They have the same number of electrons
A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two.
Without any ionization, the element (Cn) would have 112 electrons because the atomic number of an element is the number of protons the element has and a neutral element has the same number of electrons as it does protons.
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⁻
Hey there!
Balance the equation:
SiCl₄ + H₂O → H₄SiO₄ + HCl
Balance H.
2 on the left, 5 on the right. Add a coefficient of 3 in front of H₂O and a coefficient of 2 in front of HCl.
SiCl₄ + 3H₂O → H₄SiO₄ + 2HCl
Balance O.
3 on the left, 4 on the right. Change the coefficient of 3 in front of H₂O to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 2HCl
This unbalanced our H, so change the coefficient of 2 in front of HCl to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Balance Cl.
4 on the left, 4 on the right. Already balanced.
Balance Si.
1 on the left, 1 on the right. Already balanced.
Our final balanced equation:
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Hope this helps!
