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Karolina [17]
3 years ago
6

How does the value of Ka and Kb compare with the concentrations of hydronium ion and hydroxide ion?

Chemistry
2 answers:
Andrews [41]3 years ago
8 0
Assuming that we are given the values of Ka and Kb, these can be rewritten and represented by the concentration of the Hydronium ion and Hydroxide Ion:

Hydronium ion: [H3O+]
Hydroxide ion: [OH-]

Ka is the equilibrium constant of an acid which dissociates to a hydronium ion and its cation.
Kb is the equilibrium constant of a base which dissociates to a hydroxide ion and its anion. 

So,

Ka x Kb = [H3O+][OH-]
Nadusha1986 [10]3 years ago
8 0

Answer:

The correct answer is A, Ka × Kb = [H₃0⁺][OH⁻]

I hope this helps, have a good day and good luck! :)

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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac
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Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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Electron is one of the sub-atomic particle present around the nucleus of an atom which is negatively charged.

In an atomic model, it is assumed that the electron revolves around the nucleus in discrete orbits having fixed energy levels.

These electrons when jumping from one energy level to another, some amount of radiation is either emitted or absorbed.

These fixed energy levels are given by the Bohr model and thus, the electrons are quantized.

Hence, the correct option is A) They have fixed energy values.

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