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Lyrx [107]
3 years ago
9

The smell of fresh cut pine is due in part to the cyclic alkene called pinene. Given the following data of pinene: Vapor pressur

e (torr) Temperature (K) 760 429 515 415 Calculate the heat of vaporization, ΔHvap, of pinene.
Chemistry
1 answer:
Sveta_85 [38]3 years ago
4 0

Answer: Heat of vaporization is 41094 Joules

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at  429 K = 760 torr

P_2 = final pressure at 415 K  = 515 torr

= enthalpy of vaporisation = ?

R = gas constant = 8.314 J/mole.K

T_1= initial temperature =  429 K

T_2 = final temperature = 515 K

Now put all the given values in this formula, we get

\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]

\Delta H=41094J

Thus the heat of vaporization is 41094 Joules

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An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at
lana [24]

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

6 0
2 years ago
A 4.27 g sample of a lab solution contains 1.81g of acid. What is the concentration of the solution as a mass percentage?
vaieri [72.5K]

Answer:

42.38875878%

Explanation:

i divided 4.27 g from 1.81g using a percentage calculator but im not sure if its correct

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2 years ago
Increasing the ? of a solvent increases the solubility of the solute
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7 0
3 years ago
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How many moles are in 39.5 grams of Lithium?
Blizzard [7]

Answer:

185.05 g.

Explanation

Firstly, It is considered as a stichiometry problem.

From the balanced equation: 2LiCl → 2Li + Cl₂

It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.

We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.

n = (30.3 g) / (6.941 g/mole) = 4.365 moles.

Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.

Using cross multiplication:

2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.

??? moles of LiCl → 4.365  moles of Li.

The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.

Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).

Molar mass of LiCl = 42.394 g/mole.

mass = n x molar mass = (4.365 x 42.394) = 185.05 g.

7 0
2 years ago
Read 2 more answers
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