Tüm alıştırma soruları

13931 soru

Soru 8281Soru

A compound has an empirical formula of CH2CH_2 and a relative molecular mass of 4242. What is the molecular formula of the compound? [C=12,H=1][C = 12, H = 1]

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Cevap: C3H6C_3H_6

Cevap

The molecular formula of the compound is C3H6C_3H_6.
The empirical formula mass of CH2CH_2 is 14 g/mol14\text{ g/mol}. Dividing the given relative molecular mass (4242) by 1414 gives an integer factor n=3n = 3. Multiplying the empirical formula indices by 33 gives C3H6C_3H_6.

Adım Adım Çözüm

1
Calculate the empirical formula mass of CH2CH_2.
Empirical formula mass =12+2(1)=14 g/mol= 12 + 2(1) = 14\text{ g/mol}.
The empirical formula mass is required to find the integer ratio multiplier nn.
2
Determine the integer multiplier nn by dividing the relative molecular mass by the empirical formula mass.
n=4214=3n = \frac{42}{14} = 3.
The integer multiplier indicates how many empirical units constitute the molecular formula.
3
Multiply the subscripts of the empirical formula CH2CH_2 by n=3n = 3.
Molecular formula =(CH2)3=C3H6= (CH_2)_3 = C_3H_6.
Applying the integer multiplier yields the true formula of the compound.

Anahtar Kavram

Molecular Formula Derivation from Empirical Formula and Molar Mass
Soru 8282Soru

Match each redox testing reagent or indicator on the left with its characteristic diagnostic observation on the right when reacting with an oxidizing or reducing agent.

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Öğeler

Acidified KMnO4\text{KMnO}_4 solution
Moist starch-iodide paper
Acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 solution
Freshly prepared FeSO4\text{FeSO}_4 solution

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Cevap

Acidified potassium tetraoxomanganate(VII) turns from purple to colorless with reducing agents; starch-iodide paper turns blue-black with oxidizing agents; acidified potassium heptaoxodichromate(VI) turns from orange to green with reducing agents; and freshly prepared iron(II) sulfate changes from pale green to reddish-brown with oxidizing agents.
Each testing reagent displays a distinct diagnostic color change depending on whether it reacts with an oxidizing or reducing agent. Acidified KMnO4\text{KMnO}_4 changes from purple to colorless in the presence of a reducing agent due to reduction of MnO4\text{MnO}_4^- to Mn2+\text{Mn}^{2+}. Starch-iodide paper turns blue-black in the presence of an oxidizing agent as iodide is oxidized to iodine. Acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 changes from orange to green in the presence of a reducing agent as Cr2O72\text{Cr}_2\text{O}_7^{2-} is reduced to Cr3+\text{Cr}^{3+}. Freshly prepared FeSO4\text{FeSO}_4 changes from pale green to reddish-brown when an oxidizing agent oxidizes Fe2+\text{Fe}^{2+} to Fe3+\text{Fe}^{3+}.

Adım Adım Çözüm

1
Identify the role of acidified KMnO4\text{KMnO}_4 solution in redox testing.
Acidified KMnO4\text{KMnO}_4 contains manganese in the +7+7 oxidation state (purple). When it oxidizes a reducing agent, manganese is reduced to Mn2+\text{Mn}^{2+} (colorless).
This is the classic quantitative and qualitative test for reducing agents.
2
Determine the response of moist starch-iodide paper to oxidizing gases.
Oxidizing agents liberate free iodine (I2I_2) from iodide ions (II^-). Free iodine reacts with starch to yield a distinctive blue-black color.
This tests specifically for oxidizing agents such as chlorine or ozone.
3
Identify the color change associated with acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7.
Dichromate ions (orange, chromium oxidation state +6+6) are reduced to Cr3+\text{Cr}^{3+} ions (green, oxidation state +3+3) by reducing agents.
The reduction of dichromate(VI) to chromium(III) causes the orange-to-green color change.
4
Determine the oxidation behavior of iron(II) sulfate solution.
Iron(II) ions (pale green) act as a reducing agent and are oxidized to iron(III) ions (reddish-brown/yellow) by oxidizing agents.
The oxidation of Fe2+\text{Fe}^{2+} to Fe3+\text{Fe}^{3+} shifts the solution color from green to brown/yellow.

Anahtar Kavram

Laboratory Diagnostic Tests for Oxidizing and Reducing Agents
Soru 8283Soru

Match each chemical mixture recovery requirement on the left with the most appropriate physical separation technique on the right.

