Chemistry Test 231

CONCEPT:

Empirical Formula Calculation

• The empirical formula of a compound represents the simplest whole-number ratio of the elements present in the compound.
• To determine the empirical formula, we use the mass percentage composition to find the mole ratios of the elements.
• This typically involves the following steps:
  • Convert mass percentages to grams (assuming 100 g of the compound).
  • Convert grams to moles using atomic masses.
  • Find the simplest whole-number ratio of moles.

EXPLANATION:

The mass of oxygen in the compound:

• Total mass of compound = 64g
• Mass of carbon = 24g
• Mass of hydrogen = 8g
• Mass of oxygen = Total mass - Mass of carbon - Mass of hydrogen
• Mass of oxygen = 64g - 24g - 8g = 32g

CONCEPT:

Electron Transitions and Spectral Lines

• When an electron transitions between energy levels in an atom, it emits or absorbs photons corresponding to the energy difference between the levels.
• The spectral lines observed correspond to these energy transitions and are characteristic of the element.
• For the hydrogen atom, visible spectral lines are seen in the Balmer series, which corresponds to transitions ending at n = 2.
• When an electron falls from a higher energy level (n > 2) to n = 2, multiple transitions can occur, each resulting in a spectral line.

EXPLANATION:

• Consider the transitions from n = 5 to n = 1:
  • Direct transitions to n = 2 are part of the Balmer series and fall in the visible region.
  • Possible transitions:
    • n = 5 → n = 2
    • n = 4 → n = 2
    • n = 3 → n = 2
  • Each of these transitions will produce a spectral line in the visible region.
  • Therefore, three spectral lines are observed in the visible region for the transitions from n = 5 to n = 1.

The correct answer is  3

CONCEPT:

Ionization Energy

• Ionization energy is the energy required to remove an electron from an isolated gaseous atom or ion.
• The first ionization energy is the energy required to remove the first electron, the second ionization energy for the second electron, and so on.
• Each successive ionization energy is higher than the previous one because it is more difficult to remove an electron from a positively charged ion.
• Thus, the third ionization energy (3rd I.P.) is greater than the second ionization energy (2nd I.P.), which in turn is greater than the first ionization energy (1st I.P.).

EXPLANATION:

• We need to compare the energy required for the given transitions:
  • (a) M^-(g) → M(g) + e: This involves the release of an electron from a negatively charged ion, which generally requires less energy.
  • (b) M^-(g) → M⁺(g) + 2e: This involves the removal of two electrons from a negatively charged ion, which requires more energy than (a) but less than the ionization energies of higher positive charges.
  • (c) M⁺(g) → M²⁺(g) + e: This is the second ionization energy, which is higher than the first but lower than the third.
  • (d) M²⁺(g) → M³⁺(g) + e: This is the third ionization energy, which is the highest among the given options as it requires removing an electron from a doubly positive ion.
• Therefore, the transition that involves the maximum amount of energy is (d) M²⁺(g) → M³⁺(g) + e.

The correct answer is) M²⁺(g) → M³⁺(g) + e

CONCEPT:

Dipole Moment and Melting Point in Dichlorobenzenes

• The dipole moment of a molecule depends on the arrangement of atoms and the electronegativity differences between them.
• In disubstituted benzenes, the relative positions of substituents (ortho, meta, para) affect the dipole moment.
• The melting point is influenced by the symmetry and intermolecular forces within the crystal lattice.
• Higher symmetry generally leads to a higher melting point due to more efficient packing.

• Highest Dipole Moment:
  • The dipole moment is the vector sum of the individual bond dipoles.
  • Ortho (a) and meta (b) configurations typically have higher dipole moments due to the angles between the substituents.
  • Para (c) configuration often has a lower dipole moment as the dipoles may cancel each other out.
  • Thus, configuration (a) has the highest dipole moment.
• Highest Melting Point:
  • Symmetry plays a crucial role in determining the melting point.
  • The para (c) configuration is the most symmetrical, allowing for better packing in the solid state.
  • Thus, configuration (c) has the highest melting point.

