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Question 8101Question

Match each of the following oxides on the left with its correct acid-base classification on the right.

Click a left item, then click its matching right item

Items

Potassium oxide (K2O\text{K}_2\text{O})
Sulfur dioxide (SO2\text{SO}_2)
Lead(II) oxide (PbO\text{PbO})
Nitrogen(II) oxide (NO\text{NO})

Matches

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Answer

Potassium oxide (K2O\text{K}_2\text{O}) matches Basic oxide; Sulfur dioxide (SO2\text{SO}_2) matches Acidic oxide; Lead(II) oxide (PbO\text{PbO}) matches Amphoteric oxide; Nitrogen(II) oxide (NO\text{NO}) matches Neutral oxide.
Potassium oxide is a basic metallic oxide; sulfur dioxide is an acidic non-metal oxide; lead(II) oxide reacts with both acids and alkalis (amphoteric); and nitrogen(II) oxide does not react with either acids or alkalis (neutral).

Step-by-Step Solution

1
Identify the nature of potassium oxide (K2O\text{K}_2\text{O}).
As an alkali metal oxide, it dissolves in water to form potassium hydroxide (KOH\text{KOH}) and reacts with acids to yield salts, classifying it as a basic oxide.
Basic oxides are metallic oxides that neutralize acids.
2
Identify the nature of sulfur dioxide (SO2\text{SO}_2).
It is a covalent non-metal oxide that reacts with alkalis to form trioxosulfate(IV) salts, classifying it as an acidic oxide.
Acidic oxides (acid anhydrides) react with bases or water to yield acidic solutions/salts.
3
Identify the nature of lead(II) oxide (PbO\text{PbO}).
It shows dual reactivity, dissolving in acids like HNO3\text{HNO}_3 and bases like concentrated NaOH\text{NaOH}, classifying it as an amphoteric oxide.
Amphoteric oxides can behave as either acids or bases depending on the reactant.
4
Identify the nature of nitrogen(II) oxide (NO\text{NO}).
It displays no acid-base properties with aqueous acids or bases, classifying it as a neutral oxide.
Neutral oxides do not form salts when treated with acids or alkalis.

Key Concept

Classification of oxides based on acid-base character (basic, acidic, amphoteric, neutral)
Question 8102Question

An alternating current (AC) circuit contains an inductor of inductance L=0.2π2 HL = \frac{0.2}{\pi^2}\ \text{H}, a resistor of resistance R=40 ΩR = 40\ \Omega, and a variable capacitor CC connected in series across a 50 Hz50\ \text{Hz} voltage supply. What capacitance CC, in microfarads (μF\mu\text{F}), is required for the circuit to operate at electrical resonance?

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Answer: 500

Answer

The capacitance required to achieve electrical resonance is 500 μF.
At electrical resonance in a series RLC circuit, the inductive reactance (XLX_L) equals the capacitive reactance (XCX_C). Setting 2πfL=12πfC2\pi f L = \frac{1}{2\pi f C} yields f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}. Substituting f0=50 Hzf_0 = 50\ \text{Hz} and L=0.2π2 HL = \frac{0.2}{\pi^2}\ \text{H} into this equation gives C=5×104 FC = 5 \times 10^{-4}\ \text{F}, which equals 500 μF500\ \mu\text{F}.

Step-by-Step Solution

1
Recall the resonant frequency formula for a series RLC circuit.
f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}
At resonance, inductive reactance equals capacitive reactance (XL=XCX_L = X_C).
2
Substitute the given numerical parameters into the equation.
50=12π0.2π2C50 = \frac{1}{2\pi \sqrt{\frac{0.2}{\pi^2} \cdot C}}
Given frequency f0=50 Hzf_0 = 50\ \text{Hz} and inductance L=0.2π2 HL = \frac{0.2}{\pi^2}\ \text{H}.
3
Isolate the square root term and simplify.
0.2C=0.01\sqrt{0.2 C} = 0.01
Simplifying 2π1π=22\pi \cdot \frac{1}{\pi} = 2 and rearranging 20.2C=150=0.022 \sqrt{0.2 C} = \frac{1}{50} = 0.02.
4
Square both sides and solve for CC in farads.
C=5×104 FC = 5 \times 10^{-4}\ \text{F}
0.2C=(0.01)2=1040.2 C = (0.01)^2 = 10^{-4}, so C=1040.2=5×104 FC = \frac{10^{-4}}{0.2} = 5 \times 10^{-4}\ \text{F}.
5
Convert capacitance from farads to microfarads.
C=500 μFC = 500\ \mu\text{F}
Multiply farads by 10610^6 to express the result in microfarads.

Key Concept

Resonant Frequency in Series AC Circuits
Question 8103Question

An unknown solid metallic oxide XX dissolves in cold dilute tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4) to form a solution that turns acidified potassium iodide (KI\text{KI}) solution brown due to the liberation of iodine, without releasing any gas during the reaction. In contrast, lead(IV) oxide (PbO2\text{PbO}_2) treated with cold dilute acid does not produce hydrogen peroxide (H2O2\text{H}_2\text{O}_2). Which of the following correctly classifies oxide XX and accounts for its distinct behavior from lead(IV) oxide?

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Answer: Oxide XX is a peroxide containing the O22\text{O}_2^{2-} ion which yields H2O2\text{H}_2\text{O}_2 with dilute acid, whereas PbO2\text{PbO}_2 is a dioxide containing simple O2\text{O}^{2-} ions.