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Öğeler

Separating insoluble chalk (CaCO3\text{CaCO}_3) powder suspended in an aqueous medium
Obtaining pure hydrated copper(II) tetraoxosulfate(VI) crystals (CuSO45H2O\text{CuSO}_4\cdot 5\text{H}_2\text{O}) from an aqueous solution
Recovering thermally stable, dry sodium chloride (NaCl\text{NaCl}) completely from sea water
Separating potassium trioxonitrate(V) (KNO3\text{KNO}_3) from a solution containing a soluble sodium chloride (NaCl\text{NaCl}) impurity based on temperature-solubility dependence

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Cevap

Insoluble chalk suspended in water is separated by filtration. Hydrated copper(II) tetraoxosulfate(VI) crystals require evaporation to saturation followed by cooling to avoid thermal decomposition. Dry sodium chloride is recovered by evaporation to complete dryness. Potassium trioxonitrate(V) is purified from sodium chloride impurity using fractional crystallization.
Each technique is uniquely suited to the physical properties of the mixture: filtration removes insoluble suspensions; evaporation to saturation followed by cooling yields hydrated crystals without thermal decomposition; direct evaporation to dryness efficiently recovers anhydrous thermally stable salts; fractional crystallization separates multiple soluble solutes using solubility-temperature gradients.

Adım Adım Çözüm

1
Analyze the solubility and thermal stability of each solute/mixture component.
Chalk is insoluble; hydrated copper sulfate is heat-sensitive; sodium chloride is heat-stable; potassium nitrate and sodium chloride mixture contains two soluble salts with different solubility-temperature slopes.
Physical properties determine which technique yields pure product without thermal breakdown.
2
Match solid-liquid heterogeneous suspensions to filtration.
Chalk powder suspended in water matches filtration.
Filtration separates undissolved solid particles from liquid filtrates.
3
Distinguish between thermal evaporation to dryness and crystallization for soluble salts.
Thermally stable salt (NaCl) matches evaporation to complete dryness, while hydrated salt (CuSO4·5H2O) matches evaporation to saturation followed by cooling.
Evaporating hydrated salts to dryness destroys water of crystallization, yielding anhydrous powder rather than crystals.
4
Match multi-solute solution separation based on temperature-dependent solubility to fractional crystallization.
Separating KNO3 from NaCl impurity matches fractional crystallization.
Cooling a hot saturated solution of both salts causes KNO3 (steep solubility rise with temperature) to crystallize out first.

Anahtar Kavram

Selection of separation techniques (filtration, evaporation, crystallization, fractional crystallization) based on solubility and thermal stability.
Soru 8284Soru

When a mechanical stress is applied to a solid metal, the material deforms without shattering. Which of the following structural features of metallic bonding is directly responsible for this malleability?

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Cevap: The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea

Cevap

The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea
In metallic lattices, delocalized valence electrons move freely throughout the array of positive metal cations. When a mechanical force is applied, layers of cations slide past one another. The mobile electron sea adapts immediately to the shifted cations, maintaining the non-directional electrostatic attraction throughout the lattice so that the metal deforms (malleability) instead of fracturing.

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1
Identify the atomic-scale structure of a metallic lattice.
Solid metals consist of a giant lattice of positive metal cations immersed in a fluid sea of delocalized valence electrons.
Understanding the non-directional nature of metallic bonds is necessary to explain physical properties.
2
Analyze how applied mechanical force alters the lattice structure.
Under mechanical stress, planes of positive cations slide past one another.
Applied mechanical forces induce shear stress across crystal lattice planes.
3
Determine why the metallic structure deforms rather than breaking.
Because the delocalized electrons are mobile and non-directional, they adjust immediately to the shifted cation layers, maintaining attractive electrostatic forces throughout the lattice and preventing repulsive cleavage.
Non-directional electrostatic attraction preserves structural cohesion during deformation.

Anahtar Kavram

Metallic Bonding and Malleability
Soru 8285Soru

Ammonium chloride (NH4ClNH_4Cl) is synthesized by reacting ammonia gas with hydrogen chloride gas. Which types of chemical bonding exist within solid ammonium chloride?

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Cevap: Covalent, coordinate (dative), and ionic bonds

Cevap

Covalent, coordinate (dative), and ionic bonds
Ammonium chloride (NH4ClNH_4Cl) exhibits three distinct types of chemical bonds: (1) Covalent bonds between nitrogen and three hydrogen atoms in the original ammonia molecule, (2) A coordinate (dative) bond formed when nitrogen donates its lone pair to a hydrogen ion (H+H^+) to form the ammonium ion (NH4+NH_4^+), and (3) An ionic (electrovalent) bond between the NH4+NH_4^+ cation and the ClCl^- anion.

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1
Analyze the structure of the ammonia molecule (NH3NH_3).
Nitrogen shares electron pairs with three hydrogen atoms, forming three polar covalent bonds and retaining one unshared lone pair of electrons.
Covalent bonding occurs when non-metal atoms share pairs of valence electrons.
2
Analyze the formation of the ammonium ion (NH4+NH_4^+).
Nitrogen donates its lone pair of electrons to an electron-deficient hydrogen ion (H+H^+), forming one coordinate (dative) covalent bond.
A coordinate bond is formed when one atom provides both electrons for the shared pair.
3
Analyze the interaction between the ammonium ion (NH4+NH_4^+) and the chloride ion (ClCl^-).
Electrostatic attraction between the positively charged NH4+NH_4^+ cation and negatively charged ClCl^- anion forms an ionic bond.
Oppositely charged ions attract each other to form a stable crystal lattice.

