Interactive Chemistry Exam — Revised Version

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🔋 First: Electrochemistry (galvanic and electrolytic cells)

1Which of the following describes what happens when a sulfate solution is electrolyzed? Copper II using pure copper electrodes?

The mass of the anode increases and the concentration of copper ions in the solution does not change.
The mass of the cathode increases and the concentration of copper ions in the solution does not change.
The mass of the anode decreases and the concentration of copper ions in the solution decreases.
The mass of the cathode decreases and the concentration of copper ions in the solution increases.

✅ Correct answer: The mass of the cathode increases and the concentration of copper ions in the solution does not change.

When two pure copper electrodes are used in the electrolysis of a copper sulphate solution, it occurs at the same time:

• At the anode (+): Copper is oxidized and dissolves in the solution: Cu(s) → Cu²⁺(aq) + 2e⁻

• At the cathode (-): copper ions are reduced and precipitated: Cu²⁺(aq) + 2e⁻ → Cu(s)

Since the number of moles of Cu²⁺ that dissolve from the anode = the number of moles of Cu²⁺ that deposits on the cathode, Ion concentration Copper in solution remains stableWhile the cathode mass increases (due to precipitation) and the anode mass decreases (due to dissolution) In the same amount.

2Electrolytic cell for purifying a metal from impurities, virtual symbols (W,Y,X) expresses the metal and impurities in no order and their efforts are as follows:

X(s) → X²⁺(aq) + 2e⁻ , E° = +0.34V

Y(s) → Y²⁺(aq) + 2e⁻ , E° = -0.34V

W(s) → W²⁺(aq) + 2e⁻ , E° = -1.42V

Which of the following describes the metal to be purified and its impurities?

The metal to be purified (Y) and the impurities that fall out (X, W).
The metal to be purified (W) and the impurities to be dissolved (Y,X).
The metal to be purified (W) and the impurities that fall out (Y,X).
The metal to be purified (X) and the impurities to be dissolved (Y, W).

✅ The correct answer: the metal to be purified (W) and the impurities to be dissolved (Y, X).

the right answer: Option (2) The metal to be purified is (W) and the impurities that dissolve are (Y, X).

In an electrorefining cell, the ore (impure metal) is placed as an anode, and the pure metal to be obtained is deposited on Cathode.

The metal given with the lowest standard oxidation potential (which is... W With effort -1.42V, which means that it is the most Tendency to reduce and easiest to precipitate as a pure metal at the cathode) is the metal to be purified. While other impurities accompanying him (X With effort +0.34V and Y With effort -0.34V) They have higher oxidation potentials and therefore dissolve easily in solution at the anode and remain dissolved as ions without precipitating.

Rule: The metal to be purified that is deposited at the cathode is the metal with the lowest oxidation potential (highest in oxidation potential). reduction), while the more active impurities (higher oxidation potential) dissolve and remain in solution.

3Of the following reactions, which of the following represents compounds (X,Z)?

CH2Br-CF3 + Cl-Cl --400°C--> X + HCl

C2H2 + H2O --HgSO4/H2SO4 (40%-60%)--> Y (مركب ثابت) --أكسدة--> Z

(Z): An acid used to preserve fruits and vegetables.
(X): A hydrocarbon used as a safe anesthetic.
(Z): An acid used in the preparation of marukh oil.
(X): A hydrocarbon derivative used as a safe anesthetic.

✅ Correct answer: (X) A hydrocarbon derivative used as a safe anesthetic

The compound CH2Br-CF3 It reacts with chlorine Cl2 at 400°C in a substitution (halogenation) reaction, dissolving an atom Chlorine replaces a remaining hydrogen atom:

CH2Br-CF3 + Cl2 --400°C--> CHClBr-CF3 (X) + HCl

The output (X) is Halothane CHBrClCF3, a halogenated hydrocarbon derivative used in medicine As a safe anesthetic Alternative About toxic chloroform.

As for the second equation: ethene reacts with water (catalytic hydration) to produce acetaldehyde (Y), then it is oxidized to produce Acetic acid (Z) used in food preservation.

4The molecular formula C6H12 has two unsaturated isomers in it The basic carbon chain is made up of only four carbon atoms. Which of the following choices expresses the number of these isomers? And the IUPAC name for one of them?

Four isomers, 3,3-dimethyl-1-butene.
Four isomers, 3-ethyl-1-butene.
Three isomers, 2-ethyl-1-butene.
Three isomers, 3,3-dimethyl-1-butene.

✅ The correct answer: four isomers, 3,3-dimethyl-1-butene

Scientific interpretation (in Arabic):

What is required is to find the number of isomers of (unsaturated) alkenes for the formula C₆H₁₂ Provided that it is the continuous and longest carbon chain containing The double bond consists of only 4 carbon atoms (i.e. the butene chain).

Since the main chain has 4 carbon atoms, and 2 atoms are left out of 6, they are added as branches (methyl or ethyl) on The chain is no more than 4 carbon atoms in length. The correct isomers are:

1- Biotin as a branch carrier:
1. 2,3-dimethyl-1-butene (2,3-dimethyl-1-butene)
2. 3,3-dimethyl-1-butene (3,3-Dimethyl-1-butene)
3. 2-ethyl-1-butene (2-ethyl-1-butene)
2- Biotin as a branch carrier:
4. 2,3-dimethyl-2-butene (2,3-dimethyl-2-butene)

Thus, the total number of isomers is 4 isomersThe correct name mentioned for one of them in the options is 3,3-Dimethyl-1-butene.

note: The designation "3-ethyl-1-butene" in the second option is incorrect, because the longest chain in it becomes 5 carbon atoms (Two girls).

5From the following figure, if you know that (A) reduces (B⁺), which of the following Is it true?

