Copper (I) oxidation state is 1 Cu2So4
copper (II) oxidation state is +2 CuSo4
copper (i) also give up one electron so you need two of them to react with the sulfate ion (which has charge of -2)
and also all metallic ions have an multiple oxidation levels corresponding to the number of electrons they can exchange or loose
Hope this helps
Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.
Explanation:
The reaction equation is as follows.

Initial : 0.3 0.4 0
Change: -x -x x
Equilbm: (0.3 - x) (0.4 - x) x
We know that, relation between standard free energy and equilibrium constant is as follows.

Putting the given values into the above formula as follows.


x = 0.1417
Hence, at equilibrium
= 0.1583 M
= 0.2583 M
Answer:
<h2>∞∞∞∞∞∞║⊕║Hello Person║⊕║∞∞∞∞∞∞</h2>
Your answer should be:
<em>1. No </em>
<em>2. Yes</em>
<em>3. No</em>
<em>4. Yes</em>
<em>5. No</em>
Explanation:
I hope this answered your question.
<em>Brainliest would be appreciated! </em>
<em>Have a wonderful day!</em>
If I made no mistake in calculation, the given answer must be correct...(tried my best)
elements : carbon hydrogen oxygen Fluorine
composition [C] 24 3 16 57
M r 12 1 16 19
(divide C by Mr) 2 3 1 3
(Divide by smallest value) 2 3 1 3
(smallest value = 1...so all value remained constant)
Empirical formula : C2H3OF3
if molar mas = 100 g per mole, then
first step calculate Mr. of empirical formula: [= 100]
Them molecular formula = empirical formula