Anahtar Kavram

Coexistence of covalent, dative, and ionic bonding in polyatomic salts
Tahmini Süre:1m 0s
Soru 8286Soru

A student dilutes 100 cm3100\text{ cm}^3 of a 0.05 mol dm30.05\text{ mol dm}^{-3} nitric acid (HNO3\text{HNO}_3) solution with distilled water to achieve a total volume of 500 cm3500\text{ cm}^3 at 25C25^\circ\text{C}. What is the pOH of the resulting diluted solution?

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Cevap: 12.012.0

Cevap

The pOH of the resulting diluted solution is 12.012.0.
Diluting 100 cm3100\text{ cm}^3 of 0.05 mol dm3 HNO30.05\text{ mol dm}^{-3}\text{ HNO}_3 to 500 cm3500\text{ cm}^3 decreases the hydrogen ion concentration to 0.01 mol dm30.01\text{ mol dm}^{-3}. The pH of this solution is log10(0.01)=2.0-\log_{10}(0.01) = 2.0. Since pH+pOH=14\text{pH} + \text{pOH} = 14 at 25C25^\circ\text{C}, the pOH is 142.0=12.014 - 2.0 = 12.0.

Adım Adım Çözüm

1
Calculate the hydrogen ion concentration of the diluted solution using the dilution formula C1V1=C2V2C_1 V_1 = C_2 V_2.
C2=0.05 mol dm3×100 cm3500 cm3=0.01 mol dm3=1.0×102 mol dm3C_2 = \frac{0.05\text{ mol dm}^{-3} \times 100\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}.
Nitric acid is a strong monoprotic acid that fully dissociates in water, so [H+]=C2[\text{H}^+] = C_2.
2
Determine the pH of the diluted solution.
pH=log10[H+]=log10(1.0×102)=2.0\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-2}) = 2.0.
The negative logarithm of the hydrogen ion concentration defines the pH.
3
Calculate the pOH using the water ion product relationship at 25C25^\circ\text{C}.
pOH=14pH=142.0=12.0\text{pOH} = 14 - \text{pH} = 14 - 2.0 = 12.0.
The sum of pH and pOH equals 1414 at standard room temperature.

Anahtar Kavram

Dilution effect on hydrogen ion concentration and the relation pH+pOH=14\text{pH} + \text{pOH} = 14
Tahmini Süre:1m 30s
Soru 8287Soru

During the municipal treatment of water for public consumption, alum, Al2(SO4)3Al_2(SO_4)_3, is added to untreated river water. What is the primary chemical function of adding alum in this process?

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Cevap: To cause coagulation and sedimentation of fine suspended impurities

Cevap

The primary function of adding alum in municipal water treatment is to cause coagulation and sedimentation of fine suspended impurities.
Alum (Al2(SO4)3Al_2(SO_4)_3) serves as a coagulating agent during water treatment. It destabilizes colloidal particles suspended in water, allowing them to join together into heavy flocs that settle out easily during sedimentation.

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1
Identify the role of alum (Al2(SO4)3Al_2(SO_4)_3) in water purification
Alum provides trivalent aluminum ions (Al3+Al^{3+}) which neutralize the negative surface charges on microscopic suspended particles.
Neutralizing these surface charges causes the tiny particles to clump together into larger, visible aggregates called flocs.
2
Distinguish coagulation from other treatment stages
The heavy flocs created by coagulation settle to the bottom of sedimentation tanks by gravity.
Disinfection (destroying pathogens) is performed later using chlorine gas or sodium hypochlorite, while softening removes dissolved calcium and magnesium ions.

Anahtar Kavram

Coagulation and Flocculation in Municipal Water Treatment
Soru 8288Soru

Match each structural feature of the electron sea model on the left with the macroscopic metal property it directly accounts for on the right.

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Öğeler

Movement of delocalized electrons toward a positive terminal under an applied voltage
Layers of positive metal cations sliding past one another while maintaining electrostatic attraction with mobile electrons
Absorption and immediate re-emission of incident light by free surface electrons
Strong electrostatic attraction extending uniformly throughout the 3D lattice between metal cations and delocalized electrons

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Cevap

1. Movement of delocalized electrons toward a positive terminal matches High electrical conductivity. 2. Layers of cations sliding past each other matches Malleability and ductility. 3. Absorption and re-emission of light by free electrons matches Lustrous (shiny) appearance. 4. Strong non-directional electrostatic attraction matches High melting and boiling points.
Each structural feature in the electron sea model directly dictates a specific macroscopic behavior: delocalized electron motion provides electrical conductivity, cation layer flexibility allows deformation (malleability/ductility), light oscillation by surface electrons causes shiny luster, and extensive electrostatic forces produce high thermal melting thresholds.

