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Leviafan [203]
2 years ago
8

must undergo addition because they have easily broken π bonds. ____________ rule states in the addition of HX to an unsymmetrica

l alkene, the H atom bonds to the less substituted carbon atom. ____________ are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon. ____________ have good leaving groups and therefore readily undergo substitution and elimination reactions. In hydroboration, the boron atom bonds to the ____________ substituted carbon.
Chemistry
1 answer:
erastova [34]2 years ago
3 0

Answer:

Alkenes, Markovnikov's, Alkenes, Alkyl halides, and less.

Explanation:

Alkenes must undergo addition because they have easily broken π bonds.

Markovnikov's rule states in the addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon atom.

Alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.

Alkyl halides have good leaving groups and therefore readily undergo substitution and elimination reactions.

In hydroboration, the boron atom bonds to the less substituted carbon.

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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
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Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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