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>.
C. losing one electron
Explanation:
It is because the potassium atom electronic configuration is 2,8,8,1 where by if it loses one electron it becomes stable
The answer is B charged particles
We can use a variety of formulas to determine our answers here.
Our formula for pOH is -log(mol), and we can plug it in as -log(0.010). Take note that OH- is a base, not an acid.
So, the pOH of OH- is 2.
To find pH we can set up this simple equation:
pH + pOH = 14
All we need to do is subtract 2 from 14, therefore the pH is 12.
This makes sense since acids range in the pH of 1-6, and we are dealing with a base. Hope I could help!
Answer:
Its the first second and the fourth