Adım Adım Çözüm

1
Relate electric charge transport to metallic conduction
Free electrons moving toward a positive potential corresponds to high electrical conductivity.
Electric current in solid metals consists of a net flow of delocalized valence electrons.
2
Analyze deformation behavior of metallic lattices under pressure
Sliding layers of cations buffered by the electron sea corresponds to malleability and ductility.
Metals deform without shattering because metallic bonding is non-directional.
3
Connect light interaction with free electron oscillations
Free surface electrons absorbing and re-emitting light photons corresponds to luster.
Unbound electrons respond dynamically to electromagnetic waves, reflecting light.
4
Examine thermal stability of the lattice bonding
Strong omnidirectional electrostatic attraction corresponds to high melting and boiling points.
Separating metallic particles requires inputting significant thermal energy to overcome electrostatic bonds.

Anahtar Kavram

Electron Sea Model of Metallic Bonding
Soru 8289Soru

A solid mixture consists of insoluble silicon(IV) oxide (SiO2\text{SiO}_2) and soluble potassium tetraoxosulfate(VI) (K2SO4\text{K}_2\text{SO}_4). After adding distilled water to the mixture and stirring thoroughly, which procedure should be used to separate the insoluble solid and obtain pure crystals of potassium tetraoxosulfate(VI)?

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Cevap: Filter the mixture to remove SiO2\text{SiO}_2, then heat the filtrate to concentrate it before allowing it to cool and crystallize.

Cevap

Filtration of the mixture to remove the insoluble silicon(IV) oxide (SiO2\text{SiO}_2), followed by partial evaporation of the filtrate to concentrate it and subsequent cooling to allow potassium tetraoxosulfate(VI) (K2SO4\text{K}_2\text{SO}_4) crystals to form.
Filtration successfully separates the insoluble silicon(IV) oxide (SiO2\text{SiO}_2) as the residue on the filter paper from the soluble potassium tetraoxosulfate(VI) (K2SO4\text{K}_2\text{SO}_4) in the filtrate. Heating the filtrate to concentrate it and then letting it cool slowly yields pure crystals of the salt.

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1
Dissolve the mixture in water and filter the suspension.
Insoluble silicon(IV) oxide (SiO2\text{SiO}_2) is retained on the filter paper as the residue, while dissolved potassium tetraoxosulfate(VI) (K2SO4\text{K}_2\text{SO}_4) passes through as the clear filtrate.
Filtration separates insoluble solids from liquids based on particle size and solubility.
2
Evaporate the filtrate partially until it reaches its crystallization point (saturated solution).
Excess solvent is driven off without decomposing the salt or trapping impurities.
Controlled evaporation concentrates the solution to saturation without heating to complete dryness.
3
Allow the hot saturated solution to cool slowly at room temperature.
Pure crystals of K2SO4\text{K}_2\text{SO}_4 form and deposit out of solution.
Solubility decreases with decreasing temperature, causing the solute to crystallize.

Anahtar Kavram

Separation of insoluble and soluble solid mixtures using filtration and crystallization
Tahmini Süre:1m 0s
Soru 8290Soru

Three unlabelled liquid mixtures, PP, QQ, and RR, are tested in a laboratory. Mixture PP passes completely through both filter paper and semi-permeable membranes without scattering light. Mixture QQ passes through filter paper but is retained by semi-permeable membranes and scatters a narrow beam of light. Mixture RR leaves a solid residue on filter paper upon gravity filtration. Which mixture will undergo precipitation upon the addition of an electrolyte, and what physical property accounts for this behavior?

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Cevap: Mixture QQ, because its dispersed particles carry surface charges and have diameters between 1 nm1\text{ nm} and 100 nm100\text{ nm}.

Cevap

Mixture QQ, because its dispersed particles carry surface charges and have diameters between 1 nm1\text{ nm} and 100 nm100\text{ nm}.
The experimental observations identify Mixture QQ as a colloid because it passes through ordinary filter paper, is retained by a semi-permeable membrane, and displays light scattering (Tyndall effect). Colloidal particles range in size between 1 nm1\text{ nm} and 100 nm100\text{ nm} and carry surface electrical charges. Addition of an electrolyte neutralizes these charges, leading to coagulation.

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1
Analyze the filtration and optical properties of each mixture to classify them.
Mixture PP is a true solution (particle size <1 nm< 1\text{ nm}). Mixture QQ is a colloidal system (particle size 1100 nm1\text{--}100\text{ nm}, exhibits Tyndall scattering, retained by semi-permeable membrane). Mixture RR is a suspension (particle size >100 nm> 100\text{ nm}, retained by filter paper).
Differentiation among true solutions, colloids, and suspensions is based on particle size and membrane permeability.
2
Determine which system undergoes coagulation upon adding an electrolyte.
Colloidal particles (Mixture QQ) carry electric charges that stabilize them. Adding an electrolyte supplies ions of opposite charge, neutralizing the colloidal particles and causing them to coagulate/precipitate.
True solutions do not coagulate on electrolyte addition, and suspensions settle naturally by gravity.

