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gtnhenbr [62]
1 year ago
13

The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the

energy needed to break only the π bond of the double bond of 2-butene. Express your answer numerically in joules per molecule.
Chemistry
1 answer:
AysviL [449]1 year ago
7 0

Answer:

4.42x10⁻¹⁹ J/molecule

Explanation:

At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:

E = 614 - 348 = 266 kJ/mol

Knowing that 1 kJ = 1000 J, E = 266,000 J/mol

By Avogadro's number, 1 mol = 6.02x10²³ molecules, thus:

E = 266,000 J/mol * 1mol/6.02x10²³ molecules

E = 4.42x10⁻¹⁹ J/molecule

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A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200
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Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

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2 years ago
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Answer: The balanced equation is

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HCl which is called hydrochloric acid, is an acid that does not react with gold, but it react with every other substance, like your skin, metals etc. it is used to clean a gold, by dipping the gold inside it, all the metals on the surface of the gold will dissolve.

When dissolving a gold in aqua regia solution, HCL is added to prepare this solution because it will help to dissolve all other substance on the surface of the gold.

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