The number of moles oxidized equals the number of moles reduced.
The mass of the electrode (A) decreases, electrons move from (A) to (B).
The number of moles oxidized is three times the number of moles reduced.
The mass of electrode (B) decreases, electrons move from (B) to (A).

✅ The correct answer: The mass of (A) decreases, and electrons move from (A) to (B).

Since (A) reduces (B⁺), this means that (A) is the reducing agent and is itself oxidized, while (B⁺) is reduced to (B).

A(s) → A³⁺(aq) + 3e⁻ (oxidation at A - anode) 3B⁺(aq) + 3e⁻ → 3B(s) (reduction at B - cathode) -------------------------------- A(s) + 3B⁺(aq) → A³⁺(aq) + 3B(s)

So electrode A loses electrons (its mass decreases), and electrons move from A to B through the external wire.

6While charging a lithium-ion battery, which of the following is true?

The anode mass increases and Co³⁺ ions are oxidized to Co⁴⁺.
The mass of the cathode decreases and the Co³⁺ ions are reduced to Co⁴⁺.
The mass of the anode decreases and the concentration of lithium ions in the electrolyte increases.
The mass of the cathode increases and the concentration of lithium ions in the electrolyte decreases.

✅ Correct answer: The mass of the anode increases and Co³⁺ ions are oxidized to Co⁴⁺.

During operation shipping The reactions go in the opposite direction:

LiCoO2 + C6 --charging--> LiC6 + CoO2

When charging, lithium ions migrate from the cathode to the anode (graphite) and enter it, making the anode mass increase, While cobalt is oxidized from +3 to +4 at the other electrode.

During shipping: LiCoO₂ → CoO₂ + Li⁺ + e⁻ C₆ + Li⁺ + e⁻ → LiC₆

7A galvanic cell (X) consists of two poles (A and B), and electrons move in the outer wire from (B) to (A).

When the following reaction occurs, which of the following choices is represented by cell (X)?

A(s) + B²⁺(aq) → A²⁺(aq) + B(s)

Electrolytic and emf are negative, and electrode (A) is negative.
Galvanic and emf are positive, and electrode (A) is negative.
Electrolytic and emf are negative, and electrode (B) is negative.
Galvanic and emf are positive, and electrode (B) is positive.

✅ Correct answer: The cell is electrolytic, the emf is negative, and electrode (B) is negative.

Correct answer: Option (C).

In the original galvanic cell, electrons are transferred from... B to A; Therefore, B is the anode and A is the cathode, and the direction of the spontaneous reaction is opposite to the given reaction.

The given reaction requires:

A(s) → A²⁺(aq) + 2e⁻ → oxidation at A B²⁺(aq) + 2e⁻ → B(s) → reduction at B

So this trend does not happen automatically in the galvanic cell, but rather it requires an external source of electricity. The cell turns into... Electrolytic cell The emf is negative. In the electrolytic cell, the cathode (the site of reduction) is negative, i.e Pole B is negative.

8Three metals (A, B, C): The metal (C) acts as cathodic protection for the metal (A), metal (B) acts as cathodic protection for metal (A), metal (B) acts as cathodic protection for metal (C)

The emf of the cell consisting of (A,C) is equal to the emf of the cell consisting of (B,A).
The emf of the cell consisting of (C,B) is greater than the emf of the cell consisting of (B,A).
The emf of the cell consisting of (B,A) is less than the emf of the cell consisting of (C,A).
The emf of the cell consisting of (B, A) is greater than the emf of the cell consisting of (C, A).

✅ The correct answer: The emf of the cell consisting of (B, A) is greater than the emf of the cell consisting of (C, A).

Cathodic protection means that the more active metal protects the less active metal. Order of activity in descending order: B > C > A.

The greater the difference in activity between the two poles of the cell, the greater the emf value. The difference between B and A is greater than the difference between C and A:

emf(B,A) > emf(C,A)

🧪 Second: Chemical reactions and inorganic compounds

9(A) and (B) Carboxylic acids

(A): The simplest aromatic carboxylic acid reacts substitutively with some strong metallic oxyacids.

(B): A carboxylic acid whose sodium salt is used in the preparation of the simplest organic compounds.

Which of the following choices expresses each of the two acids?

(A) is stronger than (B) and reduces to toluene, and (B) is more volatile than (A) and reduces to ethanol.
(A) is weaker than (B) and is used in the preparation of gasoline, and (B) is less volatile than (A) and is used in the preparation of methane gas.
(A) is poorly soluble in water and is produced by the oxidation of toluene, and (B) is dissolved in water and reduced to methanol.
(A) It is poorly soluble in water and is produced by oxidation Ar–CH₂OH(B) dissolves in water and reduces to... R–CH₂OH.

✅ The correct answer: (A) It is poorly soluble in water and is produced by the oxidation of toluene. (B) It dissolves in water and is reduced to ethanol.

Correct answer: Option (D).

Acid (A) is... Benzoic acid C₆H₅COOH; It is poorly soluble in water and is produced by the oxidation of benzyl alcohol Ar–CH₂OH. It can also participate in substitution reactions on the aromatic ring with some strong oxygenic acids.

Acid (B) is Ethanoic (acetic) acid CH₃COOH; It dissolves in water, and is reduced to primary alcohol of the form R–CH₂OH, i.e. ethanol. Its sodium salt is used to prepare methane.

C₆H₅CH₂OH —(oxidation)→ C₆H₅COOH CH₃COOH —(reduction)→ CH₃CH₂OH

10Upon acid hydrolysis of the compound C6H5CH2OOCC6H5, two compounds are produced (A,B). Which of the following is true?

(A): benzoic acid, (B): aromatic alcohol.
(A): Benzic acid, (B): Aliphatic alcohol.
(A): carbolic acid, (B): aromatic alcohol.
(A): carbolic acid, (B): aliphatic alcohol.

