Tüm alıştırma soruları

Soru 1321Soru →

Match each solubility phenomenon or term on the left with its correct thermodynamic or practical description on the right. Which pair correctly connects each term to its appropriate description?

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Fractional crystallization pairs with separation technique using solubility differences; Endothermic dissolution pairs with heat absorption increasing solubility with temperature; Exothermic dissolution pairs with heat evolution decreasing solubility with temperature; Supersaturated solution pairs with holding excess dissolved solute beyond equilibrium limit.

Each term directly aligns with its fundamental thermodynamic property or chemical application. Fractional crystallization isolates salts using temperature-dependent solubility variations; endothermic dissolution absorbs heat so solubility increases with temperature; exothermic dissolution gives off heat so solubility decreases with temperature; and a supersaturated solution holds excess dissolved solute beyond standard equilibrium capacity.

Adım Adım Çözüm

1

Identify the separation technique based on solubility curve variance.

Fractional crystallization separates solutes based on their distinct solubility curves across temperatures.

Differences in solubility curves allow one component to crystallize out of solution before another when cooled.

2

Analyze how enthalpy changes govern temperature dependence on solubility.

Endothermic processes absorb heat leading to increased solubility with temperature rise, whereas exothermic processes release heat leading to decreased solubility with temperature rise.

According to Le Chatelier's principle, adding heat favors the endothermic direction of a solution equilibrium.

3

Determine the saturation state definition for excess solute concentration.

A supersaturated solution contains a higher concentration of dissolved solute than a saturated solution at the specified temperature.

It represents a metastable condition created by careful cooling without crystallization.

Anahtar Kavram

Temperature effects on solubility, enthalpy of solution, and fractional crystallization

Soru 1322Soru →

Match each of the following chemical species on the left with its corresponding central atom hybridization state and molecular geometry on the right, as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. Which set of pairs accurately connects each species to its geometry and hybridization?

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Iodine trifluoride (

IF_3

) matches with

sp^3d

hybridization and T-shaped geometry; Pentafluoroxenate ion (

XeF_5^-

) matches with

sp^3d^3

hybridization and pentagonal planar geometry; Sulfite ion (

SO_3^{2-}

) matches with

sp^3

hybridization and trigonal pyramidal geometry; Nitronium ion (

NO_2^+

) matches with

sp

hybridization and linear geometry.

Each chemical species is correctly paired according to its total steric number (sum of bonding electron domains and non-bonding lone pairs).

IF_3

has steric number 5 (

sp^3d

, T-shaped),

XeF_5^-

has steric number 7 (

sp^3d^3

, pentagonal planar),

SO_3^{2-}

has steric number 4 (

sp^3

, trigonal pyramidal), and

NO_2^+

has steric number 2 (

sp

, linear).

Adım Adım Çözüm

1

Determine valence electron count and steric number for

IF_3

Iodine has 7 valence electrons + 3 from fluorine = 10 electrons (5 pairs). Steric number = 5 (3 bonding pairs, 2 lone pairs).

Steric number 5 corresponds to

sp^3d

hybridization. Two equatorial lone pairs force the 3 terminal fluorines into a T-shaped arrangement.

2

Determine valence electron count and steric number for

XeF_5^-

Xenon has 8 valence electrons + 5 from fluorine + 1 from overall negative charge = 14 electrons (7 pairs). Steric number = 7 (5 bonding pairs, 2 lone pairs).

Steric number 7 corresponds to

sp^3d^3

hybridization. The two lone pairs occupy axial positions above and below the equatorial plane, creating a pentagonal planar molecular shape.

3

Determine valence electron count and steric number for

SO_3^{2-}

Sulfur has 6 valence electrons + 2 from charge = 8 valence shell electrons. It forms 3 sigma bonds with oxygen and retains 1 lone pair. Steric number = 4.

Steric number 4 corresponds to

sp^3

hybridization. Three bonding domains and 1 lone pair produce a trigonal pyramidal molecular shape.

4

Determine valence electron count and steric number for

NO_2^+

Nitrogen has 5 valence electrons - 1 from positive charge = 4 electrons. It forms two double bonds with oxygen atoms and has 0 lone pairs. Steric number = 2.

Steric number 2 corresponds to

sp

hybridization, resulting in a linear geometry with a

180^\circ

bond angle.

Anahtar Kavram

VSEPR Theory, Steric Numbers, and Central Atom Hybridization States

Tahmini Süre:2m 30s

Soru 1323Soru →

Match each thermodynamic condition or combination on the left with its corresponding reaction spontaneity description on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

\Delta G < 0

matches with 'The reaction is spontaneous under the given conditions';

\Delta G = 0

matches with 'The system has reached dynamic equilibrium';

\Delta H < 0

and

\Delta S > 0

matches with 'The reaction is spontaneous at all temperatures';

\Delta H > 0

and

\Delta S < 0

matches with 'The reaction is non-spontaneous at all temperatures'.

Matching

\Delta G < 0

to spontaneity under specified conditions,

\Delta G = 0

to dynamic equilibrium,

\Delta H < 0

and

\Delta S > 0

to spontaneity at all temperatures, and

\Delta H > 0

and

\Delta S < 0

to non-spontaneity at all temperatures follows directly from the Gibbs-Helmholtz relation

\Delta G = \Delta H - T\Delta S

.