Anahtar Kavram

Classification and properties of true solutions, colloidal systems, and suspensions
Soru 8291Soru
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 8292Soru

A sealed rigid glass flask contains carbon dioxide gas at a temperature of 7C7^\circ\text{C} and a pressure of 1.4 atm1.4\text{ atm}. If the flask is heated to a final temperature of 77C77^\circ\text{C} at constant volume, what is the new pressure exerted by the gas?

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Cevap: 1.75 atm1.75\text{ atm}

Cevap

The new pressure exerted by the gas is 1.75 atm1.75\text{ atm}.
According to Gay-Lussac's Pressure Law, the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature (P1/T1=P2/T2P_1/T_1 = P_2/T_2). Converting the temperatures to Kelvin (T1=7+273=280 KT_1 = 7 + 273 = 280\text{ K} and T2=77+273=350 KT_2 = 77 + 273 = 350\text{ K}) and solving for P2P_2 gives P2=1.4×350280=1.75 atmP_2 = 1.4 \times \frac{350}{280} = 1.75\text{ atm}.

Adım Adım Çözüm

1
Convert initial and final temperatures from Celsius to Kelvin.
T1=7+273=280 KT_1 = 7 + 273 = 280\text{ K} and T2=77+273=350 KT_2 = 77 + 273 = 350\text{ K}.
Gas laws require absolute temperature in Kelvin for proportional relationship calculations.
2
Apply the Pressure Law (Gay-Lussac's Law).
P1T1=P2T2    P2=P1×T2T1\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}.
For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.
3
Substitute the values to calculate the final pressure P2P_2.
P2=1.4×350280=1.4×1.25=1.75 atmP_2 = 1.4 \times \frac{350}{280} = 1.4 \times 1.25 = 1.75\text{ atm}.
Simplifying the fraction 350280=54=1.25\frac{350}{280} = \frac{5}{4} = 1.25 gives the final answer.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law)
Soru 8293Soru

Under identical conditions of temperature and pressure, a neutral oxide of nitrogen diffuses 1.211.21 times faster than carbon(IV) oxide (CO2CO_2). Which of the following is a characteristic chemical property of this oxide of nitrogen? (Relative atomic masses: C=12,N=14,O=16\text{Relative atomic masses: } C = 12, N = 14, O = 16)

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Cevap: It reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.

Cevap

The gas reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.
Using Graham's law of diffusion, the molar mass of the unknown gas is calculated as 441.21230.0 g mol1\frac{44}{1.21^2} \approx 30.0\text{ g mol}^{-1}, which corresponds uniquely to nitrogen(II) oxide (NONO). A defining chemical test for NONO is its immediate oxidation by air to give reddish-brown fumes of nitrogen(IV) oxide (NO2NO_2).

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1
Calculate the molar mass of carbon(IV) oxide (CO2CO_2).
M(CO2)=12+2(16)=44 g mol1M(CO_2) = 12 + 2(16) = 44\text{ g mol}^{-1}.
Graham's law requires the molar mass of the reference gas.
2
Apply Graham's law of diffusion to determine the molar mass of the unknown nitrogen oxide (MxM_x).
RxRCO2=M(CO2)Mx    1.21=44Mx    (1.21)2=44Mx    Mx=441.464130.0 g mol1\frac{R_x}{R_{CO_2}} = \sqrt{\frac{M(CO_2)}{M_x}} \implies 1.21 = \sqrt{\frac{44}{M_x}} \implies (1.21)^2 = \frac{44}{M_x} \implies M_x = \frac{44}{1.4641} \approx 30.0\text{ g mol}^{-1}.
The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
3
Identify the formula of the nitrogen oxide with a molar mass of 30 g mol130\text{ g mol}^{-1}.
For nitrogen(II) oxide (NONO), M(NO)=14+16=30 g mol1M(NO) = 14 + 16 = 30\text{ g mol}^{-1}.
Matching the calculated molar mass to known oxides of nitrogen (N2O=44N_2O = 44, NO=30NO = 30, NO2=46NO_2 = 46).
4
Determine the key chemical property of nitrogen(II) oxide (NONO).
Nitrogen(II) oxide (NONO) is a colorless neutral gas that reacts spontaneously with oxygen in air to form brown nitrogen(IV) oxide (2NO+O22NO22NO + O_2 \rightarrow 2NO_2).
To select the correct property corresponding to NONO.

Anahtar Kavram

Graham's Law of Diffusion and Chemical Properties of Oxides of Nitrogen
Tahmini Süre:2m 0s
Soru 8294Soru

An industrial gas cylinder is equipped with a pressure safety relief valve calibrated to open when the internal gas pressure reaches 3.60 atm3.60\text{ atm}. At an ambient temperature of 27C27^\circ\text{C}, the enclosed gas exerts a pressure of 2.40 atm2.40\text{ atm}. Assuming the volume of the container remains strictly constant, what is the maximum temperature in degrees Celsius (C^\circ\text{C}) to which the cylinder can be heated before the safety valve opens?

