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bixtya [17]
3 years ago
9

Please help-- 15 pts!

Chemistry
2 answers:
Leona [35]3 years ago
5 0

Answer:

A liquid with weak  intermolecular forces and a high temperature

Explanation:

I just took the test and got it right

harkovskaia [24]3 years ago
4 0

Answer:

a liquid with weak intermolecular forces and a high temperature

Explanation:

According to encyclopedia Britannica, "Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature. The temperature at which the vapour pressure at the surface of a liquid becomes equal to the pressure exerted by the surroundings is called the boiling point of the liquid."

This implies that the vapour pressure of a liquid tends to increase as the temperature of the liquid increases. Recall that vapour pressure also depends on the magnitude of intermolecular forces. The greater the magnitude of intermolecular forces, the lower the vapour pressure hence water has very much higher vapour pressure than ethane.

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In a mixture of hydrogen and nitrogen gases, the mole fraction of nitrogen is 0.333. If the partial pressure of hydrogen in the
notka56 [123]

Answer:

P_T=112.4torr

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

P_{H_2}=x_{H_2}*P_T

And can be solved for the total pressure as follows:

P_T=\frac{P_{H_2}}{x_{H_2}}

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667

Then, we can plug in to obtain the total pressure:

P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr

Regards!

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Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th
Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M

The given equilibrium reaction is,

                           N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially conc.      0        0           0.1576

At eqm.               (x)       (x)        (0.1576-2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}

By solving the term, we get:

x=0.0742,0.0839

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

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3 years ago
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