✅ The correct answer: (A) benzoic acid, (B) benzyl alcohol (aromatic)

The compound is an ester (benzyl benzoate). Upon acid hydrolysis of the ester, a carboxylic acid and an alcohol are produced:

C6H5COOCH2C6H5 + H2O → C6H5COOH (A) + C6H5CH2OH (B)

Benzoic acid is an aromatic acid, and benzyl alcohol is considered an aromatic alcohol because it is bonded to benzene.

11Study the following chart and decide which of the following is true

(B): 2-chloro-2-methylpropane, (C): 3-methyl-2-butanol.
(A): 2-methylpropene, (B): isobutyl chloride.
(A): 2-methylpropene, (C): 2-methyl-1-propanol.
(A): 2-methyl-2-butene, (B): 2-chloro-2-methylbutane.

✅ Correct answer: (A) 2-methyl-2-butene, (B) 2-chloro-2-methylbutane, (C) 2-methyl-2-butanol

Correct answer: Option (D).

Compound (A) is 2-methyl-2-butene. By adding HCl to the double bond according to Markovnikov's rule, compound (B) is produced: 2-Chloro-2-methylbutane.

The acid hydrolysis of (A) gives tertiary alcohol (C), which is... 2-methyl-2-butanol; It is transformed by Lucas's detector HCl/ZnCl₂ To the halide itself (B).

CH₃–C(CH₃)=CH–CH₃ + HCl → CH₃–C(Cl)(CH₃)–CH₂–CH₃

CH₃–C(CH₃)=CH–CH₃ + HCl → CH₃–C(Cl)(CH₃)–CH₂–CH₃ CH₃–C(CH₃)=CH–CH₃ + H₂O → CH₃–C(OH)(CH₃)–CH₂–CH₃

12Which of the following expresses the correct order of chemical processes necessary to obtain a compound used as a raw material in the preparation of disinfectants?

Ethene polymerization - acylation - oxidation.
Ethanoic acid reduction - dehydration - reaction with H2O2.
Ethene polymerization - halogenation - alkaline hydrolysis.
Chlorination of benzene - halogenation - alkaline hydrolysis - polymerization.

✅ The correct answer: ethene polymerization - halogenation - alkaline hydrolysis.

3 CH≡CH (ethyne) --polymerization--> C6H6 (benzene) C6H6 --halogenation/chlorination--> C6H5Cl (chlorobenzene) C6H5Cl --alkaline hydrolysis (conc. NaOH, high pressure)--> C6H5OH (phenol)

🧬 Third: Organic compounds - isomers and functional groups

13Two organic compounds (X,Y) have the same molecular formula C4H8O

(X): Decolorizes acidified potassium permanganate solution
(Y): Reduced to a secondary alcohol
Which of the following expresses (X) and (Y)?

X = 2-Methylpropanal;Y = 2-methyl-1-propanol
X = 2-Methylpropanal;Y = Butanone
X = 2-Methylpropanol;Y = 2-Butanol
X = Butanal;Y = Diethyl ether

✅ Correct answer: X = 2-methylpropanal (aldehyde), Y = butanone (ketone)

Correct answer: Option (B).

Compound (X) is 2-Methylpropanal; It is an aldehyde that is easily oxidized, so it is discolored by acidified potassium permanganate.

Compound (Y) is Butanone; It is a ketone, and upon reduction it turns into a secondary alcohol, 2-butanol:

CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃

(CH₃)₂CHCHO + [O] → (CH₃)₂CHCOOH CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃

14From the following diagram if you know that (X) contains only hydrocarbon On two hydrogen atoms

(Y) Produced from the oxidation of a secondary alcohol.
(Y) is produced by the complete oxidation of compound (Z).
(Z) The simplest alkenes are prepared from it.
(W) Isomer of compound (Z).

✅ The correct answer: (Z) The simplest alkenes are prepared from it.

Correct answer: Option (C).

Since (X) is a hydrocarbon whose molecule contains two hydrogen atoms, X = C₂H₂ (ethene). By acid hydrolysis, (Y) ethanal is formed, then by reduction, (Z) ethanol is formed:

C₂H₂ —(acid hydration)→ CH₃CHO —(reduction)→ CH₃CH₂OH

Ethanol can be removed from water to produce the simplest alkene, ethene:

CH₃CH₂OH —(dehydration)→ CH₂=CH₂ + H₂O

So the correct statement is that (Z) The simplest alkene is prepared from it. As for (W), it is ethene and its formula is C₂H₄, so it is not an isomer of ethanol (Z).

15Study the following compound, which of the following is true?

1-chloro-3-methylbenzene

It is prepared by alkylation of benzene and then reaction with chlorine in the presence of a catalyst.
It is prepared by reacting benzene with chlorine in the presence of a catalyst, then the Friedel-Crafts reaction.
It is called according to the IUPAC system [1-chloro-3-methylbenzene].
Its common name is orthochlorotoluene.

✅ The correct answer: It is called according to the IUPAC system [1-chloro-3-methylbenzene]

The compound is a benzene ring bearing the CH3 and Cl groups in the meta (1,3) position. According to IUPAC nomenclature: 1-chloro-3-methyl Gasoline. The correct common name for this situation is "metachlorotoluene" and not "orthochlorotoluene".

16Cyclic hydrocarbons (A and B) each contain the same number of atoms carbon; (A) Saturated aliphatic, (B) aromatic

(A) cyclohexane, (B) toluene.
(A) It is produced by the hydrogenation of benzene, (B) It has a higher boiling point than gasoline.
(A) Methylcyclohexane, (B) Benzene.
(A) Produced from the hydrogenation of toluene, (B) Produced from the catalytic reformation of heptane.