Answer

Oxide X is a peroxide containing the O22\text{O}_2^{2-} ion which yields H2O2\text{H}_2\text{O}_2 with dilute acid, whereas PbO2\text{PbO}_2 is a dioxide containing simple O2\text{O}^{2-} ions.
Peroxides (such as sodium peroxide or barium peroxide) contain the peroxide ion O22\text{O}_2^{2-} with oxygen in the 1-1 oxidation state. When treated with cold dilute acids, they form hydrogen peroxide (H2O2\text{H}_2\text{O}_2), which oxidizes I\text{I}^- to brown I2\text{I}_2. In contrast, dioxides like lead(IV) oxide (PbO2\text{PbO}_2) contain standard oxide ions (O2\text{O}^{2-}) with lead in the +4+4 oxidation state, so they do not produce H2O2\text{H}_2\text{O}_2 under cold dilute acid conditions.

Step-by-Step Solution

1
Analyze the chemical test result for Oxide X.
Oxide XX reacts with cold dilute H2SO4\text{H}_2\text{SO}_4 to form a solution that oxidizes KI\text{KI} to I2\text{I}_2 (brown solution).
This behavior is characteristic of peroxides (such as Na2O2\text{Na}_2\text{O}_2 or BaO2\text{BaO}_2), which react with dilute acids to produce hydrogen peroxide (H2O2\text{H}_2\text{O}_2). Hydrogen peroxide is a powerful oxidizing agent that liberates iodine from acidified potassium iodide: H2O2+2H++2I2H2O+I2\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{I}^- \rightarrow 2\text{H}_2\text{O} + \text{I}_2.
2
Distinguish between peroxides and dioxides.
Peroxides contain the peroxide linkage [OO]2[-\text{O}-\text{O}-]^{2-} (O22\text{O}_2^{2-} ion with oxygen in oxidation state 1-1), while true dioxides contain O2\text{O}^{2-} ions with the metal in oxidation state +4+4.
Dioxides like PbO2\text{PbO}_2 and MnO2\text{MnO}_2 do not contain peroxide ions and therefore do not form H2O2\text{H}_2\text{O}_2 when treated with cold dilute acids.
3
Select the correct classification matching the observed chemistry.
Oxide XX is a peroxide, whereas PbO2\text{PbO}_2 is a dioxide.
This structural difference explains why oxide XX yields H2O2\text{H}_2\text{O}_2 capable of liberating iodine from KI\text{KI}, while PbO2\text{PbO}_2 does not.

Key Concept

Distinction between peroxides (containing O22\text{O}_2^{2-}) and dioxides (containing O2\text{O}^{2-}) based on acid reactions and peroxide test
Estimated Time:2m 0s
Question 8104Question

Based on the kinetic molecular theory of matter, which of the following best explains why liquids have a fixed volume but take the shape of their container?

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Answer: The particles are close together with moderate intermolecular attractive forces, allowing them to slide past one another while remaining bound in volume.

Answer

The particles are close together with moderate intermolecular attractive forces, allowing them to slide past one another while remaining bound in volume.
According to the kinetic theory, liquid particles are held closely by intermolecular forces, maintaining a definite volume, but possess enough kinetic energy to slide over one another, enabling the liquid to flow and take the shape of its container.

Step-by-Step Solution

1
Analyze the microscopic properties of the liquid state according to kinetic molecular theory.
Liquid particles possess intermediate kinetic energy compared to solids and gases.
Understanding the balance between kinetic energy and intermolecular forces explains liquid behavior.
2
Relate particle arrangement to volume and shape characteristics.
Moderate intermolecular forces keep the particles close together (definite volume), but allow them to slide past each other dynamically (indefinite shape).
Fluidity requires movement between particles, while fixed volume requires persistent intermolecular contacts.

Key Concept

Kinetic Molecular Postulates and States of Matter
Question 8105Question

A sample of 90 cm390\text{ cm}^3 of pure oxygen gas (O2\text{O}_2) is passed through an ozonizer where 20%20\% by volume of the oxygen is converted into ozone (O3\text{O}_3). If the resulting gas mixture is reacted completely with excess carbon(II) oxide (CO\text{CO}) to yield carbon(IV) oxide (CO2\text{CO}_2), what is the total volume of carbon(IV) oxide gas produced under the same conditions of temperature and pressure?

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Answer: 180 cm3180\text{ cm}^3

Answer

The total volume of carbon(IV) oxide gas produced is 180 cm3180\text{ cm}^3.
The total volume of carbon(IV) oxide produced is 180 cm3180\text{ cm}^3 because the total number of oxygen atoms available for reaction remains conserved regardless of the conversion into ozone. 72 cm372\text{ cm}^3 of diatomic oxygen yields 144 cm3144\text{ cm}^3 of carbon(IV) oxide and 12 cm312\text{ cm}^3 of triatomic ozone yields 36 cm336\text{ cm}^3 of carbon(IV) oxide.