Adım Adım Çözüm

1

Recall the fundamental thermodynamic criterion for spontaneity involving Gibbs free energy change

(\Delta G)

.

A reaction is spontaneous when

\Delta G < 0

, non-spontaneous when

\Delta G > 0

, and at dynamic equilibrium when

\Delta G = 0

.

Gibbs free energy combines enthalpy and entropy factors to determine direction of feasible chemical changes.

2

Analyze the Gibbs-Helmholtz equation

\Delta G = \Delta H - T\Delta S

for sign combinations.

If

\Delta H < 0

and

\Delta S > 0

,

\Delta G = (-\text{value}) - T(+\text{value})

, which is always negative at any absolute temperature

T > 0\text{ K}

.

Exothermic enthalpy releases energy while positive entropy increases disorder, driving spontaneity unconditionally.

3

Evaluate the opposing sign combination where

\Delta H > 0

and

\Delta S < 0

.

\Delta G = (+\text{value}) - T(-\text{value}) = +\text{value} + T(\text{value})

, which is always positive.

Endothermic process with decreasing entropy is thermodynamically unfavorable at all temperatures.

Anahtar Kavram

Gibbs Free Energy Equation and Reaction Spontaneity

Soru 1324Soru →

Match each water treatment chemical or industrial effluent contaminant on the left with its correct chemical action or environmental impact on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The correct pairings connect calcium hydroxide and sodium carbonate to the precipitation of hardness-causing cations; non-biodegradable detergents to persistent foam formation that inhibits aeration; high BOD effluent to microbial oxygen depletion; and heavy metal ions to trophic bioaccumulation and neurological toxicity.

The correct pairings accurately link chemical reagents (

Ca(OH)_2, Na_2CO_3

) to precipitation of hardness-causing metal ions, non-biodegradable synthetic detergents to persistent surface foam formation, high BOD organic waste to microbial depletion of dissolved oxygen, and heavy metal ions (

Pb^{2+}, Hg^{2+}

) to biological biomagnification.

Adım Adım Çözüm

1

Analyze the chemical function of lime and soda ash in water treatment

Calcium hydroxide (

Ca(OH)_2

) converts soluble

Ca(HCO_3)_2

into insoluble

CaCO_3

, and sodium carbonate (

Na_2CO_3

) precipitates

Ca^{2+}

and

Mg^{2+}

ions as insoluble carbonates (

CaCO_3

and

MgCO_3

).

Chemical precipitation using lime-soda softening removes both temporary and permanent water hardness.

2

Evaluate the environmental impact of branched alkylbenzene sulfonate detergents

Because soil and aquatic microbes lack enzymes to degrade branched-chain alkylbenzene sulfonates, these surfactants accumulate as stable surface foams.

Surface foam restricts light transmission required for photosynthesis and reduces re-aeration at the water-air boundary.

3

Relate Biochemical Oxygen Demand (BOD) to organic pollution effects

High BOD signifies extensive organic pollution, fueling exponential growth of decomposer bacteria that rapidly absorb dissolved oxygen.

Aerobic respiration by decomposers lowers the dissolved oxygen concentration, creating hypoxic aquatic conditions.

4

Assess the biological toxicity of heavy metal effluents

Heavy metal ions such as

Pb^{2+}

and

Hg^{2+}

are non-biodegradable toxins that accumulate in fatty tissues and biomagnify up food chains.

Inhibition of key metabolic enzymes by heavy metals causes chronic neurological and physiological damage.

Anahtar Kavram

Chemical Methods of Water Softening, Effluent Pollution Mechanisms, and Ecological Impacts

Soru 1325Soru →

Match each aqueous salt solution to its characteristic effect on litmus paper at $25^\circ\text{C}$ .

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Ammonium chloride solution (

NH_4Cl(aq)

) matches with 'Turns blue litmus paper red (

pH < 7

)'; Sodium ethanoate solution (

CH_3COONa(aq)

) matches with 'Turns red litmus paper blue (

pH > 7

)'; Sodium chloride solution (

NaCl(aq)

) matches with 'Has no effect on either red or blue litmus paper (

pH = 7

)'.

Salt hydrolysis determines the acidity or alkalinity of an aqueous salt solution based on the strengths of the parent acid and base. Ammonium chloride (

NH_4Cl

) yields acidic solutions (

pH < 7

) turning blue litmus red due to

NH_4^+

cation hydrolysis. Sodium ethanoate (

CH_3COONa

) produces alkaline solutions (

pH > 7

) turning red litmus blue due to

CH_3COO^-

anion hydrolysis. Sodium chloride (

NaCl

) consists of spectator ions from a strong acid and strong base, undergoing no hydrolysis and remaining neutral (

pH = 7

).

Adım Adım Çözüm

1

Analyze the parent acid and base for Ammonium chloride (

NH_4Cl

).

NH_4Cl

forms from

HCl

(strong acid) and

NH_3

(weak base). Cation hydrolysis occurs:

NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)

. Excess

H_3O^+

turns blue litmus red.

Salts of strong acids and weak bases yield acidic solutions.

2

Analyze the parent acid and base for Sodium ethanoate (

CH_3COONa

).

CH_3COONa

forms from

CH_3COOH

(weak acid) and

NaOH

(strong base). Anion hydrolysis occurs:

CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)

. Excess

OH^-

turns red litmus blue.

Salts of weak acids and strong bases yield alkaline solutions.

