles of COx = \text{moles of CO} and y=moles of CO2y = \text{moles of CO}_2.
Equation 1 (Mass): 28x+44y=10.028x + 44y = 10.0
Equation 2 (Oxygen atoms): x+2y=0.40x + 2y = 0.40
CO contains 1 O atom per molecule and CO₂ contains 2 O atoms per molecule.
4
Solve the system of linear equations for yy (moles of CO2\text{CO}_2).
From Equation 2, x=0.402yx = 0.40 - 2y. Substituting into Equation 1 gives 28(0.402y)+44y=10.0    11.256y+44y=10.0    12y=1.2    y=0.10 mol28(0.40 - 2y) + 44y = 10.0 \implies 11.2 - 56y + 44y = 10.0 \implies 12y = 1.2 \implies y = 0.10\text{ mol}.
Algebraic substitution yields the mole quantity of carbon dioxide.
5
Calculate the mass of CO2\text{CO}_2 present in the sample.
Mass of CO2=y×Molar mass=0.10 mol×44.0 g mol1=4.4 g\text{Mass of CO}_2 = y \times \text{Molar mass} = 0.10\text{ mol} \times 44.0\text{ g mol}^{-1} = 4.4\text{ g}.
Multiplying moles of CO₂ by its molar mass gives the required mass in grams.

Anahtar Kavram

Mole Concept, Avogadro's Constant, and Gas Mixture Stoichiometry
Soru 970Soru
During the catalytic decomposition of hydrogen peroxide according to the equation:
2H2O2(aq)2H2O(l)+O2(g)2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)
a student reacts 100 cm3100\text{ cm}^3 of a 0.40 mol dm30.40\text{ mol dm}^{-3} solution of H2O2\text{H}_2\text{O}_2. If the peroxide decomposes completely in 80 s80\text{ s}, what is the average rate of formation of oxygen gas in cm3 s1\text{cm}^3\text{ s}^{-1} at STP? (Molar volume of gas at STP = 22.4 dm3 mol122.4\text{ dm}^3\text{ mol}^{-1})
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Cevap: 5.6

Cevap

The average rate of formation of oxygen gas at STP is 5.6 cm3 s15.6\text{ cm}^3\text{ s}^{-1}.
First, the amount of hydrogen peroxide in moles is calculated as 0.40 mol dm3×0.100 dm3=0.040 mol0.40\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.040\text{ mol}. From the reaction stoichiometry, 2 moles of H2O22\text{ moles of H}_2\text{O}_2 produce 1 mole of O21\text{ mole of O}_2, meaning 0.020 mol of O20.020\text{ mol of O}_2 is evolved. At STP, 0.020 mol×22400 cm3 mol1=448 cm30.020\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 448\text{ cm}^3 of oxygen. Dividing by the total time of 80 s80\text{ s} yields 5.6 cm3 s15.6\text{ cm}^3\text{ s}^{-1}.

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1
Calculate the total number of moles of H₂O₂ present in the reaction solution
Moles of H2O2=0.40 mol dm3×0.100 dm3=0.040 mol\text{H}_2\text{O}_2 = 0.40\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.040\text{ mol}
Concentration and volume determine the total amount of reactant available.
2
Determine the total moles of O₂ gas formed using stoichiometric ratios
Moles of O2=0.040 mol2=0.020 mol\text{O}_2 = \frac{0.040\text{ mol}}{2} = 0.020\text{ mol}
The balanced chemical equation shows a 2:1 mole ratio between H2O2\text{H}_2\text{O}_2 and O2\text{O}_2.
3
Convert moles of O₂ into volume in cm³ at STP
Volume of O2=0.020 mol×22400 cm3 mol1=448 cm3\text{O}_2 = 0.020\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 448\text{ cm}^3
1 mole of any ideal gas occupies 22.4 dm322.4\text{ dm}^3 (22400 cm322400\text{ cm}^3) at STP.
4
Divide total volume of O₂ produced by the elapsed reaction time
Rate of O2\text{O}_2 formation =448 cm380 s=5.6 cm3 s1= \frac{448\text{ cm}^3}{80\text{ s}} = 5.6\text{ cm}^3\text{ s}^{-1}
Reaction rate with respect to gas product evolution is change in volume per unit time.

