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2 years ago

If another hydrogen of c2h5cl is replaced by a chlorine atom to yield c2h4cl2, how many isomers would result?

Chemistry

1 answer:

Mandarinka [93]2 years ago

4 0

One isomer is formed

1,1- Dichloroethane is the isomer.

If another hydrogen of c2h5cl is replaced by a chlorine atom to yield c2h4cl2, it would result in one isomer.

In contrast to 1,2-dichloroethane, which has two chlorine atoms connected to distinct carbon atoms, 1,1-dichloroethane has two chlorine atoms bound to the same carbon atom.

Isomers are each of two or more compounds having the same formula but various atom arrangements in the molecule and unique characteristics.

<h3>What three types of isomers are there?</h3>

Chain isomers
Functional group isomers
Positional isomers

These are the three different categories of structural isomers.

<h3>How is an isomer recognized?</h3>

Their bonding patterns and the way they occupy three-dimensional space can be used to distinguish them.
Determine the bonding patterns of structural (constitutional) isomers.
Although the atoms in the compounds are the same, their connections create various functional groups.

<h3>What makes isomers significant?</h3>

Because two isomers might have the same chemical formula but different chemical structures, they are significant.
The molecule's properties are influenced by its structure.

To learn more about isomers visit:

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The correct answer is c. solution

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3 years ago

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Estimate the solubility of m(oh)2 in a solution buffered at ph = 7.0, 10.0, and 14.0.

kherson [118]

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4 0

3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Agata [3.3K]

<u>Answer:</u> The molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$PbI_2\rightleftharpoons Pb^{2+}+2I^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Pb^{2+}][I^-]^2$

We are given:

$K_{sp}=7.9\times 10^{-9}$

Putting values in above equation, we get:

$7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L$

Hence, the molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

3 0

3 years ago