3

Analyze the parent acid and base for Sodium chloride (

NaCl

).

NaCl

forms from

HCl

(strong acid) and

NaOH

(strong base). Neither ion undergoes hydrolysis. The solution remains neutral (

pH = 7

) and does not change litmus color.

Salts of strong acids and strong bases do not undergo hydrolysis.

Anahtar Kavram

Salt Hydrolysis and Solution Acidity/Alkalinity

Tahmini Süre:45s

Soru 1326Soru →

Match each chemical system in Column A with its correct outcome and rationale based on the electrochemical series in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Potassium with zinc sulfate matches spontaneous displacement due to a more negative reduction potential; Copper with hydrochloric acid matches no reaction due to a positive reduction potential relative to hydrogen; Fluorine with sodium chloride matches spontaneous halide oxidation due to a higher reduction potential; Silver with gold(III) nitrate matches spontaneous metal displacement because silver acts as a stronger reducing agent than gold.

Each chemical system correctly pairs with its electrochemical behavior: Potassium displaces Zinc because it possesses a more negative reduction potential; Copper does not react with dilute acid because its reduction potential is positive relative to hydrogen; Fluorine displaces chloride ions because it has a higher reduction potential and thus greater oxidizing power; Silver displaces gold ions because silver has a lower reduction potential than gold, making it the stronger reducing agent.

Adım Adım Çözüm

1

Analyze position of Potassium and Zinc in the electrochemical series.

Potassium (

E^\circ = -2.93\text{ V}

) has a more negative reduction potential than Zinc (

E^\circ = -0.76\text{ V}

), making it a stronger reducing agent.

Metals with more negative standard reduction potentials spontaneously displace ions of metals below them in the series.

2

Evaluate the reactivity of Copper in non-oxidizing acid.

Copper (

E^\circ = +0.34\text{ V}

) lies below Hydrogen (

E^\circ = 0.00\text{ V}

) in the electrochemical series.

Metals with positive reduction potentials cannot spontaneously reduce hydrogen ions to evolve

\text{H}_2

gas.

3

Compare oxidizing strengths of Fluorine and Chlorine.

Fluorine (

E^\circ = +2.87\text{ V}

) has a higher reduction potential than Chlorine (

E^\circ = +1.36\text{ V}

).

A halogen with a higher reduction potential acts as a stronger oxidizing agent and displaces halide ions with lower reduction potentials.

4

Determine feasibility of displacement between Silver and Gold ions.

Silver (

E^\circ = +0.80\text{ V}

) is more easily oxidized than Gold (

E^\circ = +1.50\text{ V}

).

The metal with the smaller reduction potential acts as the reducing agent, resulting in a positive standard cell potential (

E^\circ_{\text{cell}} > 0

).

Anahtar Kavram

Electrochemical Series and Reaction Spontaneity

Tahmini Süre:1m 30s

Soru 1327Soru →

Match each environmental contaminant or waste management process in Column A with its corresponding chemical characteristic or primary effect in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Lead (

\text{Pb}

) matches with 'Toxic heavy metal from battery manufacturing and old paints causing neurological harm'. DDT (Organochlorine) matches with 'Persistent synthetic pesticide that bioaccumulates up the agricultural food chain'. Incineration matches with 'High-temperature thermal waste disposal process that converts combustible refuse into ash'.

Each item correctly matches its specific chemical, environmental, or waste management role: Lead is a heavy metal associated with battery waste and nervous system toxicity; DDT is a non-biodegradable pesticide causing bioaccumulation; Incineration is thermal combustion reducing waste to ash.

Adım Adım Çözüm

1

Identify the primary environmental impact of Lead (

\text{Pb}

).

Lead is a non-biodegradable heavy metal commonly sourced from lead-acid batteries and paints, causing neurological dysfunction.

Heavy metal contamination in soil primarily stems from industrial effluents and battery production.

2

Identify the characteristic behavior of DDT in soil and ecosystems.

DDT is an organochlorine pesticide notorious for environmental persistence and fat-solubility, causing bioaccumulation in higher trophic levels.

Organochlorine pesticides resist rapid chemical and biological decomposition.

3

Identify the operational principle of waste Incineration.

Incineration thermal treatment burns solid waste at high temperatures, turning organic matter into gaseous products and non-combustible ash.

It is a thermal waste volume reduction method commonly employed in municipal waste management.

Anahtar Kavram

Soil Pollution, Heavy Metal Toxicity, Pesticide Persistence, and Waste Disposal Methods

Soru 1328Soru →

Match each chemical system modification on the left with the primary kinetic mechanism on the right that accounts for the observed increase in reaction rate.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Ultraviolet irradiation corresponds to supplying photon energy for bond cleavage and free-radical generation; pulverizing calcium carbonate corresponds to maximizing reactant contact area and collision frequency; raising temperature by 10 K corresponds to elevating kinetic energy to expand the fraction of collisions exceeding activation energy; introducing a catalyst corresponds to establishing a lower activation energy pathway for both forward and reverse processes.

Each factor influences the rate through a distinct physical or chemical mechanism: light supplies photochemical activation energy; surface area controls collision site availability; temperature dictates the population of molecules with

E \ge E_a

; and catalysts lower the energy barrier for both reaction directions.

Adım Adım Çözüm

1

Analyze the photochemical process (UV light application)

Light acts as an energy source for specific photochemical reactions, breaking bonds to yield radicals.