Anahtar Kavram

Stoichiometric rate of reaction and gas volume calculations
Soru 971Soru

For the thermal decomposition of a compound, the standard enthalpy change (ΔH\Delta H^\circ) is +117.0 kJ mol1+117.0\text{ kJ mol}^{-1} and the standard entropy change (ΔS\Delta S^\circ) is +180.0 J K1 mol1+180.0\text{ J K}^{-1}\text{ mol}^{-1}. What is the minimum temperature, in kelvin (K\text{K}), above which the reaction becomes spontaneous?

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Cevap: 650

Cevap

The minimum temperature above which the reaction becomes spontaneous is 650 K.
For a reaction with positive ΔH\Delta H^\circ and positive ΔS\Delta S^\circ, spontaneity depends on temperature. Spontaneity occurs when ΔG=ΔHTΔS<0\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ < 0, which simplifies to T>ΔHΔST > \frac{\Delta H^\circ}{\Delta S^\circ}. Converting ΔH\Delta H^\circ to joules gives 117000 J mol1117000\text{ J mol}^{-1}, so T=117000180.0=650 KT = \frac{117000}{180.0} = 650\text{ K}.

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1
Convert enthalpy change units from kilojoules per mole to joules per mole
ΔH=117.0 kJ mol1×1000 J/kJ=117000 J mol1\Delta H^\circ = 117.0\text{ kJ mol}^{-1} \times 1000\text{ J/kJ} = 117000\text{ J mol}^{-1}
Enthalpy and entropy must be in consistent energy units (joules) before performing thermodynamic calculations.
2
Apply the condition for reaction spontaneity threshold
ΔG=ΔHTΔS=0\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0
A reaction is spontaneous when ΔG<0\Delta G^\circ < 0. The threshold temperature occurs exactly when ΔG=0\Delta G^\circ = 0.
3
Calculate the threshold temperature T
T=ΔHΔS=117000 J mol1180.0 J K1 mol1=650 KT = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{117000\text{ J mol}^{-1}}{180.0\text{ J K}^{-1}\text{ mol}^{-1}} = 650\text{ K}
Solving the threshold condition for TT yields the absolute temperature in kelvin above which TΔS>ΔHT\Delta S^\circ > \Delta H^\circ, making ΔG\Delta G^\circ negative.

Anahtar Kavram

Relationship between Gibbs free energy, enthalpy, entropy, and reaction spontaneity threshold
Soru 972Soru

In a chemical reaction, the potential energy of the reactants is 45 kJ mol145\text{ kJ mol}^{-1}, while the potential energy of the activated complex at the peak of the energy profile diagram is 125 kJ mol1125\text{ kJ mol}^{-1}. What is the activation energy for the forward reaction in kJ mol1\text{kJ mol}^{-1}?

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Cevap: 80

Cevap

The activation energy for the forward reaction is 80 kJ mol^{-1}.
The activation energy (EaE_a) for a forward chemical reaction is defined as the difference in energy between the activated complex (peak of the energy profile) and the reactants. Subtracting the reactant potential energy (45 kJ mol145\text{ kJ mol}^{-1}) from the activated complex potential energy (125 kJ mol1125\text{ kJ mol}^{-1}) gives an activation energy of 80 kJ mol180\text{ kJ mol}^{-1}.

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1
Identify the potential energy levels of the reactants and the activated complex from the given data.
Energy of reactants = 45 kJ mol^{-1}; Energy of activated complex = 125 kJ mol^{-1}
Forward activation energy depends directly on the difference between these two energy states.
2
Subtract the potential energy of the reactants from the potential energy of the activated complex.
Ea = 125 kJ mol^{-1} - 45 kJ mol^{-1} = 80 kJ mol^{-1}
The activation energy represents the minimum energy barrier that reacting particles must overcome to reach the transition state.

