Hybridization refers to the mixing of atomic orbitals in an atom. The number of hybrid orbitals needs to be equal to the number of orbitals that have involved in prior to mixing.
The isolated atoms cannot prevail in a hybridized state as the atom in an isolated state do not form any kind of bond with the other atom, due to which the atomic orbitals do not go through the process of hybridization.
Answer:
1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).
2. The difference in pH values is 4.95.
Explanation:
1. The pH of a compound can be found using the following equation:
![pH = -log([H_{3}O^{+}])](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20)
First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.
<u>Trimethyl ammonium</u>:
We can calculate [H₃O⁺] using the Ka as follows:
(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺
1.0 - x x x
![Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5B%28CH_%7B3%7D%29_%7B3%7DN%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5B%28CH_%7B3%7D%29_%7B3%7DNH%5E%7B%2B%7D%5D%7D)

By solving the above equation for x we have:
x = 0.097 = [H₃O⁺]
<u>Phenol</u>:
C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺
1.0 - x x x
![Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DOH%5D%7D)


Solving the above equation for x we have:
x = 9.96x10⁻⁶ = [H₃O⁺]
![pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20%3D%20-log%289.99%20%5Ccdot%2010%5E%7B-6%7D%29%20%3D%205.00%20)
Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.
2. The difference in pH values for the two acids is:
Therefore, the difference in pH values is 4.95.
I hope it helps you!
Answer:- 64015 J
Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.
density of water is 1 g per mL.
So, the mass of water =
= 4250 g
Final temperature of water after adding the hot copper bar to it is 26.15 degree C.
So,
for water = 26.15 - 22.55 = 3.60 degree C
Specific heat for water is 4.184 
The heat gained by water is calculated by using the formula:

where, q is the heat energy, m is mass and c is specific heat.
Let's plug in the values in the formula and do the calculations:

q = 64015 J
So, 64015 J of heat is gained by the water.
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce
g of (NH₄)₂S
25g of NH₃ will produce
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
Answer:
8 Silicon atom are present in unit cell.
16 oxygen atoms are present unit cell.
Explanation:
Number of atoms in unit cell = Z =?
Density of silica = tex]2.32 g/cm^3[/tex]
Edge length of cubic unit cell = a = 0.700 nm = 

Molar mass of Silica = 
Formula used :
where,
= density
Z = number of atom in unit cell
M = atomic mass
= Avogadro's number
a = edge length of unit cell
On substituting all the given values , we will get the value of 'a'.
1 silicon is 2 oxygen atoms. then 8 silicon atoms will be 16 oxygen atoms.