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Cevap: 177C177^\circ\text{C}

Cevap

The maximum temperature before the safety valve opens is 177C177^\circ\text{C}.
According to the Pressure Law, for a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature (P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}). Converting 27C27^\circ\text{C} to absolute temperature yields T1=300 KT_1 = 300\text{ K}. Substituting P1=2.40 atmP_1 = 2.40\text{ atm}, P2=3.60 atmP_2 = 3.60\text{ atm}, and T1=300 KT_1 = 300\text{ K} gives T2=450 KT_2 = 450\text{ K}. Converting 450 K450\text{ K} back to Celsius gives 450273=177C450 - 273 = 177^\circ\text{C}.

Adım Adım Çözüm

1
Convert the initial temperature from Celsius to Kelvin
T1=27C+273=300 KT_1 = 27^\circ\text{C} + 273 = 300\text{ K}
Gas laws require thermodynamic temperature expressed in absolute units (Kelvin).
2
Apply the Pressure Law (Gay-Lussac's Law) formula at constant volume
P1T1=P2T2    2.40300=3.60T2\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies \frac{2.40}{300} = \frac{3.60}{T_2}
For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.
3
Solve for the final absolute temperature T2T_2
T2=300×3.602.40=450 KT_2 = 300 \times \frac{3.60}{2.40} = 450\text{ K}
Cross-multiplying gives the threshold temperature in Kelvin.
4
Convert the absolute temperature back to degrees Celsius
t2=450 K273=177Ct_2 = 450\text{ K} - 273 = 177^\circ\text{C}
The question specifically requests the maximum temperature in degrees Celsius.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)
Tahmini Süre:1m 30s
Soru 8295Soru

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
Soru 8296Soru

According to Hückel's rule, a planar cyclic conjugated compound exhibits aromatic stability if the number of delocalized π\pi-electrons in its conjugated ring system is equal to which expression, where nn is a non-negative integer?

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Cevap: 4n+24n + 2

Cevap

The expression for the number of delocalized π\pi-electrons in an aromatic compound is 4n+24n + 2.
Hückel's rule states that a cyclic, planar, fully conjugated molecule possesses aromatic stability when it contains (4n+2)(4n + 2) delocalized π\pi-electrons, where nn is an integer (0,1,2,3,0, 1, 2, 3, \dots). For example, benzene has 66 π\pi-electrons, satisfying the rule for n=1n = 1.

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1
Recall the structural criteria for aromaticity.
An aromatic molecule must be cyclic, planar, fully conjugated, and possess a specific number of delocalized π\pi-electrons.
These structural conditions allow complete ring delocalization of π\pi-electrons.
2
Apply Hückel's rule to determine the required π\pi-electron count.
The total number of delocalized π\pi-electrons must equal (4n+2)(4n + 2), where n=0,1,2,3,n = 0, 1, 2, 3, \dots
This formula predicts closed-shell electronic stability for planar conjugated rings.

Anahtar Kavram

Hückel's Rule of Aromaticity
Tahmini Süre:45s
Soru 8297Soru

Calculate the relative molecular mass of hydrated magnesium tetraoxosulfate(VI), MgSO47H2OMgSO_4 \cdot 7H_2O. [Mg=24,S=32,O=16,H=1][Mg = 24, S = 32, O = 16, H = 1]

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Cevap: 246

Cevap

246
The relative molecular mass of MgSO47H2OMgSO_4 \cdot 7H_2O is calculated by summing the atomic masses of all constituent atoms. The anhydrous part MgSO4MgSO_4 contributes 24+32+(4×16)=12024 + 32 + (4 \times 16) = 120. The water of crystallization 7H2O7H_2O contributes 7×(2+16)=1267 \times (2 + 16) = 126. The total relative molecular mass is 120+126=246120 + 126 = 246.

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1
Calculate the mass of the anhydrous salt component, MgSO4MgSO_4
24+32+(4×16)=12024 + 32 + (4 \times 16) = 120
Sum the relative atomic masses of one magnesium atom, one sulfur atom, and four oxygen atoms.
2
Calculate the mass of the water of crystallization component, 7H2O7H_2O
7×(2+16)=1267 \times (2 + 16) = 126
Multiply the molecular mass of one water molecule (18) by the coefficient 7.
3
Combine the mass of MgSO4MgSO_4 and 7H2O7H_2O
120+126=246120 + 126 = 246
The total relative molecular mass of a hydrated salt is the sum of the anhydrous salt mass and the water of crystallization mass.