Anahtar Kavram

Forward Activation Energy Calculation from Energy Profile Data
Soru 973Soru

For a reversible gaseous reaction A+BC+DA + B \rightleftharpoons C + D, the potential energy of the reactants is 120 kJ mol1120\text{ kJ mol}^{-1} and the potential energy of the products is 45 kJ mol145\text{ kJ mol}^{-1}. If the activation energy for the reverse reaction is 140 kJ mol1140\text{ kJ mol}^{-1}, calculate the activation energy for the forward reaction in kJ mol1\text{kJ mol}^{-1}.

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Cevap: 65

Cevap

The activation energy for the forward reaction is 65 kJ mol165\text{ kJ mol}^{-1}.
The forward activation energy (Ea,forwardE_{a,\text{forward}}) is the energy barrier measured from the energy level of the reactants (120 kJ mol1120\text{ kJ mol}^{-1}) to the peak of the activated complex. Since the products lie at 45 kJ mol145\text{ kJ mol}^{-1} and require 140 kJ mol1140\text{ kJ mol}^{-1} to reach the activated complex, the peak energy is 45+140=185 kJ mol145 + 140 = 185\text{ kJ mol}^{-1}. Subtracting the reactant energy gives 185120=65 kJ mol1185 - 120 = 65\text{ kJ mol}^{-1}. Alternatively, using ΔH=Ea,forwardEa,reverse\Delta H = E_{a,\text{forward}} - E_{a,\text{reverse}}, where ΔH=45120=75 kJ mol1\Delta H = 45 - 120 = -75\text{ kJ mol}^{-1}, gives 75=Ea,forward140    Ea,forward=65 kJ mol1-75 = E_{a,\text{forward}} - 140 \implies E_{a,\text{forward}} = 65\text{ kJ mol}^{-1}.

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1
Determine the potential energy of the activated complex (transition state).
Transition state energy Epeak=Hproducts+Ea,reverse=45+140=185 kJ mol1E_{\text{peak}} = H_{\text{products}} + E_{a,\text{reverse}} = 45 + 140 = 185\text{ kJ mol}^{-1}.
The reverse activation energy is the energy required to overcome the barrier going from products up to the peak.
2
Calculate the forward activation energy.
Ea,forward=EpeakHreactants=185120=65 kJ mol1E_{a,\text{forward}} = E_{\text{peak}} - H_{\text{reactants}} = 185 - 120 = 65\text{ kJ mol}^{-1}.
The forward activation energy is the energy difference between the peak (activated complex) and the energy level of the reactants.

Anahtar Kavram

Activation energy and potential energy relationships in energy profile diagrams
Tahmini Süre:1m 30s
Soru 974Soru

A chemical reaction has a standard enthalpy change (ΔH\Delta H^\circ) of +75.0 kJ mol1+75.0\text{ kJ mol}^{-1} and a standard entropy change (ΔS\Delta S^\circ) of +250 J K1mol1+250\text{ J K}^{-1}\text{mol}^{-1}. Calculate the minimum temperature in Kelvin (K\text{K}) at which this reaction becomes spontaneous.

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Cevap: 300

Cevap

The minimum temperature at which the reaction becomes spontaneous is 300 K.
A chemical process becomes spontaneous when the Gibbs free energy change (ΔG\Delta G^\circ) is negative (ΔG<0\Delta G^\circ < 0). Setting ΔG=0\Delta G^\circ = 0 defines the threshold temperature for spontaneity. From ΔG=ΔHTΔS\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ, re-arranging gives T=ΔHΔST = \frac{\Delta H^\circ}{\Delta S^\circ}. Expressing ΔH\Delta H^\circ as 75000 J mol175000\text{ J mol}^{-1} and ΔS\Delta S^\circ as 250 J K1mol1250\text{ J K}^{-1}\text{mol}^{-1} yields T=75000250=300 KT = \frac{75000}{250} = 300\text{ K}.

