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Serhud [2]
3 years ago
6

A postulate of the kinetic molecular theory is listed.

Chemistry
1 answer:
valina [46]3 years ago
5 0

Answer:

Postulate: Gas particles are extremely small and are far apart.

The activities can be used to demonstrate the postulate is :

<u>Observing colored gas spreading into an inverted jar placed on top of a jar containing the gas</u>

<u />

Explanation:

colored gas spreading into an inverted jar placed on top of a jar containing the gas:

This occur because of two reasons:

1. <em><u>The Gaseous particles are largely spaced . There is large distance between the gases molecule</u></em>

<em><u>2. The gases are in continuous motion . Hence they posses very high kinetic energy . This is the reason they mixes quickly if placed in a jar.</u></em>

<em><u>This occur by the process of diffusion. </u></em>

Diffusion of Gases: The intermixing of particles from the region of high concentration to low concentration.

The coloured gas goes into the space between the gaseous molecule present in the jar.(Gases are far apart)

As soon as the coloured gas is mixed in the jar , It spread quickly by diffusion because , The gaseous particles are extremely small and are far apart.

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start out as one kind of rock, but large amounts of pressure and heat change them into a different kind.
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Answer:

Metamorphic

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True or false if an object displaces 50 ml of water the object's volume is 50mL
andrew11 [14]

Answer:

True

Explanation:

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3 years ago
(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

3 0
3 years ago
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Explanation:

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