Answer:
The answer is maybe reactivity
<u>Answer:</u> The concentration of 
at equilibrium is 0.00608 M
<u>Explanation:</u>
As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.
In the second dissociation, the ions will remain in equilibrium.
We are given:
Concentration of sulfuric acid = 0.025 M
Equation for the first dissociation of sulfuric acid:
0.025 0.025 0.025
Equation for the second dissociation of sulfuric acid:
<u>Initial:</u> 0.025 0.025
<u>At eqllm:</u> 0.025-x 0.025+x x
The expression of second equilibrium constant equation follows:
![Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}](https://tex.z-dn.net/?f=Ka_2%3D%5Cfrac%7B%5BH%5E%2B%5D%5BSO_4%5E%7B2-%7D%5D%7D%7B%5BHSO_4%5E-%5D%7D)
We know that:
Putting values in above equation, we get:
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of sulfate ion = x = 0.00608 M
Hence, the concentration of at equilibrium is 0.00608 M
Answer:
Ok so, b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...
i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced
ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.
iii. Calculate the standard potential (voltage) of the cell
Look up the reduction potential,
E
⁰
red
, for the reduction half-reaction in a table of reduction potentials
Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,
E
⁰
ox
=
-
E
⁰
red
.
iv. What kind of electrochemical cell is this? Explain your answer.
All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells
1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery
Explanation:
It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.
The energy required to remove an electron from an atom is called ionization energy.
The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.
Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.
Learn more: brainly.com/question/16243729
<span>In a solid the atoms are tightly packed together and vibrate in place, in a liquid the atoms are loosely packed together and can move past each other,
extra: and in a gas the atoms are far apart and move freely and </span><em>
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