Answered

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Consider the reaction8H₂S(g) + 402(g) →8H₂O(g) + S8 (g)AH₂S/At = -0.021 M/s

Consider The Reaction8HSg 402g 8HOg S8 GAHSAt 0021 Ms class=

Sagot :

[tex]8H₂S_{(g)}+4O_{2(g)\text{ }}→\text{ }8H₂O_{(g)}+S_{8(g)}[/tex][tex]v=\frac{-1}{8}\frac{\Delta\lbrack H_2S\rbrack}{\Delta t}=\frac{-1}{4}\frac{\Delta\lbrack O_2\rbrack}{\Delta t}=\frac{1}{8}\frac{\Delta\lbrack H_2O\rbrack}{\Delta t}=\frac{1}{1}\frac{\Delta\lbrack S_8\rbrack}{\Delta t}[/tex][tex]\frac{-1}{8}\frac{\Delta\lbrack H_2S\rbrack}{\Delta t}=\frac{1}{1}\frac{\Delta\lbrack S_8\rbrack}{\Delta t}[/tex][tex]\frac{\Delta\lbrack S_8\rbrack}{\Delta t}=\text{ }\frac{-1}{8}\frac{\Delta\lbrack H_2S\rbrack}{\Delta t}=\frac{-1}{8}(-0.021\text{ M/s\rparen = 0.0026 M/s}[/tex][tex]v=\frac{\Delta\lbrack S_8\rbrack}{\Delta t}=\text{ 0.0026 M/s}[/tex]

Part C the answer is Δ[S8]/Δt = 0.0026 M/s

Parte D the answer is v = 0.0026 M/s

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