Conclusion:-

• The correct answer is a and c

CONCEPT:

Formation of Diatomic Gases

• Certain chemical reactions produce diatomic gases, which are molecules composed of two atoms.
• Common diatomic gases include nitrogen (N₂), oxygen (O₂), and hydrogen (H₂).
• The reactions that produce these gases often involve decomposition or reduction processes.

EXPLANATION:

The correct answer is  KNO₃ + K → (with heat)

CONCEPT:

Calculation of K_a for a Weak Acid from its Salt

• When a salt of a weak acid and a strong base is dissolved in water, it dissociates completely, releasing the conjugate base of the weak acid.
• The concentration of the conjugate base can be determined from the dissociation of the salt.
• The pH or pOH of the solution can be used to find the concentration of hydrogen or hydroxide ions.
• The K_a of the weak acid can then be calculated using the relationship between the ion concentrations and the K_w (ion product of water).

EXPLANATION:

• Given:
  • Concentration of Ca(Lac)₂ = 0.13 mol in 0.50 L solution
  • pOH = 5.60
• Calculate the concentration of lactate ion (Lac^-):
  • Ca(Lac)₂ dissociates as:
    • Ca(Lac)₂ → Ca²⁺ + 2 Lac^-
  • Moles of Ca(Lac)₂ = 0.13 mol
  • Volume of solution = 0.50 L
  • Concentration of Ca(Lac)₂ = 0.13 mol / 0.50 L = 0.26 M
  • Concentration of Lac^- = 2 × 0.26 M = 0.52 M
• Calculate pH from pOH:
  • pH = 14 - pOH
  • pH = 14 - 5.60 = 8.40
• Use the formula for the salt of a weak acid and a strong base:
  • pH = 1/2 (log K_w + log K_a - log C)
• Substitute the known values:
  • K_w = 10^-14 or log K_w = -14
  • Concentration of lactate ion (C) = 0.52 M
  • pH = 8.40
  • 8.40 = 1/2 (-14 + log K_a - log 0.52)
• Simplify and solve for K_a:
  • 16.80 = -14 + log K_a - log 0.52
  • 16.80 + 14 = log K_a - (-0.284)
  • 30.80 + 0.284 = log K_a
  • log K_a ≈ 8.26
  • K_a = 10^8.26 ≈ 8.26 × 10^-4
• Therefore, the K_a of lactic acid is 8.26 × 10^-4.

The correct answer is: K_a = 8.26 × 10^-4

CONCEPT:

Determination of Atomic Weights

• To determine the atomic weights of elements in a compound, we use the molecular weight and the given mass of the compound.
• The molecular weight of a compound can be found using the formula:
  • Molecular weight = (Number of atoms of X) x (Atomic weight of X) + (Number of atoms of Y) x (Atomic weight of Y)
• Given the mass and moles of the compounds, we can set up equations to solve for the atomic weights.

EXPLANATION:

• First, set up the equations using the given data:

• So, solve these simultaneous equations:
  • From 2x + 3y = 160
  • From 3x + 4y = 232
  • Multiply the first equation by 3 and the second by 2 to align coefficients of x:
    • 6x + 9y = 480
    • 6x + 8y = 464
  • Subtract the second from the first:
    • 6x + 9y - (6x + 8y) = 480 - 464
    • y = 16
  • Substitute y = 16 back into 2x + 3y = 160 :
    • 2x + 3(16) = 160
    • 2x + 48 = 160
    • 2x = 112
    • x = 56
• Therefore, the atomic weights of X and Y are 56.0 and 16.0, respectively.

The correct answer is 56.0 and 16.0

Explanation:

To determine the pH of each solution, we first need to find the degree of dissociation using the given concentrations and dissociation constants (Ka values):

• Oxalic Acid (H₂C₂O₄): Oxalic acid is a diprotic acid with two dissociation steps. The dissociation constants are:
  • Ka1 = 5.6 × 10^-2
  • Ka2 = 5.42 × 10^-5
• Acetic Acid (CH₃COOH): Acetic acid is a weak monoprotic acid with the dissociation constant:
  • Ka = 1.8 × 10^-5

Oxalic Acid Calculation: The primary dissociation (Ka1) contributes the most to the pH of the oxalic acid solution.

