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oksian1 [2.3K]
3 years ago
6

Compounds X and Y both have the formula C7H14. Both X and Y react with one molar equivalent of hydrogen in the presence of a pal

ladium catalyst to form 2-methylhexane. The heat of hydrogenation of X is greater than that of Y. Both X and Y react with HCl to give the same single C7H15Cl compound as the major product. What is the structure of X?
Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Answer:

See explanation and image attached

Explanation:

Alkenes undergo hydrogenation to give the corresponding alkanes. Where the structure of the original alkene is unknown, we can deduce the structure of the alkene from the structure of the products obtained when it undergoes various chemical reactions.

Now, the fact that we obtained 2-methylhexane upon hydrogenation and the two compounds had different heats of hydrogenation means that the two compounds were geometric isomers. The original compounds must have been cis-2-methyl-3-hexene and trans-2-methyl-3-hexene.

When reacted with HCl, the same compound C7H15Cl is formed because the stereo chemistry is removed.

However, we know that the trans isomer is more stable than the cis isomer hence the cis isomer always has a higher heat of hydrogenation than the trans isomer. Thus X is cis-2-methyl-3-hexene.

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Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
3 years ago
The radioactive atom R 88 210 a is an alpha emitter. What nucleus does it produce?
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Answer:

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Answer:

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We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

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