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⁻
Capture all of the smoke and weight it. it will weigh exactly the same before and after you burn it but will just be CO2 and H2O gas.
Answer:
k(+)
NH4(+)
Al3(+)
Explanation:
cations are those elements who donate their electrons and we put a positive charge on it and receivers get negative charge and they are called anion.....Thank you so,much Please let me know how you feel and is helps you or not....
Answer=3
<span>Decomposition, double replacement, and synthesis are 3 types of chemical reactions.</span>
To calculate the number of atoms of Cr, we first find the number of moles per unit of cubic centimeter of Cr. Then, use avogadros number for the number of atoms. Calculations are as follows:
1 cm^3 (7.15 g/cm^3) (1 mol / 51.996 g Cr) = 0.14 mol Cr
0.14 mol Cr ( 6.022 x 10^23 atoms Cr / 1 mol Cr ) = 8.28 x 10^22 atoms Cr