For 0.1 M oxalic acid (using Ka1):

[H⁺] ≈ √(Ka1 × C) = √(5.6 × 10^-2 × 0.1) ≈ √(5.6 × 10^-3)

≈ 7.48 × 10^-2 M

pH = -log[H⁺] = -log(7.48 × 10^-2) ≈ 1.13

Acetic Acid Calculation:

For 0.1 M acetic acid:

[H⁺] ≈ √(Ka × C) = √(1.8 × 10^-5 × 0.1) = √(1.8 × 10^-6)

≈ 1.34 × 10^-3 M

pH = -log[H⁺] = -log(1.34 × 10^-3) ≈ 2.87

From these calculations, we see:

Oxalic Acid Solution: pH ≈ 1.13

Acetic Acid Solution:: pH ≈ 2.87

Conclusion:

Therefore, the correct statement is The pH of the oxalic acid solution is less than that of acetic acid solution

Concept:

When hydrogen sulfide (H₂S) gas is passed through an acidic solution containing various metal cations, the solubility product (K_sp) values and sulfide solubility play key roles in determining which sulfides precipitate:

• Cu²⁺ (Copper ion): Forms CuS, which is insoluble in acidic solutions (very low K_sp).

• Pb²⁺ (Lead ion): Forms PbS, which is also insoluble in acidic solutions (very low K_sp).

• Zn²⁺ (Zinc ion): Forms ZnS, but ZnS is soluble in acidic solutions due to a relatively higher K_sp.

Explanation:

The solubility products of different metal sulfides help determine their precipitation in acidic medium:

• CuS: Copper(II) sulfide has a very low K_sp and precipitates even in acidic conditions.

• PbS: Lead(II) sulfide also has a very low K_sp and precipitates in acidic conditions.

• ZnS: Zinc sulfide has a higher K_sp, meaning it is more soluble in acidic solutions and does not readily precipitate when H₂S is passed through the acidic solution.

Conclusion:

Given the solubility factors in acidic conditions, the precipitate will contain: CuS and PbS

Concept:

Water Gas

• Composition: Equimolecular mixture of carbon monoxide (CO) and hydrogen (H₂).

• Uses: Used as a fuel gas and a reducing agent in metallurgical processes.

• Color: Typically colorless.

• Toxicity: Carbon monoxide is highly toxic if inhaled; hydrogen is non-toxic but highly flammable.

• Production: Produced by passing steam over red-hot coke or coal.

Producer Gas

• Composition: Mixture of carbon monoxide (CO) and nitrogen (N₂).

• Uses: Used as a fuel in industrial furnaces and for power generation.

• Color: Colorless.

• Toxicity: Contains carbon monoxide which is highly toxic if inhaled.

• Production: Produced by blowing air (sometimes with steam) through red-hot coke or coal.

Coal Gas

• Composition: Mixture of hydrogen (H₂), methane (CH₄), carbon monoxide (CO), and carbon dioxide (CO₂).

• Uses: Historically used for lighting and heating in gas lamps, now mainly used as a fuel.

• Color: Colorless.

• Toxicity: Contains carbon monoxide, which is highly toxic; hydrogen and methane are highly flammable.

• Production: Produced by the destructive distillation of coal.

Natural Gas

• Composition: Primarily methane (CH₄) with small amounts of other hydrocarbons such as ethane (C₂H₆).

• Uses: Used as a fuel for heating, cooking, and electricity generation; also serves as a feedstock for the production of chemicals.

• Color: Colorless.

• Toxicity: Non-toxic but highly flammable.

• Production: Extracted from underground deposits; often found associated with petroleum deposits.

Explanation:

• (a) Water gas: Known for being an equimolecular mixture of CO and H₂.

• (b) Producer gas: Typically consists of CO and N₂.

• (c) Coal gas: Contains a mixture of CO, H₂, CH₄, and CO₂.

• (d) Natural gas: Mainly composed of methane (CH₄).

Conclusion:

Based on the above analysis, the correct matches between the fuels and their compositions are: a - 3, b - 1, c - 4, d - 2