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3 years ago

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You are standing on the rim of a canyon, you drop a rock and in 7.0 seconds hear the sound pof it hitting the bottom, how deep i

s the canyon ? what assumptionsdid you make? examine how each assump[tion affects the answer. Does it lead to a larger or smaller depth than the calculated depth

1 answer:

sp2606 [1]3 years ago

8 0

This should help you with your question:

Gravitational pull = 9.8 m/s^2

1/2 x 9.8 x 7^2 = 240.1 m

The rock has fallen 240.1 m.

I cannot help you with the rest, but this should give you a headstart.

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In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher

trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

From the question we are told that

The necessary elementary step is

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Let Z be the reactant , Y be the first product and X the second product

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Where k is the rate constant , $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

$\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of $C_Z_o = 0.5M$

From the question we are told that it take 9 hours for the concentration of the reactant to become

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$\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

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=> $k = 0.044\ L/ mol \cdot hr^{-1}$

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$\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

$2.38 - 0.154 = \frac{1}{C_Z_o}$

$2.226 = \frac{1}{C_Z_o}$

$C_Z_o = \frac{1}{2.226}$

$C_Z_o = 0.4492M$

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Read 2 more answers