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1
Convert standard enthalpy change ΔH\Delta H^\circ from kilojoules per mole to Joules per mole.
ΔH=75.0×103 J mol1=75000 J mol1\Delta H^\circ = 75.0 \times 10^3\text{ J mol}^{-1} = 75000\text{ J mol}^{-1}.
Units of enthalpy and entropy must be compatible (Joules) before performing calculation.
2
Apply the Gibbs free energy relationship ΔG=ΔHTΔS\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ at the spontaneity boundary condition ΔG=0\Delta G^\circ = 0.
0=ΔHTΔS    T=ΔHΔS0 = \Delta H^\circ - T\Delta S^\circ \implies T = \frac{\Delta H^\circ}{\Delta S^\circ}.
A reaction is spontaneous when ΔG<0\Delta G^\circ < 0, making ΔG=0\Delta G^\circ = 0 the exact minimum temperature threshold.
3
Divide the enthalpy change in Joules per mole by the entropy change in Joules per Kelvin-mole.
T=75000 J mol1250 J K1mol1=300 KT = \frac{75000\text{ J mol}^{-1}}{250\text{ J K}^{-1}\text{mol}^{-1}} = 300\text{ K}.
Dividing Joules by Joules per Kelvin yields the temperature in Kelvin.

Anahtar Kavram

Gibbs free energy and temperature dependence of reaction spontaneity
Tahmini Süre:1m 15s
Soru 975Soru
Consider the unbalanced redox reaction taking place in an acidic medium:
a MnO4(aq)+b SO32(aq)+c H+(aq)d Mn2+(aq)+e SO42(aq)+f H2O(l)\text{a MnO}_4^-(\text{aq}) + \text{b SO}_3^{2-}(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d Mn}^{2+}(\text{aq}) + \text{e SO}_4^{2-}(\text{aq}) + \text{f H}_2\text{O}(\text{l})
When this chemical equation is balanced using the smallest set of whole-number coefficients, what is the value of the stoichiometric coefficient cc for H+(aq)\text{H}^+(\text{aq})?
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Cevap: 6

Cevap

The value of the stoichiometric coefficient c for H+(aq) is 6.
Balancing the reduction half-reaction (2MnO4+16H++10e2Mn2++8H2O2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}) and oxidation half-reaction (5SO32+5H2O5SO42+10H++10e5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-) gives a combined total of 16H+16\text{H}^+ on the reactant side and 10H+10\text{H}^+ on the product side. Subtracting 10H+10\text{H}^+ from both sides leaves a net coefficient of 6 for H+(aq)\text{H}^+(\text{aq}) on the reactant side.

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1
Write the balanced reduction half-reaction for permanganate ion in acidic medium.
MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})
Manganese goes from oxidation state +7 to +2, requiring 5 electrons, 8 H+ ions to balance oxygen atoms, forming 4 H2O molecules.
2
Write the balanced oxidation half-reaction for sulfite ion to sulfate ion.
SO32(aq)+H2O(l)SO42(aq)+2H+(aq)+2e\text{SO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{SO}_4^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 2e^-
Sulfur goes from oxidation state +4 to +6, releasing 2 electrons and 2 H+ ions while consuming 1 H2O molecule.
3
Equalize the number of transferred electrons by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5.
2MnO4+16H++10e2Mn2++8H2O2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}
5SO32+5H2O5SO42+10H++10e5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-
The least common multiple of 5 and 2 transferred electrons is 10.
4
Combine the half-reactions and subtract common species (10e10e^-, 10H+10\text{H}^+, and 5H2O5\text{H}_2\text{O}) from both sides.
2MnO4(aq)+5SO32(aq)+6H+(aq)2Mn2+(aq)+5SO42(aq)+3H2O(l)2\text{MnO}_4^-(\text{aq}) + 5\text{SO}_3^{2-}(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{SO}_4^{2-}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})
Subtracting 10H+10\text{H}^+ from 16H+16\text{H}^+ leaves 6H+6\text{H}^+ on the reactant side, giving c=6c = 6.

