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GenaCL600 [577]
3 years ago
6

The gas phase decomposition of hydrogen iodide at 700 K HI(g)½ H2(g) + ½ I2(g) is second order in HI with a rate constant of 1.2

0×10-3 M-1 s-1. If the initial concentration of HI is 1.09 M, the concentration of HI will be M after 2.36×103 seconds have passed.
Chemistry
1 answer:
Simora [160]3 years ago
4 0

Answer:

<em>C</em> HI = 0.2667 M

Explanation:

  • HI(g) → 1/2H2(g) + 1/2I2(g)

decomposition rate HI(g):

∴ a: HI(g)

  • - ra = (K)*(<em>C</em>a)∧α

∴ K = 1.20 E-3 M/s.........rate constant

∴ α = 2.....second order

∴ Cao = 1.09 M.........initial concentration of HI(g)

⇒ Ca = ? .....t = 2.36 E3 s

⇒ - ra = δCa/δt = K Ca²

⇒ - ∫δCa/Ca² = K∫δt

⇒ [ 1/Ca - 1/Cao ] = K*t

⇒ 1/Ca = (K*t) + 1/Cao

⇒ 1/Ca = ((1.20 E-3/M.s)(2.36 E3 s)) + (1/1.09 M)

⇒ 1/Ca = 2.832/M + 0.9174/M

⇒ 1/Ca = 3.749/M

⇒ Ca = 0.2667 M

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Answer:

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Explanation:

We have the  trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.

The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.

In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.

On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.

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azamat

Answer:

Mass = 5.56 g

Explanation:

Given data:

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Mass of NaCl produced = ?

Solution:

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2Cl₂ + 4NaOH     →   3NaCl + NaClO₂ + 2H₂O

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Now we will compare the moles of Cl₂ with NaCl.

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