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Answer:
Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻
Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)
E° = 1.60 V
Explanation:
Let's consider the reaction taking place in a galvanic cell.
2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)
The corresponding half-reactions are:
Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻ E°red = - 0.83 V
Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq) E°red = 0.77 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V
It creates
<span>THE HALOGENATION OF ALKENES</span>
When a physical change occurs, the arrangement of particles within the substance may change, but the atoms in the molecules remain bonded together.
Explanation:
The sulfonation of the naphthalene yield 2 products under different conditions:
<u>When the reaction is carried at 80 °C, 1-naphthalenesulfonic acid is the major product because it is kinetically favoured product as arenium ion formed in the transition state corresponding to 1-naphthalenesulfonic acid is more stable due to better resonance stabilization.
</u>
<u>When the reaction is carried at 160 °C, 2-naphthalenesulfonic acid is the major product as it is more stable than 1-naphthalenesulfonic acid because of steric interaction of the sulfonic acid group in 1-position and the hydrogen in 8-position.</u>
The products are shown in image below.