Light intensity and wavelength directly influence rates of photochemical reactions like halogenation of alkanes.

2

Analyze the effect of particle size (pulverization of solid)

Breaking solid chunks into powder exposes more surface atoms to acid molecules.

Greater exposed surface area increases the total frequency of particle collisions per second.

3

Analyze thermal energy and kinetic distribution

A temperature rise increases average kinetic energy and Maxwell-Boltzmann tail population.

The rate increases exponentially with temperature because a much larger percentage of collisions meet the activation energy threshold.

4

Analyze catalytic action in reversible systems

Catalysts lower activation energy (

E_a

) for both forward and reverse reactions without altering equilibrium position.

Catalysts alter the reaction mechanism to provide an alternative pathway with lower energy requirements.

Anahtar Kavram

Collision theory principles underlying reaction rate factors (light intensity, surface area, thermal kinetic energy distribution, and catalytic pathways)

Soru 1329Soru →

Match each chemical process or reaction involving alkanes and petroleum refining in Column A with its corresponding chemical description or primary purpose in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Catalytic Cracking matches with the thermal breakdown of long-chain hydrocarbons into shorter alkanes and alkenes. Reforming matches with converting straight-chain alkanes into branched or aromatic hydrocarbons to boost octane rating. Complete Combustion matches with reacting alkanes in excess oxygen to produce

CO_2

and

H_2O

. Free-Radical Substitution matches with replacing hydrogen atoms with halogens under UV light.

Catalytic Cracking breaks larger hydrocarbon molecules into smaller, more useful molecules (alkanes and alkenes). Reforming increases fuel quality (octane rating) by isomerizing straight chains to branched chains or aromatics. Complete Combustion converts alkanes in excess oxygen to carbon dioxide and water. Free-Radical Substitution halogenates alkanes in the presence of UV light.

Adım Adım Çözüm

1

Identify the primary function of Catalytic Cracking.

Cracking involves breaking heavy petroleum fractions into smaller alkanes and alkenes.

Heavy oils have low demand, whereas lighter fractions like petrol and gases have high industrial demand.

2

Identify the structural transformation involved in Reforming.

Reforming converts straight-chain alkanes into branched-chain alkanes and aromatic compounds.

Straight-chain alkanes cause engine knocking; branched and aromatic structures improve fuel efficiency by increasing the octane rating.

3

Determine the products of Complete Combustion of alkanes.

Alkanes react completely with excess oxygen to yield

CO_2(g)

and

H_2O(g)

.

Hydrocarbon oxidation in excess

O_2

yields fully oxidized carbon dioxide and water.

4

Determine the mechanism for alkane halogenation.

Halogenation of alkanes requires ultraviolet light to generate free radicals for substitution.

Alkanes are unreactive saturated hydrocarbons (paraffins) and require UV light to initiate homeolytic fission of chlorine or bromine molecules.

Anahtar Kavram

Chemical reactions of alkanes and industrial petroleum refining processes

Soru 1330Soru →

Match each of the following chemical species with the predominant intermolecular force operating between its molecules in the liquid or solid state.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Hydrogen fluoride matches with Hydrogen bonding; Trichloromethane matches with Permanent dipole-dipole interactions; Solid iodine matches with London dispersion forces (in a non-polar crystalline lattice); Methane matches with Weak London dispersion forces (in a small non-polar molecule).

Hydrogen fluoride forms hydrogen bonds due to the extreme electronegativity difference between H and F. Trichloromethane exhibits permanent dipole-dipole attractions because of its permanent net molecular dipole. Solid iodine is non-polar but has a large polarizable electron cloud leading to substantial London dispersion forces in its solid crystal. Methane is non-polar and small, possessing only weak London dispersion forces.

Adım Adım Çözüm

1

Analyze the polarity and chemical structure of each given substance.

HF

is highly polar with

H-F

bonds;

CHCl_3

is a polar asymmetrical molecule;

I_2

is a non-polar diatomic solid;

CH_4

is a non-polar tetrahedral gas.

Intermolecular forces depend strictly on molecular polarity, presence of

N-H

,

O-H

, or

F-H

bonds, and molecular size/polarizability.

2

Identify specific conditions for hydrogen bonding.

HF

satisfies the requirement of hydrogen attached to highly electronegative fluorine, giving rise to intermolecular hydrogen bonds.

Hydrogen bonding requires a hydrogen atom covalently bonded to

N

,

O

, or

F

interacting with a lone pair on a neighbouring electronegative atom.

3

Differentiate dipole-dipole forces from dispersion forces in neutral covalent compounds.

CHCl_3

possesses a permanent dipole moment giving dipole-dipole forces, while

I_2

and

CH_4

are non-polar and rely on London dispersion forces, with

I_2

having larger dispersion forces due to greater electron cloud polarizability.

Dispersion forces scale with molecular size and electron count, while dipole-dipole forces require permanent polar bonds in asymmetrical shapes.

Anahtar Kavram

Classification and Origin of Intermolecular Forces

Soru 1331Soru →

For the endothermic steam-reforming process represented by the thermochemical equation below:

CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g) \quad \Delta H = +206\text{ kJ mol}^{-1}

Match each applied change (stress) on the system with its corresponding effect on the equilibrium position or system performance.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Increasing temperature shifts the equilibrium to the right; increasing pressure shifts the equilibrium to the left; adding a nickel catalyst increases reaction rates without changing the equilibrium position; continuous removal of carbon monoxide shifts the equilibrium to the right.