✅ Answer: (A) = methylcyclohexane, (B) = toluene, and (A) is produced from the hydrogenation of (B)

Correct answer: Option (D).

For the two compounds to contain the same number of carbon atoms, it is:

B = toluene C₆H₅CH₃ (aromatic hydrocarbon, 7 carbon atoms) A = methylcyclohexane C₆H₁₁CH₃ (saturated cycloaliphatic hydrocarbon, 7 carbon atoms)

(A) is formed by the hydrogenation of toluene, while (B) is formed by the catalytic reformation of heptane.

C₆H₅CH₃ + 3H₂ → C₆H₁₁CH₃ C₇H₁₆ —(catalytic reformation)→ C₆H₅CH₃ + 4H₂

🔬 Fourth: The periodic table, ionic bonds and interactions

17The representative element (X) is the following alloy

The representative element (X) is the following alloy:

-Penmetallic alloy with element (Y).

– An alloy used in the manufacture of MiG fighter aircraft with the element (Z).

– An alloy used in the manufacture of spacecraft with the element (L).

Which of the following represents the elements (X, Y, Z, L)?

Question 17 options
(L)	(Z)	(Y)	(X)
Au	Sc	Ti	Pb
Sc	Fe	Ti	Al
Cr	Ni	Fe	C
Ti	Sc	Ni	Al

✅ Correct answer: Fourth row (X=Al, Y=Ni, Z=Sc, L=Ti)

Aluminum (Al) forms a regular alloy with nickel, and a titanium-aluminum alloy is used in fighters with... Scandium, aluminum-titanium in spacecraft.

18Which of the following does not represent one of the products of heating a bicarbonate solution? Calcium?

A compound that does not dissolve in mineral acids.
A compound that does not dissolve in water.
A gas that does not oxidize in air.
Another salt of the same acid.

✅ The correct answer: A compound that does not dissolve in mineral acids — this statement does not express any product

Ca(HCO3)2 --heating--> CaCO3↓ + H2O + CO2↑

It produces calcium carbonate (CaCO3), which does not dissolve in water It dissolves in acidsSo the wrong statement is that It says the product "does not dissolve in mineral acids."

19Which of the following does not distinguish between magnesium sulphate and chloride? Magnesium?

Copper nitrate solution.
Bium chloride solution.
Hot concentrated sulfuric acid.
Calcium bicarbonate solution.

✅ The correct answer: copper nitrate solution (does not distinguish between them)

The copper nitrate solution does not distinguish between the two salts: No distinct precipitate is formed with both MgSO₄ and MgCl₂. As for BaCl₂, it precipitates sulphates, hot concentrated sulfuric acid affects chloride, and calcium bicarbonate distinguishes between them by forming a precipitate of calcium sulphate with sulphates.

20From the following diagram, which of the following represents the materials (X, Y, Z)?

Choose the row that matches the X, Y, and Z values in the chart.

Question options 20
Z	Y	X
Na2SO4(aq)	AgCl(s)	AgNO3(aq)
Na2CO3(aq)	CuS(s)	Na2S(aq)
Na2CO3(aq)	AgCl(s)	AgNO3(aq)
Na2SO4(aq)	CuS(s)	HCl(aq) + H2S(g)

✅ Correct answer: Z = Na2CO3, Y = CuS (black precipitate), X = Na2S

Correct answer: Option (B). Sodium sulfide reacts with copper(II) chloride to form black copper sulfide X = Na₂S andY = CuS. Sodium carbonate reacts with copper(II) chloride to form a copper carbonate precipitate Z = Na₂CO₃.

Na₂S(aq) + CuCl₂(aq) → CuS(s)↓ + 2NaCl(aq) Na₂CO₃(aq) + CuCl₂(aq) → CuCO₃(s)↓ + 2NaCl(aq)

⚖️ Fifth: Chemical balance and acidity

21Study the following equilibrium reaction in a closed container

ICl(g)Dark brown + Cl₂(g) ⇌ ICl₃(g)yellow

When the pot is placed in a basin of hot water, the brown color becomes more intense.

Which of the following represents the type of reaction (exothermic or endothermic) and change that leads to the reduction in browning?

Exothermic - reduce pressure.
Endothermic - increasing the volume of the vessel.
Exothermic - reduce pot size.
Endothermic - increased pressure.

✅ Correct answer: exothermic - reducing the volume of the container

Correct answer: (c) exothermic—reducing the volume of the container.

color ICl(g) Dark brown, while color ICl₃(g) yellow. The increase in brown color when the pot is heated means that the equilibrium shifts to the left when the temperature is raised; That is, the reverse direction is endothermic, and hence it is The forward reaction is exothermic.

To reduce the intensity of the brown color, the balance must be shifted to the right to form ICl₃(g). On the left side are two moles of gas, and on the right side is one mole; So it leads Reduce the size of the container (increasing pressure) shifts it to the right and reduces the amount of brown ICl.

22Two solutions (X) and (Y) of weak bases

The concentration of both solutions is equal 0.1 M. If the values of the basic ionization constant are as follows:

For the solution (X)K_b = 1.8 × 10⁻⁵

For the solution (Y)K_b = 3.8 × 10⁻⁸

When each of them is diluted until it becomes concentrated 0.01 MWhich of the following is true?

The concentration of hydroxide ions of the solution (X) decreases and becomes equal to the concentration of hydroxide ions of the solution (Y).
The hydroxide ion concentration of the solution (X) decreases and remains less than the hydroxide ion concentration of the solution (Y).
The value of (α) for solution (X) increases and remains less than the value of (α) for solution (Y).
The value of (α) for solution (X) increases and remains greater than the value of (α) for solution (Y).

