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frosja888 [35]
3 years ago
15

Calculate the vapor pressure at 25 ?c of hexane above the solution. the vapor pressure of pure hexane at 25 ?c is 151 torr.

Chemistry
1 answer:
Studentka2010 [4]3 years ago
8 0
One of the many ways in order to solve for the vapor pressure of pure components at a given temperature is through the Antoine's equation which is written below,
                                    P = 10^(A - B/C+T)
where A, B, and C are constants and T is the temperature in °C and P is the vapor pressure in mm Hg. 
For hexane,
                             A = 7.01
                             B = 1246.33
                             C = 232.988

Substituting the known values,
                            P = 10^(7.01 - 1246.33/232.988+25)
                             <em> P = 151.199 mm Hg</em>
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At 20 ∘C the vapor pressure of benzene C6H6 is 75 torr and that of toluene C7H8 is 22 torr. Assume that benzene and toluene fo
Olegator [25]

Answer:

a) Xbenzene = 0.283

b) Xtoluene = 0.717

Explanation:

At T = 20°C:

⇒ vapor pressure of benzene (P*b) = 75 torr

⇒ vapor pressure toluene (P*t) = 22 torr

Raoult's law:

  • Pi = Xi.P*i

∴ Pi: partial pressure of i

∴ Xi: mole fraction

∴ P*i: vapor pressure at T

a) solution: benzene (b) + toluene (t)

∴ Psln = 37 torr;   at T=20°C

⇒ Psln = Pb + Pt

∴ Pb = (Xb)*(P*b)

∴ Pt = (Xt)*(P*t)

∴ Xb + Xt = 1

⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)

⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb

⇒ 15 torr = 53 torrXb

⇒ Xb = 15 torr / 53 torr

⇒ Xb = 0.283

b) Xb + Xt = 1

⇒ Xt = 1 - Xb

⇒ Xt = 1 - 0.283

⇒ Xt = 0.717

6 0
3 years ago
Write the definition of a compound and give two examples
QveST [7]

Answer:

a thing that is composed of two or more separate elements ex. Acid and Water

3 0
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Which of the following ion is the least and the most hydrated? OH- , Li+, K+, Na+​
vodomira [7]
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Answer:

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Explanation:

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3 years ago
Consider the following reaction: 2{\rm{ N}}_2 {\rm{O(}}g)\; \rightarrow \;2{\rm{ N}}_2 (g)\; + \;{\rm{O}}_2 (g)
fredd [130]

Answer:

a. 5.9 × 10⁻³ M/s

b. 0.012 M/s

Explanation:

Let's consider the following reaction.

2 N₂O(g) → 2 N₂(g) + O₂(g)

a.

Time (t): 12.0 s

Δn(O₂): 1.7 × 10⁻² mol

Volume (V): 0.240 L

We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

r = 5.9 × 10⁻³ M/s

b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:

5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s

4 0
3 years ago
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