Anahtar Kavram

Balancing Redox Equations using the Ion-Electron Method in Acidic Medium
Soru 976Soru

A gas sealed inside a rigid container of fixed volume is connected to a pressure gauge that displays a gauge pressure of 1.50 atm1.50\text{ atm} at an initial temperature of 27C27^\circ\text{C}. If the ambient atmospheric pressure is 1.00 atm1.00\text{ atm}, calculate the temperature in degrees Celsius (C^\circ\text{C}) to which the gas must be heated for its total absolute pressure to double.

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Cevap: 327

Cevap

The final temperature required is 327C327^\circ\text{C}.
To find the required temperature, first calculate the initial absolute pressure (1.50 atm+1.00 atm=2.50 atm1.50\text{ atm} + 1.00\text{ atm} = 2.50\text{ atm}) and convert the initial temperature to Kelvin (27C+273=300 K27^\circ\text{C} + 273 = 300\text{ K}). Doubling the absolute pressure to 5.00 atm5.00\text{ atm} requires doubling the absolute temperature according to the Pressure Law (T2=600 KT_2 = 600\text{ K}). Converting 600 K600\text{ K} back to Celsius yields 600273=327C600 - 273 = 327^\circ\text{C}.

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1
Determine the initial absolute pressure of the gas.
P1=2.50 atmP_1 = 2.50\text{ atm}
Absolute pressure is the sum of gauge pressure and ambient atmospheric pressure (Pabs=Pgauge+PatmP_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}).
2
Convert the initial temperature from Celsius to Kelvin.
T1=300 KT_1 = 300\text{ K}
Gas law calculations strictly require thermodynamic temperature expressed in Kelvin (TK=tC+273T_{\text{K}} = t_{^\circ\text{C}} + 273).
3
Calculate the required final absolute pressure.
P2=5.00 atmP_2 = 5.00\text{ atm}
The total absolute pressure doubles, so P2=2×2.50 atm=5.00 atmP_2 = 2 \times 2.50\text{ atm} = 5.00\text{ atm}.
4
Apply Gay-Lussac's Pressure Law to calculate the final absolute temperature.
T2=600 KT_2 = 600\text{ K}
At constant volume, pressure is directly proportional to absolute temperature (P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}), yielding T2=T1×P2P1=300×2=600 KT_2 = T_1 \times \frac{P_2}{P_1} = 300 \times 2 = 600\text{ K}.
5
Convert the final absolute temperature back to degrees Celsius.
t2=327Ct_2 = 327^\circ\text{C}
Subtract 273 from the Kelvin temperature (tC=600273=327Ct_{^\circ\text{C}} = 600 - 273 = 327^\circ\text{C}).

Anahtar Kavram

Pressure Law (Gay-Lussac's Law) states that the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature (PTP \propto T). Calculations must strictly use absolute pressure and Kelvin temperatures.
Soru 977Soru

A solution is formed by mixing 300 cm3300\text{ cm}^3 of 0.10 mol dm30.10\text{ mol dm}^{-3} tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4) solution with 700 cm3700\text{ cm}^3 of 0.10 mol dm30.10\text{ mol dm}^{-3} sodium hydroxide (NaOH\text{NaOH}) solution at 25C25^\circ\text{C}. Assuming complete dissociation of both electrolytes, what is the pH of the resulting mixture?

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Cevap: 12

Cevap

The pH of the resulting mixture is 12.0.
The mixture contains excess hydroxide ions (0.010 mol in 1.0 dm³ solution), resulting in a pOH of 2.0. Subtracting this from 14.0 gives a pH of 12.0.