Anahtar Kavram

Relative Molecular Mass of Hydrated Salts
Soru 8298Soru

Consider the standard reduction potentials (EE^\circ) at 25C25^\circ\text{C} for the following half-reactions:

Ce4+(aq)+eCe3+(aq)E=+1.61 V\text{Ce}^{4+}(aq) + e^- \rightarrow \text{Ce}^{3+}(aq) \quad E^\circ = +1.61\text{ V}
Br2(l)+2e2Br(aq)E=+1.07 V\text{Br}_2(l) + 2e^- \rightarrow 2\text{Br}^-(aq) \quad E^\circ = +1.07\text{ V}
Fe3+(aq)+eFe2+(aq)E=+0.77 V\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\circ = +0.77\text{ V}
I2(s)+2e2I(aq)E=+0.54 V\text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \quad E^\circ = +0.54\text{ V}
Sn4+(aq)+2eSn2+(aq)E=+0.15 V\text{Sn}^{4+}(aq) + 2e^- \rightarrow \text{Sn}^{2+}(aq) \quad E^\circ = +0.15\text{ V}

Which of the following chemical species can spontaneously oxidize I(aq)\text{I}^-(aq) to I2(s)\text{I}_2(s) under standard conditions, but is UNABLE to oxidize Br(aq)\text{Br}^-(aq) to Br2(l)\text{Br}_2(l)?

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Cevap: Fe3+(aq)\text{Fe}^{3+}(aq)

Cevap

The species Fe3+(aq)\text{Fe}^{3+}(aq) is the correct choice because its standard reduction potential (+0.77 V) lies strictly between the reduction potentials of I2/I\text{I}_2/\text{I}^- (+0.54 V) and Br2/Br\text{Br}_2/\text{Br}^- (+1.07 V).
To oxidize I\text{I}^- (Ered=+0.54 VE^\circ_{\text{red}} = +0.54\text{ V}), an oxidizing agent must have a standard reduction potential greater than +0.54 V+0.54\text{ V}. To fail to oxidize Br\text{Br}^- (Ered=+1.07 VE^\circ_{\text{red}} = +1.07\text{ V}), its reduction potential must be less than +1.07 V+1.07\text{ V}. The species Fe3+(aq)\text{Fe}^{3+}(aq) has E=+0.77 VE^\circ = +0.77\text{ V}, which satisfies +0.54 V<+0.77 V<+1.07 V+0.54\text{ V} < +0.77\text{ V} < +1.07\text{ V}. Thus, Ecell(Fe3+/I)=+0.770.54=+0.23 V>0E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{I}^-) = +0.77 - 0.54 = +0.23\text{ V} > 0 (spontaneous) and Ecell(Fe3+/Br)=+0.771.07=0.30 V<0E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{Br}^-) = +0.77 - 1.07 = -0.30\text{ V} < 0 (non-spontaneous).

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1
Determine the required condition for spontaneous oxidation of I(aq)\text{I}^-(aq) to I2(s)\text{I}_2(s).
The oxidation half-reaction is 2I(aq)I2(s)+2e2\text{I}^-(aq) \rightarrow \text{I}_2(s) + 2e^- with Eox=0.54 VE^\circ_{\text{ox}} = -0.54\text{ V}. For Ecell=Ered(oxidant)+Eox>0E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} > 0, the oxidant must have Ered>+0.54 VE^\circ_{\text{red}} > +0.54\text{ V}.
A redox reaction is spontaneous if the cell potential EcellE^\circ_{\text{cell}} is positive.
2
Determine the required condition for non-spontaneous oxidation of Br(aq)\text{Br}^-(aq) to Br2(l)\text{Br}_2(l).
The oxidation half-reaction is 2Br(aq)Br2(l)+2e2\text{Br}^-(aq) \rightarrow \text{Br}_2(l) + 2e^- with Eox=1.07 VE^\circ_{\text{ox}} = -1.07\text{ V}. For Ecell=Ered(oxidant)+Eox<0E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} < 0, the oxidant must have Ered<+1.07 VE^\circ_{\text{red}} < +1.07\text{ V}.
An unfeasible (non-spontaneous) reaction corresponds to a negative overall standard cell potential.
3
Combine the potential boundary constraints and evaluate the options.
The standard reduction potential of the ideal oxidant must satisfy +0.54 V<Ered<+1.07 V+0.54\text{ V} < E^\circ_{\text{red}} < +1.07\text{ V}. Among the choices, Fe3+(aq)\text{Fe}^{3+}(aq) has Ered=+0.77 VE^\circ_{\text{red}} = +0.77\text{ V}, which falls squarely within this range.
Comparing standard electrode potentials directly establishes which species can act as selective oxidizing agents.

Anahtar Kavram

Predicting reaction spontaneity and selective oxidation using standard electrode potentials (EE^\circ).
Tahmini Süre:2m 0s
Soru 8299Soru

When excess copper turnings are reacted with dilute trioxonitrate(V) acid (HNO3\text{HNO}_3), 1.12 dm31.12\text{ dm}^3 of nitrogen(II) oxide (NO\text{NO}) gas, measured at STP, is evolved. What mass of copper metal was oxidized during this reaction?