According to Le Chatelier's principle: heating an endothermic reaction shifts equilibrium toward products; increasing total pressure shifts equilibrium toward the side with fewer gas molecules (reactants, 2 moles vs 4 moles); adding a catalyst speeds up reaching equilibrium without shifting its position; and removing a product shifts equilibrium toward products to compensate for the decrease in concentration.

Adım Adım Çözüm

1

Analyze the temperature change using enthalpy sign

The positive enthalpy change (

\Delta H = +206\text{ kJ mol}^{-1}

) indicates an endothermic forward reaction. Raising the temperature shifts the equilibrium to the right.

Le Chatelier's principle dictates that adding thermal energy favors the heat-absorbing (endothermic) direction.

2

Count gaseous moles on both sides to determine the pressure effect

Reactant gas moles = 2 (

CH_4 + H_2O

), Product gas moles = 4 (

CO + 3H_2

). Increasing pressure shifts the system to the left.

An increase in pressure shifts the equilibrium position toward the side with fewer moles of gas to reduce pressure.

3

Determine the role of a catalyst

The nickel catalyst accelerates both forward and backward reaction rates equally.

Catalysts do not alter the position of dynamic equilibrium or change product yield.

4

Determine the concentration change effect

Removing

CO(g)

causes a rightward shift in equilibrium.

Removing a component drives the system to shift in the direction that replaces it.

Anahtar Kavram

Le Chatelier's Principle

Soru 1332Soru →

Match each chemical term or phenomenon on the left with its correct defining characteristic on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Deliquescence matches the spontaneous absorption of atmospheric moisture until dissolving into a solution; Efflorescence matches the loss of water vapor to dry air forming a powdery residue; Hygroscopy matches the absorption of water vapor without dissolving; Water of crystallization matches the definite ratio of water molecules chemically bound in a crystal lattice.

Deliquescence describes solids absorbing moisture until they dissolve into a solution. Efflorescence describes hydrated crystals spontaneously losing water of crystallization into dry air to become powdery. Hygroscopy describes substances absorbing water vapor without turning into a solution. Water of crystallization is the fixed stoichiometric quantity of water built into the salt crystal structure.

Adım Adım Çözüm

1

Define deliquescence and match it with its defining process.

Deliquescence is matched with spontaneous absorption of moisture until the solid completely dissolves into a liquid solution.

Deliquescent solids absorb so much water from moist air that they form a liquid solution.

2

Define efflorescence and match it with its defining process.

Efflorescence is matched with the process in which a crystalline salt loses water vapor to dry air, forming a powdery residue.

This occurs because the vapor pressure of the hydrated crystal exceeds the ambient atmospheric water vapor pressure.

3

Define hygroscopy and distinguish it from deliquescence.

Hygroscopy is matched with the absorption of water vapor from the surrounding atmosphere without dissolving or forming a liquid solution.

Hygroscopic materials absorb water but remain in their original state without liquefying into a solution.

4

Define water of crystallization.

Water of crystallization is matched with the definite ratio of water molecules stoichiometrically locked inside a salt's crystal framework.

It represents the chemically bound water necessary for maintaining the specific crystal structure of hydrated salts.

Anahtar Kavram

Atmospheric Behavior and Hydration of Salts

Tahmini Süre:1m 15s

Soru 1333Soru →

Match each substance or chemical reagent related to iron extraction, corrosion, and qualitative analysis on the left with its corresponding chemical role or property on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Coke matches with reacting with CO2 to generate CO; Calcium silicate matches with forming molten slag to prevent re-oxidation; Hydrated iron(III) oxide matches with the chemical composition of rust; Potassium hexacyanoferrate(III) matches with testing for Fe(II) ions.

Each substance is matched directly to its chemical function: Coke supplies carbon to generate carbon(II) oxide gas; Calcium silicate acts as slag floating atop molten iron; Hydrated iron(III) oxide is the exact chemical composition of rust; Potassium hexacyanoferrate(III) is the standard bench reagent for detecting iron(II) ions in qualitative testing.

Adım Adım Çözüm

1

Analyze the blast furnace chemical reactions involving carbon input.

Coke (

\text{C}

) reacts with ascending

\text{CO}_2

(

\text{C} + \text{CO}_2 \rightarrow 2\text{CO}

) to produce carbon(II) oxide, which acts as the chief reducing agent for haematite.

Identify the role of Coke in blast furnace extraction.

2

Analyze slag formation and its function in the blast furnace.

Lime (

\text{CaO}

) combines with silica (

\text{SiO}_2

) to yield calcium silicate (

\text{CaSiO}_3

), a molten waste slag that floats above molten iron.

Identify the function of calcium silicate in molten iron isolation.

3

Identify the chemical identity of rust.

Atmospheric corrosion of iron in the presence of oxygen and water forms reddish-brown hydrated iron(III) oxide (

\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}

).

Match the rust formula with its physical phenomenon.

4

Recall qualitative analysis reagents for iron oxidation states.

Potassium hexacyanoferrate(III) reacts with

\text{Fe}^{2+}

ions to give a characteristic dark blue precipitate.

Match the analytical reagent with its specific ion test.

