Two electron pairs is the answer
Answer:
a) equivalence point
b) direct titration
c) primary standard
d) titrand
e) Back titration
f) back titration
g) standard solution
h) standard solution
I) indirect titration
j) end point
Explanation:
A volumetric analysis is one in which a solution of unknown concentration is determined from its volume. This is commonly referred to as titration.
In titration, a standard solution is reacted with another solution of unknown concentration. The point at which the concentration of the standard solution is equal to that of the analyte is known as the equivalence point (usually indicated by a colour change). An indicator may be added to the analyte solution to help identify when the reaction is complete.
Between equatorial positions, the F-P-F angle is 120°, and between axial and equatorial positions, it is 90°.
<h3>What in chemical bonding is equatorial position?</h3>
Equatorial Organic Chemistry Illustrated Glossary. Equatorial: In cyclohexane, a bond that runs parallel to the ring's axis (i.e., it follows the chair's equator), or a group joined by such a bond. Positions A are axial, and positions E are equatorial.
<h3>Why do equatorial positions have greater stability?</h3>
As was said in the preceding section, the equilibrium favors the more stable conformer because the chair conformation, in which the methyl group is equatorial, lowers steric repulsion. All monosubstituted cyclohexanes share this property.
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Answer:answers are in the explanation
Explanation:
(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.
pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.
(b). Equation of reaction;
HBr + KOH ---------> KBr + H2O
One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O
Calculating the mmol, we have;
mmol KOH = 28.0 ml × 0.50 M
mmol KOH= 14 mmol
mmol of HBr= 56 ml × 0.25M
mmol of HBr= 14 mmol
Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.
The pH here is greater than 7
(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL
= 0.10 M
Ka=Kw/kb
10^-14/ 1.8× 10^-5
Ka= 5.56 ×10^-10
Therefore, ka= x^2 / 0.20
5.56e-10 = x^2/0.20
x= (0.20 × 5.56e-10)^2
x= 1.05 × 10^-5
pH = -log [H+]
pH= - log[1.05 × 10^-5]
pH = 4.98
Acidic(less than 7)
(c). 0.5 × 20/40
= 0.25 M
Ka= Kw/kb
kb= 10^-14/1.8× 10^-5
Kb = 5.56×10^-10
x= (5.56×10^-10 × 0.5)^2
x= 1.667×10^-5 M
pH will be basic
The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>
