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almond37 [142]
3 years ago
9

At 25°C the vapour pressure of pure pentane is 511 torr and that of hexane is 150. torr.

Chemistry
1 answer:
Genrish500 [490]3 years ago
3 0

Based on the vapor pressure of the pure substances, the mole fraction of pentane is 0.45 and mole fraction of hexane at equilibrium is 0.60.

<h3>What is vapor pressure of a liquid?</h3>

Vapor pressure of of a liquid is the pressure that is excreted on the walls of a sealed vessel by the molecules of the liquid that have been converted to gaseous phase.

TO determine the mole fraction of pentane in the solution, we use the vapor pressure of the pure substances.

Let the mole fraction of pentane in the mixture be X.

The mole fraction of hexane will be 1 - X

  • The vapor pressure of the solution = (mole fraction of pentane × vapor pressure of pentane) + (mole fraction of hexane × vapor pressure of hexane)

Vapor pressure of solution = 383 torr

(mole fraction of pentane × vapor pressure of pentane) + (mole fraction of hexane × vapor pressure of hexane) = 383 torr

X × 511 + (1 - X) × 150 = 383

511X - 150X + 150 = 383

511X = 233

X = 0.45

Also , at equilibrium, the vapor pressure of pentane = 0.45 × 511 = 230 torr

mole fraction of hexane in the vapor that is in equilibrium with this solution = 230/383

mole fraction of hexane at equilibrium = 0.60

Therefore, the mole fraction of pentane is 0.45 and mole fraction of hexane at equilibrium is 0.60.

Learn more about vapor pressure at: brainly.com/question/26006166

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2. If 3 is now behind one H2, it must be behind the other.

So now it's 3H2.

3. Al2 (SO4) 3 has 2 ahead of Al which means there will be 2Al in the reactants.

1. FeCl3 has 3 ahead of Cl, and Cl2 has 2. Which means that behind FeCl3 goes 2, and behind Cl2 goes 3 so now we have equated all Cl.

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Consider the following reaction between mercury(II) chloride and oxalate ion.
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<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

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Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

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