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2 years ago

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Heptane (C7H16) and octane (C8H18) are constitutents of gasoline. At 80 degree celsius, the vapor pressure of heptane is 428mmHg

and the vapor pressure of octane is 175mmHg. What Xheptane in a mixture of heptane and octane that has a vapor pressure of 305mmHg and 80 degree celsius?

2 answers:

harkovskaia [24]2 years ago

7 0

<u>Answer:</u> The mole fraction of heptane at 80°C is 0.506

<u>Explanation:</u>

We are given:

Vapor pressure of heptane at 80°C = 428 mmHg

Vapor pressure of octane at 80°C = 175 mmHg

Total pressure at 80°C = (428 + 175) = 603 mmHg

To calculate the mole fraction of heptane at 80°C, we use the equation given by Raoult's law, which is:

$p_{heptane}=p_T\times \chi_{heptane}$

where,

$p_A$ = partial pressure of heptane = 305 mmHg

$p_T$ = total pressure = 603 mmHg

$\chi_A$ = mole fraction of heptane = ?

Putting values in above equation, we get:

$305mmHg=603mmHg\times \chi_{heptane}\\\\\chi_{heptane}=0.506$

Hence, the mole fraction of heptane at 80°C is 0.506

ss7ja [257]2 years ago

4 0

Looking at this equation P= (pa*pb)/ (pa+(pb-pa)) ya where pa=vap press a and ya= vap composition a and P= total pressure,it relates vapor pressure mixture to vapor composition. This is derived using the combination of Dalton's and Raoult's laws.

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Explanation:

Half life is simply the amount of time it takes for half of a substance to decompose.

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Balance the following redox reaction occurring in an acidic solution. The coefficient of Cr2O72−(aq) is given. Enter the coeffic

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Answer:

Cr₂O₇²⁻ (aq) + 6Ti³⁺ (aq) + 2 H⁺(aq) → 2 Cr³⁺ (aq) + 6TiO²⁺(aq) + H2O(l)

Explanation:

Cr₂O₇²⁻ (aq) + Ti³⁺ (aq) + H⁺ (aq) → Cr³⁺ (aq) + TiO²⁺ (aq) + H₂O(l)

This is the reaction, without stoichiometry.

We have to notice the oxidation number of each element.

In dicromate, Cr acts with +6 and we have Cr³⁺, so oxidation number has decreased. .- REDUCTION

Ti³⁺ acts with +3, and in TiO²⁺ acts with +4 so oxidation number has increased. .- OXIDATION

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

To change 6+ to 3+, Cr had to lose 3 e⁻, but as we have two Cr, it has lost 6e⁻. As we have 7 Oxygens in reactant side, we have to add water, as the same amount of oxygens atoms we have, in products side. Finally, we have to add protons in reactant side, to ballance the H.

H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺

Titanium has to win 1 e⁻ to change 3+ to 4+. We had to add 1 water in reactant, and 2H⁺ in products, to get all the half reaction ballanced.

Now we have to ballance the electrons, so we can cancel them.

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) .1

(H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺) .6

Wre multiply, second half reaction .6

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

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Now we can sum, the half reactions:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6H₂O + 6Ti³⁺ → 6TiO²⁺ + 6e⁻ + 12H⁺ + 2Cr³⁺ + 7H₂O

Electrons are cancelled and we can also operate with water and protons

7H₂O - 6H₂O = H₂O

14 H⁺ - 12H⁺ = 2H⁺

The final ballanced equation is:

2H⁺ + Cr₂O₇²⁻ + 6Ti³⁺ → 6TiO²⁺ + 2Cr³⁺ + H₂O

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assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

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$[C_3H_5O_3^{-} ]= c\alpha$

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$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

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$K_a = {c\alpha \times \alpha}$

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