Strong acids have a ______ value of Ka and the equlibrium for the reaction with water lies far to the _______.
1 answer:
Answer:
All strong acids have a higher value of and the equilibrium for the reaction with water lies far to the right.
Explanation:
All strong acids dissociate completely in the solution. Higher the value of dissociation constant of the acid, higher will be the dissociation of the acid.
The reaction of the acid with water will be favored in the forward direction for acids having higher dissociation constant value ().
The dissociation of a strong acid say HA in water is shown below
Higher the value of , more will be the dissociation of the acid in water. The reaction will move far to the right side.
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Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.
<h3>What is mass percentage</h3>The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.
In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
In this case, you know:
- mass of CaCO₃: 2.62 grams
- mass of limestone: 3.82 grams
Replacing in the definition of mass percentage:
<u><em>mass percentage= 68.59 %</em></u>
Finally, the mass percentage of CaCO₃ is 68.59%.
Learn more about percentage by mass:
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Answer :
(a) The initial cell potential is, 0.53 V
(b) The cell potential when the concentration of has fallen to 0.500 M is, 0.52 V
(c) The concentrations of and
when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M
Explanation :
The values of standard reduction electrode potential of the cell are:
From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,
First we have to calculate the standard electrode potential of the cell.
(a) Now we have to calculate the cell potential.
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 2
= emf of the cell = ?
Now put all the given values in the above equation, we get:
(b) Now we have to calculate the cell potential when the concentration of has fallen to 0.500 M.
New concentration of = 1.50 - x = 0.500
x = 1 M
New concentration of = 0.100 + x = 0.100 + 1 = 1.1 M
Using Nernest equation :
Now put all the given values in the above equation, we get:
(c) Now we have to calculate the concentrations of and
when the cell potential falls to 0.45 V.
Using Nernest equation :
Now put all the given values in the above equation, we get:
The concentration of = 1.50 - x = 1.50 - 1.49 = 0.01 M
The concentration of = 0.100 + x = 0.100 + 1.49 = 1.59 M