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3 years ago

02 (s) + Pb(s) — PbO2(s) find the reduced and oxidized reactants

Chemistry

1 answer:

Arlecino [84]3 years ago

7 0

Answer:

Oxygen is oxidizing gent and itself get reduced while lead is reducing agent and itself get oxidize.

Explanation:

Chemical equation with oxidation state:

O⁰₂(s) + Pb⁰ (s) → Pb⁺⁴O₂⁻²(s)

Oxidizing agent:

Oxidizing agents, oxidize the other substance and itself get reduced. They can be identify by just looking the oxidation state. If oxidation state is decreased on the right side it means substance gain the electron and get reduced.

Reducing agent:

Reducing agents, reduce the other substance and itself get oxidize. They can be identify by just looking the oxidation state. If oxidation state is increased on the right side it means substance lose the electron and oxidation state increased.

In given chemical equation, oxygen and lead were in neutral form they react and form lead dioxide. In this product oxygen is oxidizing gent and itself get reduced and oxidation state change from 0 to -2 while lead is reducing agent, itself get oxidize and oxidation state change from o to +4.

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2 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

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Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

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If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

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2 years ago