✅ Correct answer: The value of (α) increases for both solutions, but α for solution (X) remains greater than α for solution (Y)

Ostwald's Dilution Law: When any weak electrolyte is diluted, the degree of ionization (α) always increases. Since Kb is for solution (X) Greater than Kb for solution (Y), the degree of ionization of (X) remains greater than (Y) at any concentration:

α = √(Kb/C) α(X) = √(1.8×10⁻⁵/C) > α(Y) = √(3.8×10⁻⁸/C) at any C

23100mL of solution containing 0.001mol HCl was added to 100mL of pure water at 25°C

The pH of the water increases by 5.
The pH of the water decreases by 5.
The pOH of water decreases by 4.7.
The pOH of water increases by 4.7.

✅ Correct answer: The pOH of water increases by 4.7

[H3O+]1 × V1 = [H3O+]2 × V2 0.01 × 0.1 = [H3O+]2 × 0.2 [H3O+]2 = 0.005 M pH = -log(0.005) = 2.3 pOH = 14 - 2.3 = 11.7 Change in pOH = 11.7 - 7 = 4.7 (increase)

24Three equal volumes of saturated solutions of salts that are poorly soluble in water at 25°C

The dissolution constants for saturated solutions are:

(1)AgClK_sp = 1.0 × 10⁻¹⁰

(2)PbI₂K_sp = 4.0 × 10⁻¹²

(3)Ag₂SO₄K_sp = 0.0976 × 10⁻⁵

Which of the following expresses the order of these solutions according to their ability to conduct electricity?

3 > 2 > 1
3 > 1 > 2
1 > 2 > 3
2 > 1 > 3

✅ Correct answer: 3 > 2 > 1

AgCl (1): Ksp = S² S = √(1.0 × 10⁻¹⁰) = 1.0 × 10⁻⁵ M PbI₂ (2): Ksp = 4S³ S = ∛(4.0 × 10⁻¹² ÷ 4) = 1.0 × 10⁻⁴ M Ag₂SO₄ (3): Ksp = 4S³ S = ∛(0.0976 × 10⁻⁵ ÷ 4) = 6.25 × 10⁻³ M

Electrical conductivity depends on the actual ionic concentration, and the highest ionic concentration occurs in Ag2SO4 solution due to its solubility The largest number of ions produced, so the correct order is: 3 > 2 > 1.

🧮 Sixth: Chemical calculations (molar mass and concentrations)

25A sample of double hydrated salt was heated until its mass was constant

K₂SO₄ · Al₂(SO₄)₃ · xH₂O

Sample mass before heating4.62 g

The mass of the sample after heating until the mass is constant3.256 g

K₂SO₄ = 174 | Al₂(SO₄)₃ = 342 | O = 16 | H = 1

Which of the following expresses the x value and mass of potassium sulfate in the sample?

x = 24block K₂SO₄ = 1.098 g
x = 12block K₂SO₄ = 1.098 g
x = 24block K₂SO₄ = 2.158 g
x = 12block K₂SO₄ = 2.158 g

✅ Correct answer: X = 12, mass of potassium sulfate = 1.098g

Mass of water of crystallization lost:

4.620 − 3.256 = 1.364 g

Molar mass of anhydrous salt K₂SO₄·Al₂(SO₄)₃ It is equal to 516 g/mol.

n(H₂O) = 1.364 ÷ 18 = 0.07578 mol n(anhydrous salt) = 3.256 ÷ 516 = 0.00631 mol x = 0.07578 ÷ 0.00631 ≈ 12 m(K₂SO₄) = 3.256 × (174 ÷ 516) = 1.098 g

26A 2.9g sample of iron(II) chloride was dissolved in water, and an abundance of sodium hydroxide solution was added to its solution

[Fe=56, Cl=35.5, Na=23, O=16, H=1]
Which of the following expresses the mass of the resulting precipitate?

0.985 g
5.91 g
2.055 g
3.94 g

✅ The correct answer: 2.055 g

FeCl2 + 2NaOH → Fe(OH)2↓ + 2NaCl Molar mass FeCl2 = 127 g/mol Molar mass Fe(OH)2 = 90 g/mol 127 g FeCl2 -----> 90 g Fe(OH)2 2.9 g FeCl2 -----> X X = (2.9 × 90) / 127 = 2.055 g

🌡️ Seventh: Thermodynamic equilibrium

27Which of the following represents a system that does not reach equilibrium?

A saturated solution of a poorly soluble salt in a sealed container.
Dissolving carbon dioxide in water in an airtight container.
Dissolve acetic acid in water in an open container.
Dissolution of hydrogen peroxide in an open container.

✅ The correct answer: dissolution of hydrogen peroxide in an open container

The decomposition of hydrogen peroxide in an open container is an irreversible reaction (it does not reach equilibrium) because the resulting oxygen rises and escapes from the Open system.

28Study the two tubes (A and B) in the figure shown, which of the following describes What is the result of the reaction in each of the two tubes?

The speed of disappearance of the piece of zinc in tube (B) is greater than the speed of its disappearance in (A).
The amount of hydrogen gas produced at the end of the reaction in (B) is greater than in (A).
The amount of hydrogen gas produced at the end of the reaction in (B) is less than in (A).
The speed of disappearance of the piece of zinc in tube (A) is greater than the speed of its disappearance in (B).

✅ The correct answer: The speed at which a piece of zinc disappears in tube (A) is greater

Tube (A) contains a higher concentration of HCl (1M) compared to (B) (0.5M), and the reaction speed is directly proportional to the concentration. The reaction in (A) is faster.

