To counter the removal of A the equilibrium change by <u>s</u><em>hifting toward the left</em>
<em> </em><u><em>explanation</em></u>
<u><em> </em></u>If the reaction is at equilibrium and we alter the condition a new equilibrium state is created
<u><em> </em></u>The removal of A led to the shift of equilibrium toward the left since it led to less molecules in reactant side which favor the backward reaction.( equilibrium shift to the left)
Answer is: Benzene is trigonal (or triangular) planar.
VSEPR theory (The Valence Shell Electron Pair Repulsion Theory) uses the AXE notation (m and n are integers, m + n = number of regions of electron density).
For benzene molecule (C₆H₆):
m = 3; the number of atoms bonded to the central atom.
n = 0; the number of lone pairs on the central atom.
Answer:
C₂ = 0.334 M
Explanation:
Given data:
Volume of HCl = 0.0780 L
Concentration of HCl = 0.12 M
Volume of LiOH = 0.0280 L
Concentration of LiOH = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Concentration of HCl
V₁ = Volume of HCl
C₂ = Concentration of LiOH
V₂ = Volume of LiOH
Now we will put the values in formula.
C₁V₁ = C₂V₂
0.12 M × 0.0780 L = C₂ × 0.0280 L
0.00936 M.L = C₂ × 0.0280 L
C₂ = 0.00936 M.L/0.0280 L
C₂ = 0.334 M
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Answer:
The pH of the solution is 8.
Explanation:
To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:
1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:
pOH = –Log [OH-]
pOH = 6
6 = –Log [OH-]
–6 = Log [OH-]
[OH-] = Antilog (–6)
[OH-] = 1x10^–6 mol/L
2. The pH of the solution can be obtained as follow:
pH + pOH = 14
pOH = 6
pH + 6 = 14
pH = 14 – 6
pH = 8.
From the calculations made above,
[OH-] = 1x10^–6 mol/L
pH = 8.
Therefore, the correct answer is:
The pH of the solution is 8
