Answer:
Here's what I get
Explanation:
The Lewis structure of SO₃ consists of a central sulfur atom double-bonded to each of three oxygen atoms that points to the corners of an equilateral triangle.
A ball-and-stick model of SO₃ is shown below.
Follow
these steps to solve the given equation:
Multiply
the two decimal figures together and find the sum of the exponents, that is,
(1.5
* 1.89) * 10 ^4+3
(2.835)
* 10^7
10^7
can also be written as e.70
'e'
stands for exponential.
Therefore,
we have 2. 835 e 7.0 = 2.8 e 7.0.
Based on the calculations above, the correct option is A.
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⁻
B.
Explanation:
iehebrkee keen enensjsb sh sry need to get the points?
Answer:
I would help but the picture will not load for me :(
Explanation:
