- Unit 1 Worksheet 4

Luba_88 [7]

The function does residual Co2 plays big part and in maintaining the body’s homeostasis. The addition of respiratory to reserve volume or residual volume. The lung is the one that who protect the organs, so in exchange of oxygen and carbon dioxide.

6 0

3 years ago

A student plans a two-step synthesis of 1-ethyl-3-nitrobenzene from benzene. The first step is nitration of benzene to give nitr

arlik [135]

Answer:

Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

Explanation:

The benzene ring in itself does not easily undergo electrophilic substitution reaction. Some groups activate or deactivate the benzene ring towards electrophilic substitution reactions.

-NO2 ia a highly deactivating substituent therefore, Friedel-Crafts alkylation of nitrobenzene does not take place under any conditions.

This reaction scheme is therefore flawed because Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

7 0

3 years ago

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

$2BC_2 ----->4C + B_2$

The time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

The time available is $t_a = 3.5 hr$

The desired concentration to remain $C = 0.42M$

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

Generally the elementary rate law is mathematically as

$-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

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

$C_Z = C_Z_o - \frac{1}{6} C_Z_o$

$C_Z = 0.5 - \frac{0.5}{6}$

$= 0.4167 M$

So

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

$0.400 = 9 k$

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

For $C_Z = 0.42M$

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

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

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

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

$C_Z_o = 0.4492M$

4 0

3 years ago

Chopping onions makes you cry because when you cut into an onion you break its cells. this releases sulfenic acids, which lead t

ruslelena [56]

Whats the question ?

8 0

4 years ago

Which notation is used to represent a beta particle?

Lapatulllka [165]

The beta particle is represented by the Greek letter <span>β. The beta particle is one of the particles that are emitted during radioactive decay. Beta particles are emitted in the process of beta decay. A beta particle can be in form of an electron or a positron. </span><span />

7 0

3 years ago

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