(Molar mass of Cu=64 g mol1, molar volume of gas at STP=22.4 dm3 mol1)(\text{Molar mass of Cu} = 64\text{ g mol}^{-1},\text{ molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1})

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Cevap: 4.8 g4.8\text{ g}

Cevap

4.8 g4.8\text{ g} of copper metal was oxidized.
The option specifying 4.8 g4.8\text{ g} is correct because the balanced chemical reaction of copper turnings with dilute trioxonitrate(V) acid is 3Cu+8HNO33Cu(NO3)2+2NO+4H2O3\text{Cu} + 8\text{HNO}_3 \rightarrow 3\text{Cu(NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}. Converting 1.12 dm31.12\text{ dm}^3 of NO\text{NO} at STP gives 0.05 mol0.05\text{ mol}. Multiplying by the stoichiometric ratio 32\frac{3}{2} gives 0.075 mol0.075\text{ mol} of Cu\text{Cu}, which equals 0.075×64=4.8 g0.075 \times 64 = 4.8\text{ g}.

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1
Write the balanced chemical equation for the reaction of copper with dilute trioxonitrate(V) acid.
3Cu(s)+8HNO3(aq)3Cu(NO3)2(aq)+2NO(g)+4H2O(l)3\text{Cu}(s) + 8\text{HNO}_3(aq) \rightarrow 3\text{Cu(NO}_3)_2(aq) + 2\text{NO}(g) + 4\text{H}_2\text{O}(l)
Dilute HNO3\text{HNO}_3 acts as an oxidizing agent, reducing to nitrogen(II) oxide (NO\text{NO}) gas rather than hydrogen gas.
2
Calculate the moles of nitrogen(II) oxide (NO\text{NO}) gas produced at STP.
Moles of NO=1.12 dm322.4 dm3 mol1=0.05 mol\text{Moles of NO} = \frac{1.12\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.05\text{ mol}
At STP, one mole of any ideal gas occupies 22.4 dm322.4\text{ dm}^3.
3
Use the mole ratio to determine moles of copper oxidized.
Moles of Cu=0.05 mol NO×(3 mol Cu2 mol NO)=0.075 mol Cu\text{Moles of Cu} = 0.05\text{ mol NO} \times \left(\frac{3\text{ mol Cu}}{2\text{ mol NO}}\right) = 0.075\text{ mol Cu}
The stoichiometric ratio between Cu\text{Cu} and NO\text{NO} is 3:2.
4
Convert moles of copper to mass in grams.
Mass of Cu=0.075 mol×64 g mol1=4.8 g\text{Mass of Cu} = 0.075\text{ mol} \times 64\text{ g mol}^{-1} = 4.8\text{ g}
Mass is calculated by multiplying the number of moles by the molar mass.

Anahtar Kavram

Stoichiometry of redox reactions involving transition metals and oxidizing acids
Soru 8300Soru

Match each atmospheric chemical species or pollutant listed on the left with its corresponding atmospheric role or environmental impact listed on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Chlorofluorocarbons (CFCs)
Carbon(IV) oxide (CO2\text{CO}_2)
Stratospheric ozone (O3\text{O}_3)
Methane (CH4\text{CH}_4)

Eşleşmeler

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Cevap

Chlorofluorocarbons pair with solar UV photolysis releasing chlorine free radicals; Carbon(IV) oxide pairs with absorbing terrestrial infrared radiation in the lower troposphere; Stratospheric ozone pairs with filtering solar ultraviolet radiation; Methane pairs with being a potent greenhouse gas from anaerobic decomposition.
Each chemical species is accurately matched to its distinct atmospheric layer and chemical property: Chlorofluorocarbons photolyze into free radicals that destroy ozone, Carbon(IV) oxide absorbs Earth's infrared heat, Stratospheric ozone shields the surface from UV radiation, and Methane acts as a greenhouse gas from biological anaerobic processes.

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1
Identify the primary environmental mechanism associated with Chlorofluorocarbons (CFCs).
CFCs are photolyzed by UV light in the stratosphere, generating atomic chlorine radicals that deplete ozone.
CFC molecules are unreactive in the troposphere but breakdown under high-energy UV light in the stratosphere.
2
Determine the atmospheric function of Carbon(IV) oxide (CO2\text{CO}_2).
CO2\text{CO}_2 absorbs re-radiated heat (infrared radiation) emitted by Earth, causing tropospheric warming.
CO2\text{CO}_2 molecules possess vibrational modes that absorb thermal infrared wavelengths.
3
Determine the protective role of Stratospheric Ozone (O3\text{O}_3).
Stratospheric ozone absorbs biological harmful UV-B radiation.
The Chapman mechanism demonstrates how photolysis and reformation of ozone absorb solar UV light.
4
Identify the primary source and impact of Methane (CH4\text{CH}_4).
Methane is a strong greenhouse gas emitted from anaerobic habitats like wetlands and livestock digestion.
Methanogenic bacteria produce CH4\text{CH}_4 under anaerobic conditions.

Anahtar Kavram

Distinguishing the chemical roles and environmental impacts of atmospheric pollutants causing global warming versus those causing ozone layer depletion.
Tahmini Süre:1m 30s
ÖncekiSayfa 415 / 697Sonraki
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