Step-by-Step Solution

1
Calculate the volume of oxygen converted to ozone and the volume of ozone formed.
Volume of O2\text{O}_2 converted = 20%×90 cm3=18 cm320\% \times 90\text{ cm}^3 = 18\text{ cm}^3. Remaining unreacted O2=9018=72 cm3\text{O}_2 = 90 - 18 = 72\text{ cm}^3. According to 3O2(g)2O3(g)3\text{O}_2(g) \rightarrow 2\text{O}_3(g), 18 cm318\text{ cm}^3 of O2\text{O}_2 produces 23×18 cm3=12 cm3\frac{2}{3} \times 18\text{ cm}^3 = 12\text{ cm}^3 of O3\text{O}_3.
Ozonolysis causes a volume contraction where 3 volumes of O2\text{O}_2 yield 2 volumes of O3\text{O}_3.
2
Calculate the volume of CO2\text{CO}_2 produced by the unreacted O2\text{O}_2.
Reaction equation: 2CO(g)+O2(g)2CO2(g)2\text{CO}(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g). Volume of CO2\text{CO}_2 from O2=2×72 cm3=144 cm3\text{O}_2 = 2 \times 72\text{ cm}^3 = 144\text{ cm}^3.
By Gay-Lussac's Law of Combining Volumes, 1 volume of O2\text{O}_2 produces 2 volumes of CO2\text{CO}_2.
3
Calculate the volume of CO2\text{CO}_2 produced by the O3\text{O}_3.
Reaction equation: 3CO(g)+O3(g)3CO2(g)3\text{CO}(g) + \text{O}_3(g) \rightarrow 3\text{CO}_2(g). Volume of CO2\text{CO}_2 from O3=3×12 cm3=36 cm3\text{O}_3 = 3 \times 12\text{ cm}^3 = 36\text{ cm}^3.
Each molecule of O3\text{O}_3 contains 3 oxygen atoms, reacting with 3 molecules of CO\text{CO} to yield 3 molecules of CO2\text{CO}_2.
4
Sum the total volume of CO2\text{CO}_2 produced.
Total volume of CO2=144 cm3+36 cm3=180 cm3\text{CO}_2 = 144\text{ cm}^3 + 36\text{ cm}^3 = 180\text{ cm}^3.
The total volume of carbon(IV) oxide formed is the sum of the volumes produced by both oxygen allotropes in the mixture.

Key Concept

Gay-Lussac's Law of Combining Volumes and Allotropic Conversion of Oxygen to Ozone
Estimated Time:2m 0s
Question 8106Question

Match each oxygen-containing compound listed on the left with its corresponding chemical classification or characteristic reaction behavior on the right.

Click a left item, then click its matching right item

Items

P4O10\text{P}_4\text{O}_{10}
CaO\text{CaO}
CO\text{CO}
H2O2\text{H}_2\text{O}_2

Matches

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Answer

P4O10\text{P}_4\text{O}_{10} matches 'Acidic oxide that reacts with water to yield a triprotic acid'; CaO\text{CaO} matches 'Basic oxide that reacts exothermically with water to form an alkaline solution'; CO\text{CO} matches 'Neutral oxide that fails to form salts when exposed to acids or alkalis'; H2O2\text{H}_2\text{O}_2 matches 'Peroxide species that decomposes releasing oxygen gas'.
Each item matches its corresponding behavior based on oxide and oxygen compound classification: P4O10\text{P}_4\text{O}_{10} is an acidic oxide producing triprotic acid; CaO\text{CaO} is a basic oxide producing an alkaline hydroxide; CO\text{CO} is a neutral oxide; H2O2\text{H}_2\text{O}_2 is a peroxide species that liberates oxygen gas upon decomposition.

Step-by-Step Solution

1
Classify P4O10\text{P}_4\text{O}_{10} based on its reaction with water.
P4O10\text{P}_4\text{O}_{10} is a non-metal oxide (phosphorus(V) oxide) which reacts with water to yield H3PO4\text{H}_3\text{PO}_4, a triprotic acid.
Non-metal oxides in high oxidation states act as acid anhydrides.
2
Classify CaO\text{CaO} based on its acid-base character.
CaO\text{CaO} is an alkaline earth metal oxide that reacts exothermically with water to form the alkaline base Ca(OH)2\text{Ca(OH)}_2.
Metallic oxides typically display basic chemical behavior.
3
Determine the chemical reactivity of CO\text{CO}.
CO\text{CO} is a non-metal oxide that does not react with acids or alkalis to form salts, making it a neutral oxide.
Certain low-oxidation non-metal oxides are neutral.
4
Identify the nature of H2O2\text{H}_2\text{O}_2.
H2O2\text{H}_2\text{O}_2 contains the peroxide group with oxygen in the 1-1 oxidation state and decomposes to liberate oxygen gas.
Peroxides undergo decomposition to yield oxygen and water.

Key Concept

Classification of Oxides (Acidic, Basic, Neutral) and Peroxides
Question 8107Question

A major chlor-alkali chemical plant is sited near a coastal marine environment to optimize production efficiency. Which primary raw material is extracted at this location for the industrial electrolysis that yields sodium hydroxide, chlorine gas, and hydrogen gas?

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Answer: Sodium chloride (concentrated brine or sea salt)

Answer

Sodium chloride (concentrated brine or sea salt) is the primary raw material for the chlor-alkali industry.
The chlor-alkali industry relies on the electrolysis of concentrated sodium chloride solution (brine), which yields sodium hydroxide at the cathode/solution along with chlorine gas at the anode and hydrogen gas at the cathode. Siting these factories near marine coastal zones or salt deposits ensures a direct and abundant supply of the primary raw material, sodium chloride.

Step-by-Step Solution

1
Identify the chemical industry process specified in the prompt.
The process described is the chlor-alkali industry, which produces sodium hydroxide (NaOHNaOH), chlorine (Cl2Cl_2), and hydrogen (H2H_2).
Recognizing the end-products pinpoints the required industrial process.
2
Determine the necessary ionic starting material containing sodium and chloride ions.
Electrolysis of aqueous sodium chloride (NaClNaCl), commonly sourced from sea salt or brine deposits near coastal areas, provides Na+Na^+ and ClCl^- ions.
Chemical industries are frequently sited near their primary raw material sources to minimize raw material transport costs.