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1
Calculate the moles of hydrogen ions (H⁺) contributed by the acid solution.
n(H+)=0.060 moln(\text{H}^+) = 0.060\text{ mol}
Tetraoxosulfate(VI) acid is diprotic (dibasic), releasing 2 moles of H+\text{H}^+ per mole of acid: 0.300 dm3×0.10 mol dm3×2=0.060 mol0.300\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} \times 2 = 0.060\text{ mol}.
2
Calculate the moles of hydroxide ions (OH⁻) contributed by the base solution.
n(OH)=0.070 moln(\text{OH}^-) = 0.070\text{ mol}
Sodium hydroxide is a monobasic base: 0.700 dm3×0.10 mol dm3=0.070 mol0.700\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 0.070\text{ mol}.
3
Determine the unneutralized excess ions and calculate their molar concentration in the total volume.
[OH]=0.010 mol dm3[\text{OH}^-] = 0.010\text{ mol dm}^{-3}
The neutralization reaction is H++OHH2O\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}. Excess OH=0.0700.060=0.010 mol\text{OH}^- = 0.070 - 0.060 = 0.010\text{ mol}. Divided by the total mixture volume of 1.0 dm31.0\text{ dm}^3, [OH]=0.010 mol dm3[\text{OH}^-] = 0.010\text{ mol dm}^{-3}.
4
Calculate pOH and convert it to pH using the water autoionization relation.
pH=12.0\text{pH} = 12.0
pOH=log10(0.010)=2.0\text{pOH} = -\log_{10}(0.010) = 2.0. Since pH+pOH=14.0\text{pH} + \text{pOH} = 14.0 at 25C25^\circ\text{C}, pH=14.02.0=12.0\text{pH} = 14.0 - 2.0 = 12.0.

Anahtar Kavram

Neutralization stoichiometry of diprotic acids and strong bases followed by pH determination from excess hydroxide concentration.
Soru 978Soru

During a chemical reaction between zinc metal and dilute hydrochloric acid, the volume of hydrogen gas produced was recorded over time. At 10 s10\text{ s}, the total volume of hydrogen gas collected was 15.0 cm315.0\text{ cm}^3, and at 30 s30\text{ s}, the volume collected reached 45.0 cm345.0\text{ cm}^3. What is the average rate of hydrogen gas evolution in cm3 s1\text{cm}^3\text{ s}^{-1} over this 20-second20\text{-second} time interval?

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Cevap: 1.5

Cevap

The average rate of evolution of hydrogen gas over the time interval is 1.5 cm3 s11.5\text{ cm}^3\text{ s}^{-1}.
The average rate of reaction is calculated by dividing the volume of gas produced during the interval (ΔV=45.0 cm315.0 cm3=30.0 cm3\Delta V = 45.0\text{ cm}^3 - 15.0\text{ cm}^3 = 30.0\text{ cm}^3) by the elapsed time (Δt=30 s10 s=20 s\Delta t = 30\text{ s} - 10\text{ s} = 20\text{ s}), yielding 1.5 cm3 s11.5\text{ cm}^3\text{ s}^{-1}.

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1
Determine the volume of hydrogen gas evolved during the specified time interval.
ΔV=45.0 cm315.0 cm3=30.0 cm3\Delta V = 45.0\text{ cm}^3 - 15.0\text{ cm}^3 = 30.0\text{ cm}^3
The volume change represents the quantity of product formed specifically between 10 s10\text{ s} and 30 s30\text{ s}.
2
Determine the time elapsed over the interval.
Δt=30 s10 s=20 s\Delta t = 30\text{ s} - 10\text{ s} = 20\text{ s}
The reaction rate is measured over the duration between the two observations.
3
Compute the average rate of gas evolution.
\text{Rate} = \frac{\Delta V}{\Delta t} = \frac{30.0\text{ cm}^3}{20\text{ s}} = 1.5\text{ cm}^3\text{ s}^{-1}
The average rate of reaction with respect to gas volume is defined as the change in volume per unit time.

