A. Manipulating the interpreted rate law of zero-order reactions and substituting values,
"\\begin{aligned}\n[A]&=[A_o\u200b]\u2212kt\\\\\nt &= 5(60)= 300s\\\\\nk &= 0.42M\/s\\\\\n[A_o\u200b] &= 0.084M\\\\\n\\\\\n[A]&=0.084 - 0.42(300)\\\\\n[A] &= 0.084 - 126\\\\\n[A] &=-125.916M\n\\end{aligned}"
But since concentration can't be negative, we can safely infer that it's final concentration is 0M.
B. "\\begin{aligned}\nt_\\frac{1}{2} &= \\dfrac{[A_o]}{2k}\\\\\n\\\\\nt_\\frac{1}{2} &= \\dfrac{0.37}{2(0.42)}\\\\\n\\\\\nt_\\frac{1}{2} &= \\dfrac{0.37}{0.84}\\\\\n\\\\\nt_\\frac{1}{2} &= 0.44s\n\\end{aligned}"
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