According to the reaction equation:

and by using ICE table:

** CN- + H2O **↔** HCN + OH- **

**initial 0.08 0 0**

**change -X +X +X**

**Equ (0.08-X) X X**

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

= (1 x 10^-14) / (4.9 x 10^-10)

= 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013

∴ POH = -㏒[OH]

= -㏒0.0013

= 2.886

**∴ PH** = 14 - POH

= 14 - 2.886 = **11.11**

**Answer:**

See explanation

**Explanation:**

Before the advent of the wave-particle duality theory proposed by Louis de Broglie, there was a sharp distinction between mater and waves.

However, Louis de Broglie introduced the idea that mater could display wave-like properties. Erwin Schrödinger developed this idea into what is now known as the wave mechanical model of the atom.

In this model, electrons are regarded as waves. We can only determine the probability of finding the electron within certain high probability regions within the atom called orbitals.

This idea has been the longest surviving atomic model and has greatly increased our understanding of atoms.

**Answer:**

**Molarity of **** solution is 0.0928 M.**

**Explanation:**

**Balanced equation:**

Number of moles of in 10.8 mL of 0.0215 M solution

= = 0.000232 moles

**Let's assume molarity of **** solution is S(M) **

Number of moles of in 12.50 mL of S(M) solution

= = 0.0125S moles

**According to balanced equation, 5 moles of **** react with 1 mol of **** **

So, 0.0125S moles of react with moles of

Hence,

or, S = 0.0928

So, molarity of solution is 0.0928 M.

Answer:

333.3mL

Explanation:

Using the formula as follows:

C1V1 = C2V2

Where;

C1 = initial concentration (M)

C2 = final concentration (M)

V1 = initial volume (mL)

V2 = final volume (mL)

According to the information provided in this question,

C1 = 4.00M

C2 = 1.50M

V1 = 125mL

V2 = ?

Using C1V1 = C2V2

4 × 125 = 1.5 × V2

500 = 1.5V2

V2 = 500/1.5

V2 = 333.3mL

Therefore, the CuSO4 solution needs to be diluted to 333.3mL to make 1.50 M solution.