Answer:
ans is (2) 2,4- hexadiene
d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
Answer:
C
Explanation:
The higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, the described reaction is not possible due to the fact that bromine is less active
Answer:
Yes
Explanation:
Yes, A substance can be a lewis acid without being a Bronsted-Lowery acid because there are some substances which cannot donate protons(Bronsted-Lowery acid) but can accept a pair of electron.
<u><em>For Example:</em></u>
Let us take the example of BF₃
BF₃ contains no proton so it is not a Bronsted Lowery Acid
However, BF₃ has an incomplete octet with 6 electrons. It needs an electron pair to complete its octet. It accepts a pair of electron to become a Lewis Acid