🧊 Eighth: Electronic structure and oxidation states

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)↑

29The following table shows the electronic structure of some ions Items

The electronic structure of the ions of the elements in question 29
ion	W²⁺	Z⁴⁺	Y³⁺	X²⁺
Electronic composition	5s²,4d¹⁰,5p⁶	4s²,3d¹⁰,4p⁶	3s²,3p⁶	5s⁰,4d¹⁰

Which of the following is true?

Element (X) is non-transitional, element (Y) is non-transitional.
Element (X) is non-transitional, element (Z) is non-transitional.
Element (Z) is non-transitional, element (W) is non-transitional.
Element (Y) is non-transitional, element (W) is non-transitional.

✅ The correct answer: element (Z) is non-transitional, element (W) is non-transitional

W (barium), a non-transition alkaline earth metal, Z (germanium), a representative non-transition element, Y (scandium), a transition element, (Silver) A transition element.

30Which of the following compounds in their solutions can reduce an ion? Manganese (Mn⁷⁺) in acidic medium easily?

ZnCl2
Ti(NO3)4
ScCl3
FeSO4

✅ The correct answer: FeSO4

FeSO4 contains the Fe²⁺ ion which is easily oxidized to Fe³⁺, acting as a reducing agent that reduces permanganate (Mn⁷⁺) to Mn²⁺.

Simplified representation of the reduction of manganese ion by Fe²⁺ in acidic media

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

31Two transition elements (X,Y) of the first transition series

(X): One of its oxides is used as a catalyst in the manufacture of superconducting magnets.

(Y): Its atom contains a number of single electrons equal to that of the element preceding it in the transition series The first.

The density of the element (Y) is less than the density of the element preceding it in the series.
The effective nucleus charge of element (Y) is less than that of element (X).
The atomic mass of element Y is less than the next element in the series.
The magnetic moment of the ion (X³⁺) is less than the magnetic moment of the ion (Y³⁺).

✅ Correct answer: The magnetic moment of the ion (X³⁺) is less than the magnetic moment of the ion (Y³⁺)

(X), one of whose oxides (V2O5) is used as a catalyst in the manufacture of superconducting magnets is Vanadium V. The magnetic moment is directly proportional to the number of individual electrons:

μ = √n(n+2) where n = number of unpaired electrons

Since the Y³⁺ ion has more single electrons, its magnetic moment is greater than the (X³⁺) ion.

μ = √n(n + 2) BM Where n is the number of single electrons.

32(X,Y) express operations performed during the preparation of a... Iron ores

Process (X): results in increasing the size of iron ore grains.

Process (Y): results in an increase in the number of lone electrons in the iron ion.

Question options 32
Y	X
Surface tension	Cracking
Hematite roasting	Sintering
Siderite roasting	Sintering
Magnetic separation	Cracking

✅ Correct answer: (X) sintering, (Y) siderite roasting

Felting(X): It collects fine iron ore grains into larger blocks that allow gases to pass through the blast furnace.

Siderite roasting (Y): When siderite (FeCO3) is heated in air, Fe²⁺ (4 lone electrons) is oxidized to Fe³⁺ (5 single electrons):

4FeCO3 + O2 --roasting--> 2Fe2O3 + 4CO2

📊 Ninth: Calculations of oxidation, reduction, and quantitative equilibrium

33When an amount of electricity is passed through the two connected cells straight

The first contains chromium oxide magma II (CrO), and the second contains chromium oxide magma III (Cr2O3), which rises at the anode. The first cell (16g) of oxygen gas [O=16]. Which of the following represents the number of moles of chromium deposited in each of the cells?

Question options 33
The second cell	First cell
1mol	2/3 mol
2/3 mol	1mol
1mol	1mol
2/3 mol	2/3 mol

✅ The correct answer: The first cell = 1 mol, the second cell = 2/3 mol

At anode: 2O²⁻ → O2 + 4e⁻ 16g O2 = 0.5 mol O2 moles of electrons = 0.5 × 4 = 2 mol e⁻ (same charge passes through both cells - series) Cell 1 (Cr²⁺ + 2e⁻ → Cr): moles Cr = 2 ÷ 2 = 1 mol Cell 2 (Cr³⁺ + 3e⁻ → Cr): moles Cr = 2 ÷ 3 = 2/3 mol

34Three organic compounds (A, B, C)

(A)An organic acid whose molecule contains one oxygen atom.

(B): A neutral hydrocarbon derivative that has a neutral effect on litmus dye, and has a weak acidic property.

(C): An aromatic carboxylic acid that contains the same functional group as (A).

Which of the following choices represents the compounds that react with each of the substances in the table?

Question options 34
Concentrated HCl	NaOH(aq)	Speak immediately with Na₂CO₃	Gives a violet color with FeCl₃
B	A,C	C	A,C
B,C	A	C	B
C	A,C	A, B	B
B	B	C	A

✅ The correct answer: first grade.

Correct answer: Option (A).

Identify the vehicles first:

A = phenol (carbolic acid) C₆H₅OH B = Ethanol (ethyl alcohol) C₂H₅OH C = salicylic acid (2-hydroxybenzoic acid) C₆H₄(OH)COOH

Accordingly, the row corresponding to the required properties and interactions is:

Concentrated HCl → B NaOH(aq) → A, C Na₂CO₃ → C FeCl₃ → A, C

So the first row of the table is selected.

35Study the following chart. Which of the following choices is correct?

Process (1) strong decomposition, (3) ethyl methanoate, (5) oxidation, (E) ethyl alcohol.
Process (1) acidolysis, (3) acetic acid, (B) reduction, (B) methyl methanoate.
Process (1) acidolysis, (D) formic acid, (2) oxidation, (E) acetic acid.
Process (1) strong hydrolysis, (D) ethyl alcohol, (2) ethyl alcohol, dry distillation, sodium methanoate.

