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Paladinen [302]
3 years ago
14

The reaction between C2H6 and Cl2 is by addition

Chemistry
1 answer:
igomit [66]3 years ago
5 0

The reaction between C2H6 and Cl2 by addition is C2H4Cl2 or dichloroethane. It must be under the presence of sunlight because halogen such as chlorine cannot easily react with the sigma bond of alkane such as ethane.

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It is 79 - + 3 = 76 electrons.
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Can anyone check my chemistry answers?
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in question 6, the answer is the third choice. remember to find the neutrons, you take the atomic mass minus the atomic number. 38 - 18= 20
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A sample of gas is in a sealed rigid cointaner that maintains a constant volume which changes occur between the gas particles wh
sashaice [31]

In the given situation, the gas is heated under constant volume. As energy is supplied to the system in the form of heat, the frequency of collision between the gas particles increases. This increases the temperature of the gas consequently bringing about a decrease in pressure.

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8 0
3 years ago
At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:
Butoxors [25]

<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

Initial concentration of iodine gas = \frac{0.45}{2}=0.225M

Initial concentration of chlorine gas = \frac{0.45}{2}=0.225M

For the given chemical equation:

2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

                      I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091

<u>Initial:</u>         0.225      0.225

<u>At eqllm:</u>   0.225-x    0.225-x     2x

The expression of K_c for above equation follows:

K_c=\frac{[ICl]^2}{[Cl_2][I_2]}

Putting values in above equation, we get:

9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

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Yes it should, not 100% possitive though
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