Answer:
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)
Explanation:
Which ONE of the following is an oxidation–reduction reaction?
A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.
B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.
C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.
D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.
Answer:- 
.
Explanation:- Atomic number for fluorine(F) is 9 and it's electron configuration is 
. 
is formed when F loses one electron from it's valence shell.
Second shell is the valence shell for fluorine and so it loses one electron from 2p to form and the electron configuration of the ion becomes .
The chemical formula : 3HgBr₂(Mercury(II) bromide)
<h3>Further explanation</h3>
Given
The chemical formulas of Mercury and Bromine
Required
The appropriate chemical formula
Solution
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
The number of molecules is determined by the coefficient in front of the compound
the number of atoms is determined by the subscript after the atom and the coefficient
Three molecules⇒ coefficient = 3
one atom of Mercury ⇒Hg
two atoms of Bromine ⇒ Br₂
The chemical formula : 3HgBr₂
