Correct Answer: C) The two equilibrium constants are identical

Explanation

In a reversible chemical reaction, the equilibrium constant for the forward reaction is related to that of the reverse reaction. The correct answer isOption C: The two equilibrium constants are identical.

To understand this concept, let's consider a general reversible reaction:

\(A + B \rightleftharpoons C + D\)

For the forward reaction, the equilibrium constant, \(K_f\), is given by:

\(K_f = \frac{[C][D]}{[A][B]}\)

For the reverse reaction, the equilibrium constant, \(K_r\), is given by:

\(K_r = \frac{[A][B]}{[C][D]}\)

Now, let's look at the relationship between the equilibrium constants of the forward and reverse reactions. The equilibrium constant for the overall reaction, \(K\), is given by:

\(K = \frac{K_f}{K_r}\)

Since the forward and reverse reactions are equal at equilibrium, the overall reaction's equilibrium constant, \(K\), is equal to 1. Therefore, we can write:

\(1 = \frac{K_f}{K_r}\)

From the above equation, we can see that the equilibrium constants for the forward and reverse reactions are reciprocals of each other, which means they are equal when the concentrations of reactants and products are equal at equilibrium. Hence, the two equilibrium constants are identical.

Correct Answer: D. Gases are highly soluble in water at high temperatures

Explanation

The statement that is not correct about gases is Gases are highly soluble in water at high temperatures. Gases are not highly soluble in water at high temperatures. The solubility of gases in water decreases as the temperature increases. This is because the kinetic energy of the gas molecules increases as the temperature increases. The increased kinetic energy of the gas molecules makes it more difficult for them to be dissolved in water.

Correct Answer: C. \(Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}\)

Explanation

The reduction half equation for the given reaction is:

\(Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}\)

In the overall reaction, copper(II) ions (\(Cu^{2+})\) gain two electrons (2e^-) and get reduced to form solid copper (Cu) metal. This is the reduction half-reaction in the redox reaction between zinc (Zn) and copper(II) sulfate (\(CuSO_4)\).

Correct Answer: D) 8

Explanation

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Correct Answer: B. NaH

Explanation

NaH is a saline (salt-like) hydride, composed of Na+ and H? ions

Correct Answer: B) Water and Oxygen

Explanation

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Explanation

(a)(i) Fused calcium chloride and concentrated tetraoxosulphate (VI) acid.

(ii) Ca(OH)\(_2\), NaOH.

(iii) Hydrogen and carbon (II) oxide

(iv) Zn metals, zinc trioxocarbonate

(IV), zinc oxides.

(b)(i) Soft solder for welding and plumbing and type metal for printing.

(ii) a black precipitate of PbS\(_{(s)}\) is deposited

(iii)(i) Polymorphism

(ii) Transition temperature

(c)(i) 2AgNO\(_{3(s)}\) --> 2Ag\(_{(s)}\) + O\(_2\) + 2NO\(_{2}\).

Lustrous Brown white

(ii) NH\(_4\)CO\(_3\) -> 2NH\(_3\) + CO\(_2\) + H\(_2\)O.

(iii)

Correct Answer: C. Trioxonitrate (V) acid

Explanation

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Correct Answer: D) 2%

Explanation

To answer this question, we first need to determine the amount of KClO\(_3\) that can dissolve in 40.0 cm\(^3\) of water at 298K.

Given the solubility of KClO\(_3\) is 20.0 mol dm\(^{-3}\) at 298K, we can calculate the maximum amount of KClO\(_3\) that can dissolve in the given volume of water. To do this, we first convert the volume from cm\(^3\) to dm\(^3\):

\[\frac{40.0\, cm^3}{1000\, cm^3/dm^3} = 0.040\, dm^3\]

Next, we can find the moles of KClO\(_3\) that can dissolve in 0.040 dm\(^3\) of water:

\[(20.0\, mol\, dm^{-3})(0.040\, dm^3) = 0.8\, mol\]

Now, we can calculate the mass of dissolved KClO\(_3\) by using the given molar masses of K, Cl, and O:

\[0.8\, mol \times (39\, g/mol\, K + 35.5\, g/mol\, Cl + 3 \times 16\, g/mol\, O) = 72.0\, g\]

Since we initially had 100.0 g of KClO\(_3\), and 72.0 g dissolved, we can calculate the percentage of KClO\(_3\) left undisolved:

\[\frac{100.0\, g - 72.0\, g}{100.0\, g} \times 100\% = 28.0\%\]

However, the given answer is 2% which may be incorrect. Based on the calculations, the correct answer should be 28%. It is recommended to consult the relevant sections of the recommended textbooks for more information and clarification.

Correct Answer: D) calcium hydrogentrioxocarbonate (IV)

Explanation

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