✅ Correct answer: Process (1) acidolysis, (D) formic acid, (2) oxidation, (E) acetic acid

(A) HCOOC2H5 (ester - ethyl methanoate) --acidic hydrolysis (1)--> (D) HCOOH + (C) C2H5OH (C) C2H5OH --oxidation (2)--> (E) CH3COOH (D) HCOOH --reduction (3)--> methanol (D) + methanol --esterification (4)--> (B) HCOOCH3 (methyl methanoate - isomer of A)

36From the equilibrium equation: A(aq) + B(aq) ⇌ C(aq) + H2O(l)

100mL containing 0.5mol of substance (A) was added to 100mL containing 0.5mol of substance (B), and at equilibrium the number of moles of water formed was 0.2mol. Which of the following expresses the value of the equilibrium constant for the previous reaction?

0.444
0.456
2.222
2.250

✅ Correct answer: 0.444

A + B ⇌ C + H₂O Start: 0.5 mol 0.5 mol 0 mol 0 mol Change: −0.2 mol −0.2 mol +0.2 mol +0.2 mol At equilibrium: 0.3 mol 0.3 mol 0.2 mol 0.2 mol Total volume = 200 mL = 0.2 L [A] = [B] = 0.3 / 0.2 = 1.5 M [C] = 0.2 / 0.2 = 1.0 M Kc = [C] / ([A][B]) = 1.0 / (1.5 x 1.5) = 0.444

370.1mol of acetic acid was dissolved in water to form a solution Its concentration is 0.045M and its degree of ionization is 0.02

Which of the following expresses the value of (α) when the volume of the solution is increased to 2.5 L?

0.042
0.420
0.210
0.021

✅ Correct answer: 0.021

Ka ≈ Cα² = 0.045 × (0.02)² = 1.8×10⁻⁵ New concentration C′ = 0.1 mol ÷ 2.5 L = 0.04 M α′ = √(Ka/C′) = √(1.8×10⁻⁵ ÷ 0.04) ≈ 0.021

With dilution, the volume of the solution increases and thus the concentration decreases, which leads to an increase in the degree of ionization according to the law of dilution For Ostwald.

38Study the following chart, which of the following is correct?

The R group contains 3 methyl groups, and the alcohol is a secondary alcohol.
The R group contains two methyl groups, and the alcohol is a primary.
The R group contains two methyl groups, and the alcohol is a tertiary.
The R group contains one methyl group, and the alcohol is a secondary alcohol.

✅ Correction Note: Alcohol 2-methylpropanol-1 (primary alcohol) and group R contain two methyl groups

Correct answer: Option (B).

Since hydrogenation of compound (B) gives 2-methylpropane, (B) is 2-Methylpropene. This alkene is formed by removing water from the primary alcohol 2-methyl-1-propanol:

(CH₃)₂CH–CH₂OH —(dehydration)→ (CH₃)₂C=CH₂ —(hydrogenation)→ (CH₃)₃CH

So the set R contains Two methyl groupsAnd primary alcohol.

🔍 Tenth: Qualitative analysis and sedimentation

39A solution containing a mixture of anions: sulfate, chloride, and sulfide

Which of the following represents the correct arrangement of solutions used to separate these anions from each other in solid form?

Copper nitrate → filtration → calcium hydroxide → filtration → silver nitrate → filtration.
Silver nitrate → filtration → calcium hydroxide → filtration → copper nitrate → filtration.
Silver nitrate → filtration → copper nitrate → filtration → calcium hydroxide → filtration.
Calcium hydroxide → filtration → silver nitrate → filtration → copper nitrate → filtration.

✅ The correct answer: copper nitrate - filtration - calcium hydroxide - filtration - silver nitrate - filtration.

Correct answer: Option (A).

Silver nitrate cannot be added before the sulfide ion is removed; because Ag⁺ Both fail Cl⁻ andS²⁻Two precipitates form together and are not separated.

Therefore the correct sequence with filtration after each step is:

1) Addition of Cu(NO₃)₂: Cu²⁺ + S²⁻ → CuS(s)↓ (black precipitate) 2) After filtration, add Ca(OH)₂: Ca²⁺ + SO₄²⁻ → CaSO₄(s)↓ 3) After filtration, add AgNO₃: Ag⁺ + Cl⁻ → AgCl(s)↓

So the order is: Copper nitrate then filtration, calcium hydroxide then filtration, silver nitrate then filtration.

40The following operations were performed on compound (A) to convert it to compound (B), in order

Alkaline hydrolysis → reduction → Friedel-Crafts reaction → oxidation
Which of the following choices represents the two compounds (A, B)?

(A)1-chloro-3-methylbenzene,(B)Benzene dicarboxylic acid
(A)1-chloro-4-methylbenzene,(B)Benzene dicarboxylic acid
(A)Benzyl chloride,(B)Benzoic acid
(A)Chlorobenzene,(B)Benzoic acid

✅ The correct answer: (A) chlorobenzene, (B) benzoic acid.

For the operations to take place in the given order, compound (A) must begin with Chlorobenzene:

C₆H₅Cl —(alkaline hydrolysis) → C₆H₅OH C₆H₅OH —(reduction)→ C₆H₆ C₆H₆ —(Friedel Kraft)→ C₆H₅CH₃ C₆H₅CH₃ —(oxidation)→ C₆H₅COOH

So compound (B) is benzoic acid.

41XmL of 0.2M monoproton acid was titrated Using 30mL calcium hydroxide concentration 0.1M

YmL of 0.2M diproton acid was titrated using 30mL of 0.2M sodium hydroxide. Which of the following expresses... The relationship between (X,Y)?

Three times Y
Four times Y
X double Y
Y is half of X

✅ Correct answer: X times Y (Va = X = 30mL, Va = Y = 15mL)

Correct answer: Option (C) — X times Y.