Anahtar Kavram

Extraction of Iron in the Blast Furnace, Rusting Mechanism, and Qualitative Analysis of Iron Ions

Soru 1334Soru →

Match each chemical transformation involving alkanols listed on the left with the appropriate reagent, enzyme, or catalyst required on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Glucose is fermented to ethanol using the enzyme zymase; dehydration of ethanol to ethene uses excess concentrated

\text{H}_2\text{SO}_4

at

170^\circ\text{C}

; ethanol is oxidized to ethanoic acid using acidified

\text{K}_2\text{Cr}_2\text{O}_7

under reflux; and industrial synthesis of ethanol from ethene uses steam with a phosphoric acid (

\text{H}_3\text{PO}_4

) catalyst at high temperature and pressure.

Each chemical process matches its unique catalyst or reaction conditions: zymase catalyzes glucose fermentation to ethanol, excess concentrated

\text{H}_2\text{SO}_4

at

170^\circ\text{C}

dehydrates ethanol to ethene, acidified

\text{K}_2\text{Cr}_2\text{O}_7

under reflux oxidizes ethanol to ethanoic acid, and phosphoric acid (

\text{H}_3\text{PO}_4

) on silica catalyzes the industrial hydration of ethene to ethanol.

Adım Adım Çözüm

1

Identify the biological catalyst for sugar fermentation

Fermentation of glucose (

\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2

) is catalyzed specifically by the enzyme zymase.

Yeast produces zymase, which converts simple hexose sugars directly into ethanol.

2

Identify the reagent and temperature for elimination/dehydration

Heating ethanol with excess concentrated

\text{H}_2\text{SO}_4

at

170^\circ\text{C}

yields ethene via removal of a water molecule.

Concentrated tetraoxosulfate(VI) acid acts as a dehydrating agent; high temperature (

170^\circ\text{C}

) favors ethene formation over ethoxyethane formation.

3

Identify the oxidizing conditions for full alkanol oxidation

Primary alkanols undergo two-stage oxidation: first to an alkanal, then under reflux with acidified

\text{K}_2\text{Cr}_2\text{O}_7

to an alkanoic acid.

Acidified potassium heptaoxodichromate(VI) is a strong oxidizing agent capable of carrying the oxidation of ethanol fully to ethanoic acid.

4

Identify the industrial catalytic addition reaction conditions

Direct hydration of ethene (

\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}

) uses a phosphoric acid catalyst.

The reversible addition of steam across the double bond of ethene requires a solid phosphoric acid catalyst at

300^\circ\text{C}

and high pressure.

Anahtar Kavram

Reagents, enzymes, and conditions for alkanol preparation and reactions

Soru 1335Soru →

Match each reaction condition modification on the left with its corresponding microscopic kinetic mechanism under collision theory on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Replacing zinc granules with zinc dust increases the surface area and total collision frequency; increasing temperature increases average kinetic energy and the fraction of particles with energy equal to or greater than activation energy (

E \ge E_a

); ultraviolet illumination provides photon energy to form free radicals; and adding solid vanadium(V) oxide provides a catalyst that lowers the activation energy pathway.

Each condition matches its microscopic collision theory explanation directly: surface area increases total collision frequency, temperature increases the fraction of molecules with

E \ge E_a

, UV light supplies energy for bond cleavage/radical formation, and catalysts lower the activation energy pathway.

Adım Adım Çözüm

1

Analyze the particle size modification for zinc granules vs zinc dust.

Zinc dust has a much greater surface area per unit mass than a solid granule, which allows more collisions per second between

Zn(s)

and

H^+(aq)

ions.

Surface area dictates the contact zone for heterogeneous reactions.

2

Evaluate the effect of temperature increase on kinetic energy distribution.

An increase in temperature shifts the Maxwell-Boltzmann distribution toward higher kinetic energies, exponentially increasing the fraction of effective collisions (

E \ge E_a

).

Reaction rate depends exponentially on temperature via the Arrhenius relationship.

3

Examine the role of light in photochemical reactions.

Ultraviolet light absorbs photon quanta (

E = h\nu

) to homolytically split

Cl_2

into reactive chlorine radicals.

Light intensity and frequency act as energy sources to overcome bond dissociation energy.

4

Assess the function of vanadium(V) oxide (

V_2O_5

).

V_2O_5

acts as a catalyst in the Contact Process, lowering the activation energy barrier for the oxidation of

SO_2

to

SO_3

.

Catalysts alter the reaction pathway to accelerate both forward and reverse rates equally.

Anahtar Kavram

Collision Theory and Factors Affecting Rates of Reaction

Soru 1336Soru →

Match each soil pollutant or waste management technique in Column A with its corresponding chemical mechanism, bio-environmental effect, or operational principle in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Arsenic contamination in agricultural soil matches the disruption of cellular respiration by binding to sulfhydryl groups and mimicking phosphate ions. Organophosphate pesticides match the inhibition of acetylcholinesterase enzymes with shorter environmental persistence. Phytoremediation matches the extraction and bioconcentration of soil heavy metals into plant biomass. Pyrolysis matches thermal decomposition in the absence of oxygen to produce bio-oil, gases, and char.

Each soil contaminant and waste treatment procedure is correctly linked to its fundamental biochemical pathway or thermodynamic condition: Arsenic disrupts sulfhydryl enzymes and phosphate pathways; Organophosphates selectively inhibit acetylcholinesterase; Phytoremediation relies on plant bioaccumulation of soil metals; and Pyrolysis achieves thermal decomposition under anaerobic conditions.

