Help with both questions ASAP PLEASE

How to apply Raoult's law to real solutions Consider mixing a liquid with a vapor pressure of 100 torr with an equimolar amount

garri49 [273]

Answer:

$\boxed{\text{positive deviation}}$

Explanation:

$\text{If the molecules interact less favourably in the solution than in the individual}\\\text{liquids, they can break away more easily than they can in the separate liquids.}\\\text{There will be more molecules in the vapour phase than you would expect.}\\\text{The solution will show a } \boxed{\textbf{positive deviation}} \text{ from Raoult's Law.}$

4 0

3 years ago

Can someone find from which website this test is?

Nadusha1986 [10]

Answer:

Try quizlet, they have everything on there

Explanation:

8 0

3 years ago

1. A high surface tension and a high boiling point both indicate that there

enyata [817]

The liquid that is expected to have the weakest intermolecular interaction is the liquid that has the lowest boiling point and surface tension.

<h3>What is boiling point?</h3>

The term boiling point refers to the temperature at which the atmospheric pressure and the pressure of the liquid becomes equal. The greater the intermolecular forces of attraction between molecules the higher the boiling point of the solid. Also a high surface tension indicates a strong intermolecular forces of attraction between molecules.

The liquid that is expected to have the weakest intermolecular interaction is the liquid that has the lowest boiling point and surface tension.

Learn more about boiling points: brainly.com/question/2153588

3 0

2 years ago

Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo

lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, $n_i=4$

Final orbit of electrons, $n_f=2$

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

$E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

Putting all the values we get :

$E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV$

We know that : $1\ eV=1.6\times 10^{-19}\ J$

So,

$E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J$

Energy of wave in terms of frequency is given by :

$E=hf$

$f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz$

Also, $c=f\lambda$

$\lambda$ is wavelength

So,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm$

Hence, this is the required solution.

4 0

3 years ago

The combustion reaction of c3h8o with o2 is represented by this balanced chemical equation. 2c3h8o + 9o2 6co2 + 8h2o when 3.00 g

MArishka [77]

<span>According to the question- 1 mol C3H8O = 60.096 g C3H8O 2 mol C3H8O = 9 mol O2 1 mol O2 = 31.998 g O2 [(3.00 g C3H8O)/1][(1 mol C3H8O)/(60.096)][(9 mol O2)/(2 mol C3H8O)][(32.998 g O2)/(1 mol O2)] = 7.1880435 g O2 Since 7.1880435 g of O2 is needed, and 7.38 g of O2 is available, 0.199565 g of O2 will be left over and oxygen is present in excess. Next, we need to convert 0.199565 g of O2 into moles of O2: [(0.199565 g O2)/1][(1 mol O2)/(31.998 g O2)] = 0.005999 mol O2, or 0.006 mol O2</span>

3 0

3 years ago