The initial equation is <span>F = (9/5)C + 32. We are asked to get the formula to solve for C. So the idea here is to isolate C on the other side, where F is currently located.
1. Subtract 32 on both sides.
2. Divide 9/5 on both sides.
You should get </span><span>C = 5 over 9(F – 32) as your answer. </span>
2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
Chemical compounds<span> tend to </span>dissolve more quickly in hot solvent than in cold solvent, t<span>he </span>solvent<span> molecules have </span>more<span> energy to pull the solute particles apart. Give an example of two liquids that are miscible.
https://www.studyblue.com/notes/note/n/chemistry/deck/938054
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The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A