Anahtar Kavram

Average Rate of Reaction
Soru 979Soru
Consider the half-reaction representing the oxidation of thiosulfate ions to tetrathionate ions:
2S2O32(aq)S4O62(aq)+ne2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + n e^-
What is the number of electrons, nn, required to balance the charge in this half-reaction?
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Cevap: 2

Cevap

The number of electrons required to balance the charge in the half-reaction is 2.
To balance a half-reaction, both atom counts and net electric charges must be equal on both sides of the equation. The reactant side contains 2 thiosulfate ions (2S2O322\text{S}_2\text{O}_3^{2-}), giving a net charge of 2×(2)=42 \times (-2) = -4. The product side contains 1 tetrathionate ion (S4O62\text{S}_4\text{O}_6^{2-}), giving a net charge of 2-2. Adding 2 electrons (2e2 e^-) to the product side lowers its total charge to 4-4, equalizing the charge on both sides.

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1
Calculate the total charge of the reactant species.
Reactant charge = 2 * (-2) = -4.
There are 2 thiosulfate ions, each with an ionic charge of -2.
2
Calculate the net charge of the ionic product species.
Product charge (excluding electrons) = -2.
There is 1 tetrathionate ion with an ionic charge of -2.
3
Equate the overall charges on both sides to solve for the number of electrons n.
-4 = -2 - n, giving n = 2.
Adding 2 electrons (each carrying a -1 charge) to the product side brings the total product charge to -4, matching the reactant side.

Anahtar Kavram

Balancing electric charge in oxidation half-reactions
Soru 980Soru

At 25C25^\circ\text{C}, an aqueous solution of a weak monoacidic base has a concentration of 0.08 mol dm30.08\text{ mol dm}^{-3} and a degree of ionization (α\alpha) of 0.0250.025 (2.5%2.5\%). What is the concentration of hydroxide ions, [OH][\text{OH}^-], in the solution in mol dm3\text{mol dm}^{-3}?

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Cevap: 0.002

Cevap

The concentration of hydroxide ions, [OH][\text{OH}^-], in the solution is 0.002 mol dm30.002\text{ mol dm}^{-3}.
For a weak monoacidic base in aqueous solution, only a fraction (α\alpha) of the base molecules ionize to form hydroxide ions. The concentration of hydroxide ions is given by [OH]=Cα[\text{OH}^-] = C \alpha. Substituting the concentration 0.08 mol dm30.08\text{ mol dm}^{-3} and degree of ionization 0.0250.025 yields [OH]=0.08×0.025=0.002 mol dm3[\text{OH}^-] = 0.08 \times 0.025 = 0.002\text{ mol dm}^{-3}.

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1
Identify the relationship between degree of ionization and hydroxide ion concentration for a weak monoacidic base
[OH]=Cα[\text{OH}^-] = C \cdot \alpha
A weak monoacidic base BOH\text{BOH} ionizes partially according to BOH(aq)B(aq)++OH(aq)\text{BOH}_{(aq)} \rightleftharpoons \text{B}^+_{(aq)} + \text{OH}^-_{(aq)}, so the concentration of produced hydroxide ions equals the initial concentration multiplied by the degree of ionization.
2
Substitute the given values into the equation
[OH]=0.08 mol dm3×0.025[\text{OH}^-] = 0.08\text{ mol dm}^{-3} \times 0.025
The initial concentration C=0.08 mol dm3C = 0.08\text{ mol dm}^{-3} and the degree of ionization α=2.5%=0.025\alpha = 2.5\% = 0.025.
3
Perform the multiplication to find the final concentration
[OH]=0.002 mol dm3[\text{OH}^-] = 0.002\text{ mol dm}^{-3}
Multiplying 0.080.08 by 0.0250.025 yields 0.002 mol dm30.002\text{ mol dm}^{-3} (or 2.0×103 mol dm32.0 \times 10^{-3}\text{ mol dm}^{-3}).

Anahtar Kavram

Ionization equilibrium of weak bases and calculation of hydroxide ion concentration
Tahmini Süre:1m 15s
ÖncekiSayfa 49 / 77Sonraki
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