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Maurinko [17]
3 years ago
9

Biphenyl, c12h10, is a nonvolatile, nonionizing solute that is soluble in benzene, c6h6. at 25 °c, the vapor pressure of pure be

nzene is 100.84 torr. what is the vapor pressure of a solution made from dissolving 13.6 g of biphenyl in 26.4 g of benzene?
Chemistry
1 answer:
tino4ka555 [31]3 years ago
7 0
One form of Raoult's Law states that the vapor pressure of a solution of a non-volatile solute at certain temperature is equal to the vapor pressure of the pure solvent at the same temperature multiplied by the mole fraction of the solvent, this is:

p = X solvent * P pure solvent,

X solvent = number of moles of solvent / total number of moles.

Here the solute is 13.6 g of C12 H10 and the solvent is 26.4 g C6H6.

=>

moles of solvent = mass in grams / molar mass

molar mass of C6H6 = 6 * 12 g/mol + 6 * 1g/mol = 78 g/mol

moles of solvent = 26.4 g / 78 g/mol = 0.33846 mol

molar mass of C12H10 = 12 * 12g/mol + 10*1g/mol = 154 g/mol

moles of solute = 13.6 g / 154 g/mol = 0.08831 mol

=> X solvent = 0.33846 / (0.33846 + 0.08831) = 0.793

=> p = 0.793 * 100.84 torr = 79.97 torr ≈ 80.0 torr

Answer: 80.0 torr

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Heat of what? This is an unclear question.
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Suppose that 0.48 g of water at 25 ∘ C condenses on the surface of a 55- g block of aluminum that is initially at 25 ∘ C . If th
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Answer:

49^oC

Explanation:

At 25^oC, the heat of vaporization of water is given by:

\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g

The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:

Q_1 = \Delta H^o_{vap} m_w

The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:

Q_2 = c_{Al}m_{Al}(t_f - t_i)

According to the law of energy conservation, the heat lost is equal to the heat gained:

Q_1 = Q_2 or:

\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)

Rearrange for the final temperature:

\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i

We obtain:

\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f

Then:

t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC

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<h3>Further explanation</h3>

In the redox reaction, it is also known  

Reducing agents are substances that experience oxidation  

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series  

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

The electrodes which are easier to reduce than hydrogen (H), have E cells = +

The electrodes which are easier to oxidize than hydrogen have a sign E cell = -

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<h3>Combustion reaction of hydrocarbon</h3>

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Thus, we can conclude that, this is not a redox reaction. It is an example of combustion.

Learn more about combustion here: brainly.com/question/9425444

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Answer: C.)

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i got it right on a unit test!

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