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nevsk [136]
3 years ago
5

Does gases have a kinetic attraction? Yes or no.

Chemistry
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

No

Explanation:

There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.

The average kinetic energy of gas particles is proportional to the absolute temperature of the gas, and all gases at the same temperature have the same average kinetic energy.

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Alcohol, alcohol is the only liquid  since copper and aluminum are a solid ammonia is a bacteria
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Read 2 more answers
why is it essential to know the water temperature in this experiment- to find the vapor pressure of the water
Oksi-84 [34.3K]

Answer:

Vapour pressure of a liquid varies with temperature

Explanation:

The vapour pressure of any liquid is directly proportional to the temperature of the liquid. This implies that, as the temperature of the liquid increases, the vapour pressure increases likewise and vice versa.

Since the vapour pressure of liquid varies with the temperature of the liquid, it is essential to know the water temperature in the experiment to determine the vapour pressure of water.

5 0
3 years ago
A 3.0-mol sample of kclo3 was decomposed according to the equation 2kclo3(s) --> 2kcl(s) + 3o2(g) how many moles of o2 are fo
d1i1m1o1n [39]
2moles of O2 are formed.
5 0
2 years ago
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

Cl_2,p_2'=26.398 Torr

PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

6 0
3 years ago
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