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OleMash [197]
3 years ago
6

What is the best name for this molecule? Explain your naming process.

Chemistry
1 answer:
iogann1982 [59]3 years ago
7 0

Hi!


The best possible name for this molecule would be: 4 - ethyl - 1 - heptene or 4 - ethylhept - 1 - ene


We always name the molecule with respect to the longest chain (in a branched molecule), taking both atoms of carbon participating in the double bond into account too. In our case, this gives us a 7 Carbon chain - hept

ene - is the suffix that is indicative of the molecule being an alekene.

<em>So we know it is a branched heptene molecule</em>

We add the number where the double bond occurs either before ene, or before heptene as a rule.

<em>Note: We always start with the end of the chain from where the double bond is the closest, and number the carbons accordingly.</em>

The title and position of the branch always comes at the start. In our case the branch is a two carbon chain, and an alkane, so it would be an ethyl branch. This branch occurs at carbon number 4


Hence, the correct names would be:

<em>4 - ethyl - 1 - heptene</em> or <em>4 - ethylhept - 1 - ene</em>


Hope this helps!

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
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Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

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PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

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At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

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The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

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