When 2-pentanol gets oxidized you end up with 2-pentanone. In this case, since we are oxidizing a secondary alcohol you end up with a ketone. They cannot be oxidized further than that. If it was a primary alcohol you would end up with an aldehyde.
You can find the mechanism as well the molecule formula of both 2pentanol and 2pentanone attached.
Your measurement implies that the range of answers is 128.6 mL to 128.8 mL.
If you do not state explicitly the range of uncertainty (e.g., ± 0.5 mL), the <em>implied range of uncertainty</em> is ±1 in the last significant digit.
Thus, a reading of 128.7 mL implies that the volume is 128.7 mL ± 0.1 mL.
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:
it is a physical change because when you heat up again a solution of sugar and tea you gonna obtain again sugar
