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katovenus [111]
2 years ago
7

A vessel contains Ar(g) at a high pressure. Which of the following statements best helps to explain why the measured pressure is

significantly greater than the pressure calculated using the ideal gas law?
1. The molar mass of Ar is relatively large.2. A significant number of Ar2 molecules form.3. The attractive forces among Ar atoms cause them to collide with the walls of the container with less force.4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.
Chemistry
1 answer:
Luda [366]2 years ago
5 0

Answer:

4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.

Explanation:

Deviations from ideality are due to intermolecular forces and to the nonzero volume of the molecules themselves. At infinite volume, the volume of the molecules themselves is negligible compared with the infinite volume the gas occupies.

However, the volume occupied by the gas molecules must be taken into account. Each <u>molecule does occupy a finite, although small, intrinsic volume.</u>

The non-zero volume of the molecules implies that instead of moving in a given volume V they are limited to doing so in a smaller volume. Thus, the molecules will be closer to each other and repulsive forces will dominate, resulting in greater pressure than the one calculated with the ideal gas law, that means, without considering the volume occupied by the molecules.

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How many atoms of fluorine are in ZrF4
harkovskaia [24]

Answer:

Zirconium tetrafluoride has 4 atoms of flourine and 1 atom of Zirconium

3 0
2 years ago
At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)
viktelen [127]

Answer: K_c for this reaction at this temperature is 0.029

Explanation:

Moles of  HBr = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M

The given balanced equilibrium reaction is,

                            2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial conc.              0.500 M              0  M        0 M  

At eqm. conc.            (0.500-2x) M   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}

Equilibrium concentration of [Br_2] = x =  0.0955 M

Now put all the given values in this expression, we get :

K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}

K_c=0.029

Thus K_c for this reaction at this temperature is 0.029

7 0
2 years ago
How many molecules of CBr4 are in 250 grams of CBr4
Kazeer [188]

Answer:- 4.54*10^2^3 molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


8 0
1 year ago
Calculate the ratio of effusion rates of cl2 to f2 .
Lelechka [254]
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4 0
2 years ago
Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
2 years ago
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