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melamori03 [73]
3 years ago
7

Compound H is optically active and has the molecular formula C6H10 and has a five carbon ring. On catalytic hydrogenation, H is

converted to I (C6H12) and I is optically inactive. Propose structures for H and I. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)

Chemistry
1 answer:
Ksivusya [100]3 years ago
8 0

Answer:

Explanation:

Given that ;

Compound H is optically active and have a molecular formula of C6H10 and therefore undergo catalytic hydrogenation. Catalytic hydrogenation involves the use Platinum/Nickel to produce C6H12

i.e

C_6H_{10} +H_2 \to  ^{Pt/Ni}  \ \ \ C_6H_{12}

The proposed H and I structures are shown in the diagrams attached below .

compound H represents  3- methyl cyclopentene

compound I represents methyl cyclopentane

However; 3- methyl cyclopentene posses just only one chiral carbon which is optically active at the third position and it R and S enantiomers are shown in the second diagram below.

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Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

ln (\frac{C}{C_0})=-kt

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as 5.11 \times 10^{-5} \, s^{-1}

So time t is given as

ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

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3 years ago
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Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

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Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both [\mathrm{SO_2\, (g)}] and [\mathrm{O_2\, (g)}] will increase if the pressure is increased through compression. However, because \rm SO_2\, (g) and \rm O_2\, (g) have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient Q.

As a result, the increase in pressure will have no impact on the value of Q\!. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.

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<h3>Answer:</h3>

Lead-205 (Pb-205)

<h3>Explanation:</h3>

<u>We are given;</u>

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We are supposed to identify its product after an alpha decay;

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  • When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
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This website helped me understand the concept.
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