Key Concept

Chlor-alkali process raw materials and industrial siting factors
Question 8108Question

A myopic person has a far point of 52 cm52\text{ cm} from the eye. A corrective lens is placed 2 cm2\text{ cm} in front of the eye to enable the person to see distant objects clearly. A second thin converging lens of focal length +20 cm+20\text{ cm} is then placed in thin coaxial contact with this corrective lens. If an object is placed 50 cm50\text{ cm} in front of this combined lens system, what is the position and nature of the final image formed?

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Answer: 100 cm100\text{ cm} behind the combined lens (real)

Answer

100 cm100\text{ cm} behind the combined lens (real image)
The corrective lens for short-sightedness (myopia) must be a diverging lens with a negative focal length. Accounting for the 2 cm2\text{ cm} distance between the eye and the lens, the far point relative to the lens is 50 cm50\text{ cm}, giving f1=50 cmf_1 = -50\text{ cm}. Combining this lens with the converging lens (f2=+20 cmf_2 = +20\text{ cm}) yields a net power 1F=150+120=+3100 cm1\frac{1}{F} = -\frac{1}{50} + \frac{1}{20} = +\frac{3}{100}\text{ cm}^{-1}, or F=+1003 cmF = +\frac{100}{3}\text{ cm}. Placing an object at u=50 cmu = 50\text{ cm} gives 1v=3100150=+1100 cm1\frac{1}{v} = \frac{3}{100} - \frac{1}{50} = +\frac{1}{100}\text{ cm}^{-1}, which yields v=+100 cmv = +100\text{ cm}. The positive sign confirms a real image formed 100 cm100\text{ cm} behind the lens system.

Step-by-Step Solution

1
Calculate the focal length f1f_1 of the corrective lens required for myopia.
f1=50 cmf_1 = -50\text{ cm}
The far point distance from the lens is 52 cm2 cm=50 cm52\text{ cm} - 2\text{ cm} = 50\text{ cm}. A diverging (concave) lens is needed to form a virtual image of distant objects (u=u = \infty) at the far point (v=50 cmv = -50\text{ cm}).
2
Determine the focal length FF of the two lenses in thin contact.
F=+1003 cmF = +\frac{100}{3}\text{ cm}
Using the combined focal length equation 1F=1f1+1f2=150+120=2+5100=+3100 cm1\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = -\frac{1}{50} + \frac{1}{20} = \frac{-2 + 5}{100} = +\frac{3}{100}\text{ cm}^{-1}.
3
Apply the thin lens formula to locate the final image distance vv for u=50 cmu = 50\text{ cm}.
v=+100 cmv = +100\text{ cm}
Using 1F=1u+1v    3100=150+1v    1v=31002100=1100 cm1\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{50} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{2}{100} = \frac{1}{100}\text{ cm}^{-1}.
4
Determine the nature of the image from the sign of vv.
Real image formed 100 cm100\text{ cm} behind the combined lens system.
A positive value of image distance (v>0v > 0) indicates a real image formed on the opposite side (behind) the lens system.

Key Concept

Thin lens combination and sight defect correction sign conventions
Estimated Time:3m 0s
Question 8109Question

A short-sighted person cannot see objects clearly beyond a distance of 80 cm80\text{ cm}. What type of lens, of what focal length and power, is required to correct this vision defect so that the person can view distant objects clearly?

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Answer: Diverging lens of focal length 80 cm80\text{ cm} and power 1.25 D-1.25\text{ D}

Answer

Diverging lens of focal length 80 cm80\text{ cm} and power 1.25 D-1.25\text{ D}
For a myopic person with a far point at 80 cm80\text{ cm}, parallel rays from a distant object (u=u = \infty) must be diverged so they appear to come from the far point (v=80 cm=0.8 mv = -80\text{ cm} = -0.8\text{ m}). Using 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}, we obtain f=0.8 m=80 cmf = -0.8\text{ m} = -80\text{ cm}. The negative focal length corresponds to a diverging lens, and its power is P=10.8 m=1.25 DP = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}.

Step-by-Step Solution

1
Identify the optical requirements for correcting myopia (short-sightedness).
For a distant object (u=u = \infty), the corrective lens must form a virtual image at the eye's far point (v=80 cm=0.8 mv = -80\text{ cm} = -0.8\text{ m}).
Myopic eyes focus rays from infinity in front of the retina; placing a virtual image at the far point allows the eye lens to focus it correctly onto the retina.
2
Apply the thin lens formula to calculate the required focal length.
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{\infty} + \frac{1}{-0.8\text{ m}} = 0 - 1.25\text{ m}^{-1} \implies f = -0.8\text{ m} = -80\text{ cm}\).
The negative sign indicates that a concave (diverging) lens is required.
3
Calculate the power of the corrective lens in dioptres.
\(P = \frac{1}{f\text{ (in metres)}} = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}\).
Lens power in dioptres (D) is the reciprocal of the focal length expressed in meters.

Key Concept

Correction of Myopia (Short-Sightedness) using Diverging Lenses
Question 8110Question

A student heats a mixture of iron filings and sulfur powder in a test tube until a black solid, iron(II) sulfide, is formed. What property of the resulting black solid confirms that a compound, rather than a mixture, has been produced?

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Answer: It possesses chemical properties entirely different from its constituents and cannot be separated by physical methods.

Answer

The black solid (iron(II) sulfide) is a compound because it possesses distinct chemical properties from iron and sulfur and cannot be separated into its constituent elements by physical methods.
When iron and sulfur chemically react upon heating, iron(II) sulfide (FeSFeS) is formed. Compounds have entirely different properties from their constituent elements and cannot be separated into those elements by physical methods such as using a magnet.

