Answer:
NH₃ (Option A)
Explanation:
Arrhenius theory explained that the acids are the ones that have H⁺, either H in its formula. Following this, the bases are the ones that have OH⁻ , either OH and its formula.
It can be used only with compounds with H, or OH.
So the ammonia is not a base, as Arrhenius theory.
It is known that ammonia behaves as a weak base, but it does not have hydroxide ions that can yield to water
Answer:
1. hydrogen - H
2. helium - He
3. sodium - Na
4. magnesium - Mg
5. potassium - K
Explanation:
Hydrogen is the element of group 1 and first period. The atomic number of hydrogen is 1 and the symbol of the element is H.
The electronic configuration of the element hydrogen is:-

Helium is the element of group 18 and first period. The atomic number of helium is 2 and the symbol of the element is He.
The electronic configuration of the element helium is:-

Sodium is the element of group 1 and third period. The atomic number of sodium is 11 and the symbol of the element is Na.
The electronic configuration of the element sodium is:-

Magnesium is the element of group 2 and third period. The atomic number of magnesium is 12 and the symbol of the element is Mg.
The electronic configuration of the element magnesium is:-

Potassium is the element of group 1 and forth period. The atomic number of potassium is 19 and the symbol of the element is K.
The electronic configuration of the element potassium is:-

The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>