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Mrac [35]
2 years ago
5

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?
Chemistry
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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Mekhanik [1.2K]

Explanation:

Soaps attach to both water and grease molecules.

The grease molecules are attracted more strongly towards each other as compared to water molecules. Also, water molecules are smaller in size hence, strong intermolecular force is required to break the hydrogen bonds of water molecule so that grease or oil molecules can enter the water molecule.

A soap molecule goes in between water and grease molecule and helps them to bind. The force for linkage between water and grease molecule through the soap molecule is weak london dispersion force.

The soap molecule has its salt end as ionic and water soluble. When grease or oil is added to the soap and water solution then the soap acts as an emulsifier. The soap forms miscelles of the non-polar tails and grease molecules are trapped between these miscelles. This miscelle is easily soluble in water hence, the grease is washed away.

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7 0
1 year ago
Read 2 more answers
Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:
Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

               2 ICl(g)   ⇄   I₂(g)    +    Cl₂(g)

Initially      0.20              -               -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React          x              x/2               x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

           (0.20 - x)          x/2             x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0
1 year ago
What type of chemical reaction is this? Cl2(g) + 2KBr(aq) - 2KCl(aq) + Br2(l)
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If a sample containing 2.50 ml of nitroglycerin (density=1.592g/ml) is detonated, how many total moles of gas are produced?
masya89 [10]
<span>0.127 moles The formula for nitroglycerin is C3H5N3O9 so let's first calculate the molar mass of it. Carbon = 12.0107 Nitrogen = 14.0067 Hydrogen = 1.00794 Oxygen = 15.999 C3H5N3O9 = 3 * 12.0107 + 5 * 1.00794 + 3 * 14.0067 + 9 * 15.999 = 227.0829 Now calculate the number of moles of nitroglycerin you have by dividing the mass by the molar mass 2.50 ml * 1.592 g/ml / 227.0829 g/mol = 0.017527 mol The balanced formula for when nitroglycerin explodes is 4 C3H5N3O9 => 12 CO2 + 10 H2O + O2 + 6 N2 Since all of the products are gasses at the time of the explosion, there is a total of 29 moles of gas produced for every 4 moles of nitroglycerin Now multiply the number of moles of nitroglycerin by 29/4 0.017527 mol * 29/4 = 0.12707075 moles Round to 3 significant figures, giving 0.127 moles</span>
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