Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Ne is isoelectronic with Na+ ion.
Answer:
lighting a match is the answer.
Answer:
The answer would be C, in both chemical and physical change, because everything is made up of matter. So if you have a chemical or physical change, that doesent change the total mass of your product ever.
Explanation:
Complete Question
You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?
Answer:
The pK_a value is
Explanation:
From the question we are told
The volume of base is ![V_B = 26.mL = 0.0260L](https://tex.z-dn.net/?f=V_B%20%3D%2026.mL%20%3D%200.0260L)
The pH of solution is ![pH = 7.82](https://tex.z-dn.net/?f=pH%20%3D%20%207.82)
The concentration of the acid is ![C_A = 0.1M](https://tex.z-dn.net/?f=C_A%20%3D%200.1M)
From the pH we can see that the titration is between a strong base and a weak acid
Let assume that the the volume of acid is ![V_A = 18mL= 0.018L](https://tex.z-dn.net/?f=V_A%20%3D%2018mL%3D%200.018L)
Generally the concentration of base
![C_B = \frac{C_AV_A}{C_B}](https://tex.z-dn.net/?f=C_B%20%3D%20%5Cfrac%7BC_AV_A%7D%7BC_B%7D)
Substituting value
![C_B = \frac{0.1 * 0.01800}{0.0260}](https://tex.z-dn.net/?f=C_B%20%3D%20%5Cfrac%7B0.1%20%2A%200.01800%7D%7B0.0260%7D)
![C_B= 0.0692M](https://tex.z-dn.net/?f=C_B%3D%200.0692M)
When 13mL of the base is added a buffer is formed
The chemical equation of the reaction is
![HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}](https://tex.z-dn.net/?f=HA_%7B%28aq%29%7D%20%2B%20OH%5E-_%7B%28aq%29%7D%20--------%3E%20A%5E%7B%2B%7D_%7B%28aq%29%7D%20%2B%20H_2%20O_%7B%28l%29%7D)
Now before the reaction the number of mole of base is
![No \ of \ moles[N_B] = C_B * V_B](https://tex.z-dn.net/?f=No%20%5C%20of%20%5C%20moles%5BN_B%5D%20%20%3D%20%20C_B%20%2A%20V_B)
Substituting value
![N_B = 0.01300 * 0.0692](https://tex.z-dn.net/?f=N_B%20%3D%200.01300%20%2A%200.0692)
Now before the reaction the number of mole of acid is
![No \ of \ moles = C_B * V_B](https://tex.z-dn.net/?f=No%20%5C%20of%20%5C%20moles%20%20%3D%20%20C_B%20%2A%20V_B)
Substituting value
![N_A = 0.01800 *0.1](https://tex.z-dn.net/?f=N_A%20%3D%200.01800%20%2A0.1)
![= 0.001800 \ moles](https://tex.z-dn.net/?f=%3D%200.001800%20%5C%20moles)
Now after the reaction the number of moles of base is zero i.e has been used up
this mathematically represented as
![N_B ' = N_B - N_B = 0](https://tex.z-dn.net/?f=N_B%20%27%20%3D%20N_B%20-%20N_B%20%3D%200)
The number of moles of acid is
![N_A ' = N_A - N_B](https://tex.z-dn.net/?f=N_A%20%27%20%3D%20N_A%20%20-%20N_B)
![= 0.0009\ moles](https://tex.z-dn.net/?f=%3D%200.0009%5C%20moles)
The pH of this reaction can be mathematically represented as
![pH = pK_a + log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%20%3D%20pK_a%20%2B%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
Substituting values
![7.82 = pK_a +log \frac{0.0009}{0.0009}](https://tex.z-dn.net/?f=7.82%20%3D%20pK_a%20%2Blog%20%5Cfrac%7B0.0009%7D%7B0.0009%7D)