I pick B Because Chemistry is all about plants and they can use a pesticide on plants
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
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<span> I got (a). The answer was 7.53. But when I try to solve (d), I keep getting the wrong answer. I subtracted the moles of NaOH from the acid and added the moles to the base. Then I did Ka = (x*([NaClO]+x))/([HClO - x) and then I found the pH</span>
