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3 years ago

15. Discuss the various factors which affect the rate of evaporation. Latent heat

Chemistry

1 answer:

Georgia [21]3 years ago

8 0

Answer:

The correct answer is liquid B.

Explanation:

Latent heat of vaporization also known as the latent heat of evaporation. This latent heat transforms the particles of liquid into a gas without affecting its temperature. For example, the latent heat of evaporation for water is 40.8 kJ per mole, that is, 40.8 kJ per mole of heat is needed to transform water into vapor at 373 K.

It is known that latent heat of evaporation of a liquid is directly proportional to the cooling effect it generates, that is, more the latent heat of evaporation more will be its cooling effect. Thus, it is clear that liquid B will show the more cooling effect as the latent heat of evaporation of liquid B is more in comparison to liquid A. Thus, more heat will be captivated by liquid B and will generate more cooling effect in comparison to liquid A.

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Why is scientific notation important in chemistry?

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3 years ago

The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2

notsponge [240]

Answer : The unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law expression for the reaction:

$\text{Rate}=k[N_2]^a[H_2]^b$

where,

a = order with respect to $N_2$

b = order with respect to $H_2$

Expression for rate law for first observation:

$0.0021=k(0.10)^a(0.10)^b$ ....(1)

Expression for rate law for second observation:

$0.0084=k(0.10)^a(0.20)^b$ ....(2)

Expression for rate law for third observation:

$0.0672=k(0.20)^a(0.40)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2$

Dividing 3 by 1 and also put value of b, we get:

$\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[N_2]^a[H_2]^b$

$\text{Rate}=k[N_2]^1[H_2]^2$

Now, calculating the value of 'k' by using any expression.

$0.0021=k(0.10)^1(0.10)^2$

$k=2.1M^{-2}min^{-1}$

The value of the rate constant 'k' for this reaction is $2.1M^{-2}min^{-1}$

That means, the unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

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4 years ago

What must be known to determine the nature of the salt formed from a weak acid and weak base?

Umnica [9.8K]

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3 years ago

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finlep [7]

Answer: The statement is true

Explanation:

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For instance

The half-life of radium is 1622 years. This means that if we have 1000 radium atoms at the beginning, then at the end of 1622 years, 500 atoms would have disintegrated, leaving 500 undecayed radium atoms

Thus, the statement is true

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3 years ago