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NARA [144]
3 years ago
6

Both 1,2−dihydronaphthalene and 1,4−dihydronaphthalene may be selectively hydrogenated to 1,2,3,4−tetrahydronaphthalene. One of

these isomers has a heat of hydrogenation of 101 kJ/mol (24.1 kcal/mol), and the heat of hydrogenation of the other is 113 kJ/mol (27.1 kcal/mol). Match the heat of hydrogenation with the appropriate dihydronaphthalene.

Chemistry
1 answer:
pochemuha3 years ago
8 0

Answer:

1,4-dihydro = 113 kJ·mol⁻¹

1,2-dihydro = 101 kJ·mol⁻¹

Explanation:

In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.

In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.

Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).

1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.

The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.

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Answer:

Reduced species and oxidizing agent: sulfur in the form of sulfate.

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6 0
3 years ago
Calculate the total volume of gas (at 119 ?c and 731 mmhg produced by the complete decomposition of 1.71 kg of ammonium nitrate.
Shtirlitz [24]
The reaction is:

NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)

This means that 1 mol of NH4 (NO3)s produces 3 moles of gases.

Now find the number of moles in 1.71 kg of NH4 (NO3)

Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol

# moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4(NO3)

We already said that every mol of NH4(NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol

Now use the equation for ideal gases to fin the volume

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V = 2143.01 liters
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3 years ago
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