First titration (monoprotonated acid): Ca(OH)₂ provides 2OH⁻ per mole Moles OH⁻ = 0.1 x 0.030 x 2 = 0.006 mol Moles of acid = 0.006 mol X = 0.006 ÷ 0.2 = 0.030 L = 30 mL Second titration (diproton acid): Moles OH⁻ = 0.2 x 0.030 = 0.006 mol Moles of acid = 0.006 ÷ 2 = 0.003 mol Y = 0.003 ÷ 0.2 = 0.015 L = 15 mL X / Y = 30 / 15 = 2

42Which of the following choices represents an alcohol that is oxidized by ordinary oxidizing agents, and when water is removed from it, it gives the simplest alkene with an ethyl group branched off it?

2-ethyl-1-butanol.
2-methyl-3-pentanol.
2-methyl-2-butanol.
3-ethyl-2-pentanol.

✅ The correct answer: 2-ethyl-1-butanol.

Correct answer: Option (A).

The compound 2-ethyl-1-butanol A primary alcohol, therefore it is oxidized by ordinary oxidizing agents. When water is removed from it, it is formed 2-ethyl-1-buteneIt is the simplest branched alkene containing an ethyl group.

CH₃CH₂–CH(CH₂CH₃)–CH₂OH —(dehydration)→ CH₂=C(CH₂CH₃)–CH₂CH₃ + H₂O

As for 2-methyl-2-butanol, it is a tertiary alcohol, and it is not oxidized by ordinary oxidizing agents.

43Two transition elements of the first series (X, Y), the outermost principal energy level of each containing one single electron

Which of the following describes the use of the two elements in different oxidation states?

Question options 43
Item X	Y element
In the manufacture of dyes in the oxidation state X³⁺	Electrical wires in oxidation state Y²⁺
Fungicide in oxidation state X²⁺	An oxidizing agent in the oxidation state Y³⁺
Metal plating in oxidation state X⁰	Coating of metals in oxidation state Y²⁺
In leather tanning in the oxidation state X³⁺	Purification of drinking water in the oxidation state Y²⁺

✅ The correct answer: fourth grade.

Correct answer: Option (D).

The mentioned single electron is in the outermost principal energy level 4sThen the two elements are:

X = Cr : [Ar] 3d⁵ 4s¹ Y = Cu : [Ar] 3d¹⁰ 4s¹

Chromium is used in Leather tanning In case of oxidation Cr³⁺While copper(II) ions are used Cu²⁺ in Drinking water purification. So that matches the fourth row.

44Compound (X) is produced by intense heating of two iron compounds (Y and Z) in the absence of air, and the oxidation number of iron in compound (Y) is greater than (Z).

Which of the following is true?

When HCl is added to compound (X), then an alkali is added, only compound (Y) is formed.
When HCl is added to compound (X), then an alkali is added, a mixture is formed containing compounds (Y) and (Z).
When a NaOH solution is added to compound (Z), compound (Y) is produced.
When concentrated H₂SO₄ is added to compound (X), compound (Z) is produced.

✅ The correct answer: When HCl is added to (X), then an alkali is added, a mixture containing (Y) and (Z) is formed.

Correct answer: Option (B).

From the data: Y = Fe(OH)₃ andZ = Fe(OH)₂It results from heating them in the absence of air X = Fe₃O₄:

2Fe(OH)₃ + Fe(OH)₂ —(intense heating, absence of air)→ Fe₃O₄ + 4H₂O

Magnetic iron oxide reacts with HCl to form iron(III) and iron(III) salts:

Fe₃O₄ + 8HCl → FeCl₂ + 2FeCl₃ + 4H₂O

Then adding alkali results in the formation of a mixture of (Y) and (Z):

FeCl₃ + 3NaOH → Fe(OH)₃↓ + 3NaCl FeCl₂ + 2NaOH → Fe(OH)₂↓ + 2NaCl

45Essay questionWhen coating A spoonful of metal (X), the cells are connected in a certain way

Answer the following questions:
1- Write the equation for the reaction that occurs at electrode (A).
2- Write the equation for the reaction that occurs at electrode (B).
3- Write the equation for the cathode reaction of the electrolytic cell.
4- What happens to the mass of the electrode (X) if it is connected to electrode (A) and the spoon is connected to electrode (B)?

✅ Model answer:

1) At pole (A): Electrode A is the anode in the galvanic cell, and it oxidizes:

A(s) → A²⁺(aq) + 2e⁻

2) At pole (B): Electrode B is the cathode, and its ions are reduced:

B⁺(aq) + e⁻ → B(s)

3) Cathode of electrolytic cell (spoon in drawn position):

X³⁺(aq) + 3e⁻ → X(s)

4) When electrode The X³⁺ ions are reduced and deposited on it, therefore The mass of pole X increases.

46Essay questionStudy the chart Next, note that (n=3)

Answer the following questions:
1- Write the name of the process (2).
2- Write the name of the isomer of the compound (D).
3- Write the IUPAC name resulting from adding an abundance of HBr to compound (A).
4- Explain: The boiling point (C) is higher than the boiling point (D).

✅ Model answer:

When n = 3: Compound A is propane CH₃C≡CH, (B) propanone (acetone), (C) 2-propanol, and (D) propane.

1) Process (2): shorthand; Propanone turns into 2-propanol:

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃

2) Isomer of the compound (D): Cyclopropane (cyclopropane).

3) Add abundant HBr to (A): According to Markovnikov's rule, the result is: 2,2-Dibromopropane:

CH₃C≡CH + 2HBr → CH₃CBr₂CH₃

4) Boiling point (C) is higher than (D): Because (C) is an alcohol with hydrogen bonds between its molecules, while (D) an alkene does not form hydrogen bonds.