Step-by-Step Solution

1
Analyze the difference between physical mixing and chemical combination of iron and sulfur.
Before heating, iron filings and sulfur form a mixture where iron remains magnetic and sulfur remains soluble in carbon disulfide.
Components in a mixture retain their individual identity and physical properties.
2
Evaluate the effect of heating the mixture to form iron(II) sulfide (FeSFeS).
Heating causes a chemical reaction forming FeSFeS, a compound with a fixed ratio of elements, new properties, and no magnetic attraction for iron.
Chemical bonds are formed in compounds, altering constituent properties completely and preventing physical separation.

Key Concept

Distinction between compounds and mixtures based on physical separation and retention of properties
Estimated Time:50s
Question 8111Question

Arrange the following regions of the electromagnetic spectrum in order of decreasing wavelength (from longest wavelength to shortest wavelength).

Drag items to arrange them in the correct order

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Answer

The correct sequence from longest to shortest wavelength is: Microwaves, Infrared radiation, Ultraviolet radiation, and Gamma rays.
The electromagnetic spectrum ordered by decreasing wavelength (longest to shortest) follows the sequence: radio waves/microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Thus, microwaves come first with the longest wavelength, followed by infrared, ultraviolet, and finally gamma rays with the shortest wavelength.

Step-by-Step Solution

1
Recall the wave equation c=fλc = f \lambda connecting frequency (ff) and wavelength (λ\lambda) for electromagnetic waves in a vacuum.
Wavelength is inversely proportional to frequency and photon energy.
Since the speed of light cc is constant, waves with lower frequencies have longer wavelengths.
2
Identify the relative wavelengths of each specified region of the electromagnetic spectrum.
Microwaves (103 m101 m10^{-3}\text{ m} - 10^{-1}\text{ m}) > Infrared (7×107 m103 m7 \times 10^{-7}\text{ m} - 10^{-3}\text{ m}) > Ultraviolet (108 m4×107 m10^{-8}\text{ m} - 4 \times 10^{-7}\text{ m}) > Gamma rays (<1011 m< 10^{-11}\text{ m}).
Microwaves sit near the radio end of the spectrum, while gamma rays lie at the extreme high-energy end.
3
Sequence the items from longest wavelength to shortest wavelength.
Microwaves \rightarrow Infrared radiation \rightarrow Ultraviolet radiation \rightarrow Gamma rays.
This arranges the waves in strict order of decreasing wavelength.

Key Concept

Electromagnetic spectrum wavelength and frequency hierarchy
Question 8112Question

An RLC series circuit operating at resonance contains an inductor of inductance 0.1 H0.1\text{ H} and a capacitor of capacitance 10 μF10\ \mu\text{F}. What is the resonant angular frequency of the circuit in radians per second (rad/s\text{rad/s})?

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Answer: 1000

Answer

The resonant angular frequency of the circuit is 1000 rad/s1000\text{ rad/s}.
The resonant angular frequency ω0\omega_0 of an AC circuit is determined by the formula ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}. Substituting the given values L=0.1 HL = 0.1\text{ H} and C=105 FC = 10^{-5}\text{ F} gives ω0=1106=1000 rad/s\omega_0 = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}.

Step-by-Step Solution

1
Convert given parameters to standard SI units
L=0.1 HL = 0.1\text{ H} and C=10×106 F=105 FC = 10 \times 10^{-6}\text{ F} = 10^{-5}\text{ F}.
Calculations must be performed in base SI units for dimensional consistency.
2
Calculate the resonant angular frequency using ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}
ω0=10.1×105=1106=1000 rad/s\omega_0 = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}.
At resonance, inductive reactance equals capacitive reactance (XL=XCX_L = X_C), which yields ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}.

Key Concept

Resonant Angular Frequency
Question 8113Question

A U-tube open to the atmosphere at both ends contains water of density 1000 kg m31000\text{ kg m}^{-3}. An immiscible liquid is poured into one arm of the tube until it forms a column of height 15.0 cm15.0\text{ cm}. If the interface between the two liquids lies 12.0 cm12.0\text{ cm} below the surface of the water in the opposite arm, what is the density of the liquid in kg m3\text{kg m}^{-3}?

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Answer: 800

Answer

The density of the immiscible liquid is 800 kg m3800\text{ kg m}^{-3}.
At the level of the liquid interface, the hydrostatic pressure exerted by the 15.0 cm15.0\text{ cm} liquid column must equal the hydrostatic pressure exerted by the 12.0 cm12.0\text{ cm} water column above it. Equating ρ1h1=ρ2h2\rho_1 h_1 = \rho_2 h_2 yields ρliquid=1000×(12.0/15.0)=800 kg m3\rho_{\text{liquid}} = 1000 \times (12.0 / 15.0) = 800\text{ kg m}^{-3}.

Step-by-Step Solution

1
Apply the principal of equal pressure at the same horizontal level in a continuous fluid at rest.
Pressure due to liquid column equals pressure due to water column above the interface line.
Hydrostatic pressure at depth hh is given by P=ρghP = \rho g h, and points at equal depth in a connected liquid body share identical pressure.
2
Set up the density-height ratio relationship: ρliquidhliquid=ρwaterhwater\rho_{\text{liquid}} \cdot h_{\text{liquid}} = \rho_{\text{water}} \cdot h_{\text{water}}.
ρliquid=ρwater×hwaterhliquid\rho_{\text{liquid}} = \rho_{\text{water}} \times \frac{h_{\text{water}}}{h_{\text{liquid}}}.
Acceleration due to gravity gg cancels out from both sides of the pressure balance equation.
3
Substitute hliquid=15.0 cmh_{\text{liquid}} = 15.0\text{ cm}, hwater=12.0 cmh_{\text{water}} = 12.0\text{ cm}, and ρwater=1000 kg m3\rho_{\text{water}} = 1000\text{ kg m}^{-3}.
ρliquid=1000×12.015.0=800 kg m3\rho_{\text{liquid}} = 1000 \times \frac{12.0}{15.0} = 800\text{ kg m}^{-3}.
Heights can remain in centimeters since their unit ratio is dimensionless.

