Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9
An acid has a pH <7. 7 is the pH of a neutral substance, like water, and substance with a pH >7 are considered basic or alkaline.
Answer : The atoms in this compound are Copper(Cu), Chlorine(Cl), Hydrogen(H), Oxygen(O).
Explanation :
The given compound is copper chloride bi-hydrate which is also called as copper (II) chloride dihydrate as it contains two water of crystallisation.
The formula of copper chloride bi-hydrate is
.
Therefore, there are 4 atoms in this compound and they are Copper(Cu), Chlorine(Cl), Hydrogen(H) and Oxygen(O).
meaning:An acid is a chemical substance, usually a liquid, which contains hydrogen and can react with other substances to form salts. Some acids burn or dissolve other substances that they come into contact with.
Answer:
The effective nuclear charge for a valence electron in oxygen atom: 
Explanation:
Effective nuclear charge
is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.
<em>It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s). </em>

<u>For an oxygen atom</u>-
Electron configuration: (1s²) (2s² 2p⁴)
<em>The atomic number (actual nuclear charge): </em>Z = 8
The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:
⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45
<u><em>Therefore, the effective nuclear charge for a valence electron in oxygen atom is:</em></u>

<u>Therefore, the effective nuclear charge for a valence electron in oxygen atom:</u> 