Consider the dissolution of AB(s):AB(s)⇌A+(aq)+B−(aq)Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.The generic metal hydroxide M(OH)2 has Ksp = 1.05×10−18. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)What is the solubility of M(OH)2 in pure water?

Answer:

6

Explanation:

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Answer: The property which depends on the quantity of the substance is called an extensive property. The free energy change for a reaction (Δ G) depends on the quantity of the substance and is therefore an extensive property. It shows the additive nature. The extensive property Δ G is easily calculated from the formula, ΔG = -nFE cell.

Explanation:

Final answer:

An extensive property is one that changes when the size of the sample changes. One such property that can be calculated is enthalpy. Enthalpy can be calculated using the formula H = E + PV.

Explanation:

An extensive property is a property that changes when the size of the sample changes. Examples include mass, volume, length, and total charge. One extensive property that can be calculated is enthalpy.

The enthalpy of a system can be calculated using the formula H = E + PV, where H represents the enthalpy, E the internal energy of the system, P the pressure, and V the volume. Like other extensive properties, the enthalpy of a system would change with the quantity or size of the sample.

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Complete question is;

What is the frequency of light emitted when the electron in a hydrogen atom undergoes a transition from energy level n=6 to level n=3?

Answer:

Frequency = 2.742 × 10^(14) s^(-1)

Explanation:

First of all, the energy of hydrogen electron from online values is;

E_n = -2.18 × 10^(-18) × (1/n²) J

n is the principal quantum number

We are told that hydrogen atom undergoes a transition from energy levels n = 3 to n = 6.

Thus, it means we have to find the difference between the electrons energy in the energy levels n = 3 to n = 6.

Thus;

E_n = E_6 - E_3

Thus;

E_n = [-2.18 × 10^(-18) × (1/6²)] - [-2.18 × 10^(-18) × (1/3²)]

E_n = (2.18 × 10^(-18)) × [-1/36 + 1/9]

E_n = 0.1817 × 10^(-18) J

From Planck expression, we can find the frequency. Thus;

E = hf

Where h is Planck's constant = 6.626 × 10^(-34) m²kg/s

Thus;

0.1817 × 10^(-18) = 6.626 × 10^(-34) × f

f = (0.1817 × 10^(-18))/(6.626 × 10^(-34))

f = 2.742 × 10^(14) s^(-1)

Final answer:

The frequency of light emitted during an electron transition in a hydrogen atom is determined by calculating the energy difference between the two energy levels and then using this to calculate the frequency using the equation for energy of a photon.

Explanation:

The frequency of light emitted during a transition of an electron in a hydrogen atom can be calculated using the formula for the energy difference (∆E) between two energy levels n1 and n2 in the hydrogen energy level diagram.

The formula to calculate energy difference is: ∆E = E(n2) - E(n1) where E(n) represents the energy of an energy level n. The energy difference ∆E is negative when an electron goes down an energy level (i.e., emits a photon), as the energy level n1 is greater than n2.

The frequency of the emitted photon (∆E) is then given by the formula ∆E = hf where h is Planck's constant (6.63 x 10^-34 Joule seconds) and f is the frequency. Therefore, you can rearrange the equation to find the frequency: f = ∆E / h.

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Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

Answer:

pH of Buffer Solution 5.69

Explanation:

Mole of anhydrous sodium acetate =

                                                           =

                                                           = 0.18 mole

 100 ml of 0.2 molar acetic acid  means

= M x V

= 0.2 x 100

= 20 mmol

= 0.02 mole

Using Henderson equation to find pH of Buffer solution

pH = pKa + log

     = 4.74 + log

     = 4.74 + log 9

     = 5.69

So pH of the Buffer solution = 5.69