Key Concept

Hydrostatic Pressure Balance in U-Tube Manometers
Estimated Time:1m 30s
Question 8114Question

Pair each of the chemical transformation descriptions listed on the left with its corresponding classical or modern redox definition concept on the right.

Click a left item, then click its matching right item

Items

Loss of electrons by an atom or ion during a chemical reaction
Removal of hydrogen from a chemical compound
Decrease in the oxidation number of an element
Addition of oxygen to an element or compound

Matches

Show answer & explanation

Answer

Loss of electrons corresponds to the modern electronic concept of oxidation; removal of hydrogen corresponds to the classical concept of oxidation via hydrogen transfer; decrease in oxidation number corresponds to the modern oxidation-number concept of reduction; addition of oxygen corresponds to the classical concept of oxidation via oxygen transfer.
The matching pairs correctly connect historical definitions (oxygen addition and hydrogen removal) and modern definitions (electron loss and oxidation number reduction) to their corresponding oxidation or reduction classifications.

Step-by-Step Solution

1
Differentiate classical definitions from modern redox concepts.
Classical concepts focus on oxygen and hydrogen transfer, whereas modern concepts focus on electron transfer and changes in oxidation state.
Historical definitions arose before the electron was discovered, while modern definitions generalize redox processes to all chemical species.
2
Match electron transfer and oxidation state statements.
Loss of electrons is modern oxidation, and a decrease in oxidation state is modern reduction.
Oxidation increases charge/oxidation state via electron loss; reduction decreases charge/oxidation state via electron gain.
3
Match hydrogen and oxygen transfer statements.
Removal of hydrogen is classical oxidation, and addition of oxygen is classical oxidation.
Classical oxidation is defined by adding oxygen or removing hydrogen from a substance.

Key Concept

Distinguishing classical (oxygen/hydrogen transfer) from modern (electron transfer / oxidation state) definitions of oxidation and reduction.
Estimated Time:45s
Question 8115Question

An atom of potassium is represented by the nuclide symbol 1939K^{39}_{19}\text{K}. How many neutrons are present in the nucleus of this atom?

Show answer & explanation

Answer: 20

Answer

The nucleus of the potassium atom contains 20 neutrons.
The correct answer is 20 because the number of neutrons in an atom is found by subtracting the atomic number (19) from the mass number (39), giving 3919=2039 - 19 = 20.

Step-by-Step Solution

1
Identify the mass number (AA) and atomic number (ZZ) from the given nuclide notation 1939K^{39}_{19}\text{K}.
Mass number A=39A = 39, Atomic number Z=19Z = 19.
In standard nuclide notation ZAX^{A}_{Z}\text{X}, the superscript represents the mass number and the subscript represents the atomic number.
2
Calculate the number of neutrons (NN) using the relationship N=AZN = A - Z.
N=3919=20N = 39 - 19 = 20.
The mass number is the sum of protons and neutrons (A=Z+NA = Z + N), so subtracting the atomic number (protons) from the mass number gives the neutron count.

Key Concept

Mass Number and Subatomic Particles
Estimated Time:45s
Question 8116Question

The standard enthalpy of formation (ΔHf\Delta H_f^\circ) of carbon dioxide gas (CO2(g)\text{CO}_2(g)) is numerically identical to the standard enthalpy of combustion (ΔHc\Delta H_c^\circ) of diamond under standard thermochemical conditions (298 K298\text{ K} and 1 atm1\text{ atm}).

Show answer & explanation

Answer: False

Answer

False
The standard enthalpy of formation (ΔHf\Delta H_f^\circ) of a compound is defined as the enthalpy change when one mole of the substance is formed from its constituent elements in their reference standard states. For carbon at 298 K298\text{ K} and 1 atm1\text{ atm}, the reference standard state is graphite. Consequently, ΔHf[CO2(g)]\Delta H_f^\circ[\text{CO}_2(g)] represents the reaction of graphite with oxygen. In contrast, the combustion of diamond involves C(s,diamond)\text{C}(s, \text{diamond}). Because conversion of graphite to diamond is endothermic (ΔHf[diamond]+1.9 kJ mol1\Delta H_f^\circ[\text{diamond}] \approx +1.9\text{ kJ mol}^{-1}), the combustion of diamond produces more heat than the formation of CO2(g)\text{CO}_2(g) from graphite. Thus, the two values are not numerically identical.