Adım Adım Çözüm

1

Analyze the biochemical toxicity mechanism of Arsenic

Arsenic (

\text{As}

) forms covalent bonds with enzyme sulfhydryl (

-\text{SH}

) groups and acts as a structural analog to phosphate, inhibiting ATP synthesis.

Heavy metals disrupt metabolic pathways by binding to functional group residues on key metabolic enzymes.

2

Examine the mode of action and biodegradability of Organophosphates

Organophosphates inhibit acetylcholinesterase, leading to acetylcholine buildup. They are biodegradable compared to persistent chlorinated hydrocarbons like DDT.

Synthetic pesticide classification relies on chemical structure, biological targets, and environmental degradation rates.

3

Identify the eco-friendly soil cleanup method utilizing plants

Phytoremediation uses hyperaccumulators to take up heavy metal pollutants from contaminated ground into plant tissues.

Biological soil remediation relies on bio-uptake processes to remove heavy metals without chemical soil destruction.

4

Distinguish Pyrolysis from other thermal waste treatment methods

Pyrolysis is anaerobic thermal degradation yielding bio-oil, syngas, and char, contrasting with oxygen-rich incineration.

Thermal waste processing methods differ in operating atmosphere (presence vs. absence of

\text{O}_2

) and end-products.

Anahtar Kavram

Mechanisms of Soil Pollutants, Agrochemicals, Heavy Metal Toxicity, and Modern Waste Processing Techniques

Soru 1337Soru →

In electrophilic aromatic substitution, benzene reacts with strong electrophiles generated by specific catalyst-reagent combinations. Match each benzene reaction system on the left with its corresponding active electrophile species generated during the reaction mechanism on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Nitration produces the nitronium ion (

\text{NO}_2^+

); Friedel-Crafts acylation generates the acylium ion (

\text{CH}_3\text{C}^+=\text{O}

); Catalytic bromination produces the bromonium ion (

\text{Br}^+

); Sulfonation generates neutral sulfur trioxide (

\text{SO}_3

).

Each benzene electrophilic substitution reaction relies on a specific reagent and catalyst mechanism to create a powerful electrophile capable of disrupting benzene's stable aromatic system. Nitration generates

\text{NO}_2^+

via acid-base protonation of nitric acid by sulfuric acid. Friedel-Crafts acylation forms the acylium ion

\text{CH}_3\text{C}^+=\text{O}

through chloride abstraction by the Lewis acid

\text{AlCl}_3

. Bromination generates a polarized

\text{Br}^+

complex using

\text{FeBr}_3

. Sulfonation relies on

\text{SO}_3

, which features an electron-deficient sulfur atom due to polar S=O bonds.

Adım Adım Çözüm

1

Identify the electrophile in nitration

Concentrated

\text{H}_2\text{SO}_4

acts as an acid to protonate

\text{HNO}_3

. Loss of

\text{H}_2\text{O}

yields

\text{NO}_2^+

(nitronium ion).

\text{H}_2\text{SO}_4

is a stronger acid than

\text{HNO}_3

and forces

\text{HNO}_3

to act as a base.

2

Identify the electrophile in Friedel-Crafts acylation

The catalyst

\text{AlCl}_3

abstracts

\text{Cl}^-

from

\text{CH}_3\text{COCl}

, leaving the resonance-stabilized cations

\text{CH}_3\text{C}^+=\text{O}

.

\text{AlCl}_3

is an electron-deficient Lewis acid capable of coordinating chloride.

3

Identify the electrophile in bromination

\text{FeBr}_3

coordinates with a bromine atom of

\text{Br}_2

, polarising the bond to create an effective

\text{Br}^+

electrophile.

Benzene requires a Lewis acid catalyst to polarize halogen molecules sufficiently for reaction.

4

Identify the electrophile in sulfonation

Equilibrium in concentrated/fuming

\text{H}_2\text{SO}_4

produces neutral

\text{SO}_3

, which has a highly electron-deficient sulfur atom.

The three electronegative oxygen atoms in

\text{SO}_3

withdraw electron density from the central sulfur atom.

Anahtar Kavram

Generation of Electrophiles in Benzene Electrophilic Substitution

Soru 1338Soru →

Match each chemical compound on the left with its characteristic solubility trend in water as temperature increases on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Potassium trioxonitrate(V) matches with steep increase in solubility; Sodium chloride matches with nearly constant solubility; Calcium tetraoxosulfate(VI) matches with decreasing solubility as temperature rises.

Potassium trioxonitrate(V) shows a steep increase in solubility with rising temperature due to its endothermic nature. Sodium chloride exhibits minimal temperature sensitivity, keeping its curve nearly flat. Calcium tetraoxosulfate(VI) exhibits retrograde solubility, decreasing as temperature increases because its dissolution is exothermic.

Adım Adım Çözüm

1

Identify the thermodynamic enthalpy change associated with dissolving each salt in water.

Potassium trioxonitrate(V) dissolution is endothermic, sodium chloride dissolution has a near-zero enthalpy change, and calcium tetraoxosulfate(VI) dissolution is exothermic.

Le Chatelier's principle determines how temperature affects solubility equilibria based on whether heat is absorbed or released.

2

Relate enthalpy of solution to the slope of the solubility curve.

