Answer:
Transition metals can have multiple oxidation states because of their electrons. The transition metals have several electrons with similar energies, so one or all of them can be removed, depending the circumstances. This results in different oxidation states. please mark me as brainellist
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:
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Explanation:
Answer:
A.) 1
Explanation:
Propane only exists in one conformation. It does not have enough carbons to form branches, and there are only hydrogens attached to each carbon. Furthermore, there is no way to twist the carbon or change its orientation (ex. cis- and trans-) to result in a different structure of propane. There is no other way to represent the molecule without drawing a different molecule.
Rubidium or strontium have larger a larger atomic radius since the further left on the periodic table you go, the larger the sizes of the atoms are. This trend can be explained through effective nuclear charge which explains how the further left and down you go, the less the atoms nucleus is able to pull in the electrons around it.<span />