Step-by-Step Solution

1
Define the reaction equation for the standard enthalpy of formation (ΔHf\Delta H_f^\circ) of CO2(g)\text{CO}_2(g).
C(s,graphite)+O2(g)CO2(g)ΔH1=ΔHf[CO2(g)]\text{C}(s, \text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1^\circ = \Delta H_f^\circ[\text{CO}_2(g)]
By international convention (IUPAC), the standard enthalpy of formation requires elements to be in their most stable allotropic form at 298 K298\text{ K} and 1 atm1\text{ atm}, which is graphite for carbon.
2
Define the reaction equation for the standard enthalpy of combustion (ΔHc\Delta H_c^\circ) of diamond.
C(s,diamond)+O2(g)CO2(g)ΔH2=ΔHc[diamond]\text{C}(s, \text{diamond}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2^\circ = \Delta H_c^\circ[\text{diamond}]
The standard enthalpy of combustion of diamond measures the enthalpy change when 1 mole of diamond completely reacts with oxygen.
3
Apply Hess's law to relate ΔH1\Delta H_1^\circ and ΔH2\Delta H_2^\circ.
ΔHc[diamond]=ΔHf[CO2(g)]ΔHf[diamond]\Delta H_c^\circ[\text{diamond}] = \Delta H_f^\circ[\text{CO}_2(g)] - \Delta H_f^\circ[\text{diamond}]
Since diamond is less stable than graphite (ΔHf[diamond]+1.9 kJ mol1\Delta H_f^\circ[\text{diamond}] \approx +1.9\text{ kJ mol}^{-1}), ΔH2ΔH1\Delta H_2^\circ \neq \Delta H_1^\circ.

Key Concept

Standard State Conventions and Allotropy in Hess's Law
Question 8117Question

A 10 dm310\text{ dm}^3 sample of well water contains 0.002 mol dm30.002\text{ mol dm}^{-3} of dissolved calcium hydrogentrioxocarbonate(IV), Ca(HCO3)2\text{Ca(HCO}_3)_2, and 0.003 mol dm30.003\text{ mol dm}^{-3} of dissolved calcium tetraoxosulfate(VI), CaSO4\text{CaSO}_4. A student boils the water sample thoroughly to soften it and filters off any precipitate formed. What is the molar concentration of Ca2+\text{Ca}^{2+} ions remaining in the filtered solution?

Show answer & explanation

Answer: 0.003 mol dm30.003\text{ mol dm}^{-3}

Answer

0.003 mol dm30.003\text{ mol dm}^{-3}
Boiling converts soluble Ca(HCO3)2\text{Ca(HCO}_3)_2 into insoluble CaCO3\text{CaCO}_3, which is removed by filtration. Since CaSO4\text{CaSO}_4 causes permanent hardness and does not decompose upon boiling, its concentration of 0.003 mol dm30.003\text{ mol dm}^{-3} remains dissolved in the filtered water.

Step-by-Step Solution

1
Identify the types of water hardness present in the sample.
Dissolved Ca(HCO3)2\text{Ca(HCO}_3)_2 causes temporary hardness, while dissolved CaSO4\text{CaSO}_4 causes permanent hardness.
Hydrogentrioxocarbonates of calcium and magnesium cause temporary hardness, whereas tetraoxosulfates and chlorides cause permanent hardness.
2
Determine the chemical effect of boiling on temporary hardness.
Thermal decomposition occurs: Ca(HCO3)2(aq)ΔCaCO3(s)+H2O(l)+CO2(g)\text{Ca(HCO}_3)_2\text{(aq)} \xrightarrow{\Delta} \text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}.
Soluble calcium hydrogentrioxocarbonate(IV) decomposes on heating into insoluble calcium trioxocarbonate(IV), which precipitates out of solution.
3
Determine the chemical effect of boiling on permanent hardness.
CaSO4\text{CaSO}_4 remains dissolved and unaffected by boiling.
Permanent hardness cannot be removed by boiling because calcium tetraoxosulfate(VI) does not undergo thermal decomposition at 100C100^\circ\text{C}.
4
Calculate the residual concentration of Ca2+\text{Ca}^{2+} ions after filtration.
[Ca2+]=0.003 mol dm3[\text{Ca}^{2+}] = 0.003\text{ mol dm}^{-3}.
All 0.002 mol dm30.002\text{ mol dm}^{-3} of Ca2+\text{Ca}^{2+} from temporary hardness is precipitated and filtered out, leaving only the 0.003 mol dm30.003\text{ mol dm}^{-3} of Ca2+\text{Ca}^{2+} contributed by CaSO4\text{CaSO}_4.

Key Concept

Distinction between temporary and permanent water hardness and their behavior upon thermal treatment (boiling).
Question 8118Question

A cylindrical copper rod of thermal conductivity 380 Wm1K1380\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, length 0.40 m0.40\text{ m}, and uniform radius 2.0 cm2.0\text{ cm} is thermally insulated along its curved surface. One flat end is maintained at a temperature of 100C100^\circ\text{C} by steam, while the opposite end is kept in an ice bath at 0C0^\circ\text{C}. Assuming steady-state heat conduction, what is the rate of heat flow through the rod? (Take π=3.14\pi = 3.14)

Show answer & explanation

Answer: 119.3 W119.3\text{ W}

Answer

The rate of heat flow through the copper rod is 119.3 W119.3\text{ W}.
According to Fourier's law of heat conduction, the rate of thermal energy transfer Qt\frac{Q}{t} through a material of thermal conductivity kk, cross-sectional area AA, and length dd across temperature difference ΔT\Delta T is given by Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}. Substituting k=380 Wm1K1k = 380\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, A=π(0.02 m)2=1.256×103 m2A = \pi (0.02\text{ m})^2 = 1.256 \times 10^{-3}\text{ m}^2, ΔT=100 K\Delta T = 100\text{ K}, and d=0.40 md = 0.40\text{ m} yields 119.3 W119.3\text{ W}.