Potassium trioxonitrate(V) has a steep positive curve, sodium chloride has a nearly horizontal curve, and calcium tetraoxosulfate(VI) has a negative curve.

Endothermic dissolution shifts right with heat (increasing solubility), whereas exothermic dissolution shifts left with heat (decreasing solubility).

Anahtar Kavram

Solubility curves represent how solute solubility varies with temperature based on whether the dissolution process is endothermic or exothermic.

Soru 1339Soru →

Match each sulfur-related compound or allotrope on the left with its correct physical characteristic, qualitative test, or stability range on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Rhombic sulfur matches the octahedral allotrope stable up to 96°C; Hydrogen sulfide matches the rotten-egg smelling gas turning lead(II) ethanoate paper black; Sulfur(IV) oxide matches the choking gas turning acidified potassium dichromate(VI) green; Monoclinic sulfur matches the needle-shaped allotrope stable between 96°C and 119°C.

Rhombic sulfur (

\alpha

-sulfur) is octahedral and stable below

96^\circ\text{C}

. Monoclinic sulfur (

\beta

-sulfur) is needle-shaped and stable between

96^\circ\text{C}

and

119^\circ\text{C}

. Hydrogen sulfide (

H_2S

) is a gas with a rotten-egg odor that reacts with

Pb^{2+}

to form black insoluble

PbS

. Sulfur(IV) oxide (

SO_2

) has a pungent choking smell and reduces orange dichromate(VI) solutions to green

Cr^{3+}

ions.

Adım Adım Çözüm

1

Analyze the crystal structures and thermal stability range of sulfur allotropes.

Rhombic sulfur is octahedral and stable below 96°C. Monoclinic sulfur is needle-shaped and stable between 96°C and 119°C.

96°C is the transition temperature at which the two crystalline allotropes exist in equilibrium.

2

Identify the qualitative gas tests and chemical properties of

H_2S

and

SO_2

.

H_2S

forms black

PbS

precipitate with lead(II) ethanoate paper.

SO_2

reduces orange dichromate(VI) to green

Cr^{3+}

ions.

Both gases are reducing agents, but

H_2S

forms insoluble sulfides while

SO_2

undergoes specific color-change redox reactions with dichromate ions.

Anahtar Kavram

Physical and chemical properties of sulfur allotropes, hydrogen sulfide, and sulfur(IV) oxide

Soru 1340Soru →

Match each transition metal complex ion on the left with its corresponding structural, electronic, and magnetic characteristics on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

[Fe(CN)_6]^{3-}

matches Octahedral geometry with a

3d^5

low-spin central metal ion containing 1 unpaired electron;

[Ni(CN)_4]^{2-}

matches Square planar geometry with a

3d^8

central metal ion that is diamagnetic;

[Co(NH_3)_6]^{3+}

matches Octahedral geometry with a

3d^6

low-spin central metal ion that is diamagnetic;

[Cu(NH_3)_4]^{2+}

matches Square planar geometry with a

3d^9

central metal ion containing 1 unpaired electron.

Each complex ion's central metal ion exhibits a specific oxidation state, electronic configuration, coordination geometry, and spin state based on crystal field theory and ligand field strength.

[Fe(CN)_6]^{3-}

features

Fe^{3+}

(

3d^5

) in a low-spin octahedral state with 1 unpaired electron.

[Ni(CN)_4]^{2-}

features

Ni^{2+}

(

3d^8

) in a square planar diamagnetic configuration.

[Co(NH_3)_6]^{3+}

features

Co^{3+}

(

3d^6

) in a low-spin octahedral diamagnetic state.

[Cu(NH_3)_4]^{2+}

features

Cu^{2+}

(

3d^9

) in a square planar configuration with 1 unpaired electron.

Adım Adım Çözüm

1

Determine the oxidation state and d-electron count of the central metal ion in each complex ion.

For

[Fe(CN)_6]^{3-}

,

Fe^{3+}

is

3d^5

. For

[Ni(CN)_4]^{2-}

,

Ni^{2+}

is

3d^8

. For

[Co(NH_3)_6]^{3+}

,

Co^{3+}

is

3d^6

. For

[Cu(NH_3)_4]^{2+}

,

Cu^{2+}

is

3d^9

.

Ligand charges (

CN^-

=

-1

,

NH_3

=

0

) determine the oxidation state of the central transition metal ion.

2

Analyze ligand strength, coordination geometry, and crystal field splitting to determine magnetic character.

[Fe(CN)_6]^{3-}

is octahedral low-spin (

t_{2g}^5

, 1 unpaired

e^-

).

[Ni(CN)_4]^{2-}

is square planar (

dsp^2

, diamagnetic).

[Co(NH_3)_6]^{3+}

is octahedral low-spin (

t_{2g}^6

, diamagnetic).

[Cu(NH_3)_4]^{2+}

is square planar (

3d^9

, 1 unpaired

e^-

).

Strong-field ligands (

CN^-

,

NH_3

) induce electron pairing in low-spin octahedral or square planar configurations.

3

Match each complex ion to its complete set of physical, electronic, and magnetic properties.

Each complex correctly aligns with its unique d-electron configuration, geometry, and spin state.

Verifies all coordination parameters systematically.

Anahtar Kavram

Electronic Configuration, Oxidation State, Geometry, and Magnetic Properties of Transition Metal Complexes

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram