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⁻
2.34 moles titanium x (6.022 x 10^23)/1 mole titanium = 1.41 x 10^24
Answer:
Transition Element
Explanation:
Transition elements are defined as those elements which can form at least one stable ion and has partially filled d-orbitals. They are also characterized by forming complex compounds and having different oxidation states for a single metal element.
Transition metals are present between the metals and the non metals in the periodic table occupying groups from 3 to 12. There general electronic configuration is as follow,
(n-1)d
¹⁻¹⁰ns
¹⁻²
The general configuration shows that for a given metal, the d sublevel will be in lower energy level as compared to corresponding s sublevel. For example,
Scandium is present in fourth period hence, its s sublevel is present in 4rth energy level so its d sublevel will be present in 3rd energy level respectively.
Hence, we can conclude that for transition metals the electron are present in highest occupied s sublevel and a nearby d sublevel
.
Answer:
C Force of gravity acting on an object
Explanation:
Hope that helps
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