Step-by-Step Solution

1
Convert the radius from centimeters to meters and calculate the cross-sectional area of the rod.
r=2.0 cm=0.02 mr = 2.0\text{ cm} = 0.02\text{ m}. A=πr2=3.14×(0.02 m)2=1.256×103 m2A = \pi r^2 = 3.14 \times (0.02\text{ m})^2 = 1.256 \times 10^{-3}\text{ m}^2.
Fourier's law requires the area in square meters (m2m^2).
2
Determine the temperature difference across the rod.
ΔT=100C0C=100 K\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100\text{ K}.
Heat transfer rate depends on the temperature gradient along the length.
3
Apply Fourier's law of thermal conduction: Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}.
Qt=380×1.256×103×1000.40=119.32 W119.3 W\frac{Q}{t} = \frac{380 \times 1.256 \times 10^{-3} \times 100}{0.40} = 119.32\text{ W} \approx 119.3\text{ W}.
Substituting the thermal conductivity kk, cross-sectional area AA, temperature difference ΔT\Delta T, and length dd gives the steady-state heat conduction rate.

Key Concept

Fourier's Law of Heat Conduction in solids: Qt=kA(ThotTcold)d\frac{Q}{t} = \frac{k A (T_{\text{hot}} - T_{\text{cold}})}{d}
Estimated Time:2m 0s
Question 8119Question

Two rectangular slabs of equal thickness dd are mounted in parallel between a hot reservoir at 100C100^\circ\text{C} and a cold reservoir at 20C20^\circ\text{C}. Slab 1 has a thermal conductivity k1=300 Wm1K1k_1 = 300\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1} and a cross-sectional area A1=4.0×104 m2A_1 = 4.0 \times 10^{-4}\text{ m}^2. Slab 2 has a thermal conductivity k2=100 Wm1K1k_2 = 100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1} and a cross-sectional area A2=6.0×104 m2A_2 = 6.0 \times 10^{-4}\text{ m}^2. Assuming steady-state heat conduction and no lateral heat loss, what percentage of the total heat transferred per second between the reservoirs conducts through Slab 1?

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Answer: 66.7%66.7\%

Answer

The percentage of the total heat transferred per second conducted through Slab 1 is 66.7%.
Fourier's law gives the rate of heat conduction as P=kAΔTdP = \frac{k A \Delta T}{d}. Since the temperature gradient ΔTd\frac{\Delta T}{d} is identical across both parallel slabs, the heat current through each slab is directly proportional to its kAk A product. For Slab 1, k1A1=300×4.0×104=0.12k_1 A_1 = 300 \times 4.0 \times 10^{-4} = 0.12, while for Slab 2, k2A2=100×6.0×104=0.06k_2 A_2 = 100 \times 6.0 \times 10^{-4} = 0.06. The total heat current is proportional to 0.12+0.06=0.180.12 + 0.06 = 0.18. Thus, the fraction conducted through Slab 1 is 0.120.18=23\frac{0.12}{0.18} = \frac{2}{3}, which corresponds to 66.7%66.7\%.

Step-by-Step Solution

1
Express Fourier's law of heat conduction for each slab in parallel.
Rate of heat flow P1=k1A1ΔTdP_1 = \frac{k_1 A_1 \Delta T}{d} and P2=k2A2ΔTdP_2 = \frac{k_2 A_2 \Delta T}{d}.
Both slabs experience the same temperature difference ΔT=100C20C=80 K\Delta T = 100^\circ\text{C} - 20^\circ\text{C} = 80\text{ K} and have equal length dd.
2
Calculate the effective conductance factors kAk A for both slabs.
k1A1=300×(4.0×104)=0.12 WmK1k_1 A_1 = 300 \times (4.0 \times 10^{-4}) = 0.12\text{ W}\cdot\text{m}\cdot\text{K}^{-1} and k2A2=100×(6.0×104)=0.06 WmK1k_2 A_2 = 100 \times (6.0 \times 10^{-4}) = 0.06\text{ W}\cdot\text{m}\cdot\text{K}^{-1}.
Since ΔT/d\Delta T / d is identical for both parallel paths, the heat flow rate is directly proportional to kAk A.
3
Determine the total heat flow rate and the percentage carried by Slab 1.
Ptotal=P1+P20.12+0.06=0.18P_{\text{total}} = P_1 + P_2 \propto 0.12 + 0.06 = 0.18. Percentage through Slab 1 =0.120.18×100%=66.7%= \frac{0.12}{0.18} \times 100\% = 66.7\%.
In parallel conduction, individual heat currents add to form the total heat current.

Key Concept

Parallel Thermal Conduction
Question 8120Question

Which of the following oxides reacts with both dilute hydrochloric acid and aqueous sodium hydroxide to form salt and water?

Show answer & explanation

Answer: Zinc oxide (ZnO\text{ZnO})

Answer

Zinc oxide (ZnO\text{ZnO}) is the correct answer because it is an amphoteric oxide.
Zinc oxide (ZnO\text{ZnO}) is an amphoteric oxide. It exhibits both basic and acidic properties, reacting with acids like hydrochloric acid to form zinc chloride and with strong bases like sodium hydroxide to form sodium zincate.

Step-by-Step Solution

1
Classify the given oxides based on their acid-base behavior
Calcium oxide is basic, Sulfur(IV) oxide is acidic, Carbon(II) oxide is neutral, and Zinc oxide is amphoteric.
Amphoteric oxides exhibit dual properties, reacting with both acids and bases.
2
Identify the oxide that reacts with both dilute hydrochloric acid and sodium hydroxide
Zinc oxide reacts with HCl\text{HCl} to form zinc chloride and water, and with NaOH\text{NaOH} to form sodium zincate and water.
Only amphoteric oxides (such as ZnO\text{ZnO}, Al2O3\text{Al}_2\text{O}_3, and PbO\text{PbO}) form salts with both acids and strong alkalis.

Key Concept

Classification of Oxides (Amphoteric Oxides)
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