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3 years ago

Which of the following has a mass of 10.0 g?

Chemistry

1 answer:

Serjik [45]3 years ago

5 0

Answer:

a. 0.119mol Kr

Explanation:

To solve this problem, we must understand that;

Mass = number of moles x molar mass

Molar mass of Kr = 83.3g/mol

Ar = 40g/mol

He = 4g/mol

Ne = 20.18g/mol

a0.119 mol Kr mass = 0.119 x 83.3 = 9.9g

b 0.400 mol Ar mass = 0.4 x 40 = 16g

C 1.25 mol He mass = 1.25 x 4 = 5g

d 2.02 mol Ne mass = 2.02 x 20.18 = 40.8

Krypton is the answer

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Which of the following results in an increase in the entropy? (4 points)

Dima020 [189]

Dissolving sugar in water results in an increase in the entropy.

Hence, Option (D) is correct answer.

<h3>What is Entropy ?</h3>

Measurement of randomness of a system is called entropy. It is an extensive property. It is a state function. Unit of entropy is JK⁻¹ mol⁻¹.

Now lets check all options one by one

Option (A): Freezing water

Freezing water decreases the entropy because here second law of thermodynamics does not violate.

So it is incorrect option

Option (B): Cooling water

Cooling water does not increases entropy because entropy increases when solid melts to give liquid.

So it is incorrect option

Option (C): Condensing water vapour

In Condensing water vapour the temperature of liquid phase decreases and thus kinetic energy decreases. The randomness will decrease and hence entropy will also decrease.

So it is incorrect option.

Option (D): Dissolving sugar in solute

In dissolving sugar in solute the solid dissociates to ions and the randomness will increase and hence entropy will also increase.

So it is correct option

Thus from the above conclusion we can say that Dissolving sugar in water results in an increase in the entropy.

Hence, Option (D) is correct answer.

Learn more about the Entropy here: brainly.com/question/1477087

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6 0

2 years ago

Of the following gases, ________ will have the greatest rate of effusion at a given temperature. Of the following gases, _______

kolezko [41]

<u>Answer:</u> From the given gases, the greatest rate of effusion is of $CH_4$

<u>Explanation:</u>

Rate of effusion of a gas is determined by a law known as Graham's Law.

This law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

It is visible that molar mass is inversely related to rate of effusion. So, the gas having lowest molar mass will have the highest rate of effusion.

For the given gases:

Molar mass of $NH_3=17g/mol$

Molar mass of $HCl=36.5g/mol$

Molar mass of $CH_4=16g/mol$

Molar mass of $Ar=40g/mol$

Molar mass of $HBr=81g/mol$

The molar mass of methane gas is the lowest. Thus, it will have the greatest rate of effusion.

Hence, the greatest rate of effusion is of $CH_4$

8 0

4 years ago

Please help me with this problem.

brilliants [131]

1 μL = 10^-6 L by definition

So...
82 μL = 82x10^-6 L = 8.2x10^-5 L

3 0

3 years ago

A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch

ivanzaharov [21]

Answer:

The reactions free energy $\Delta G = -49.36 kJ$

Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

The pressure of (Cl) gas is $P_{Cl} = 9.15 \ atm$

The pressure of nitrosly chloride (NOCl) is $P_{(NOCl)} = 7.70 \ atm$

The reaction is

$2NO_{(g)} + Cl_2 (g)$ ⇆ $2 NOCl_{(g)}$

From the reaction we can mathematically evaluate the $\Delta G^o$ (Standard state free energy ) as

$\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}$

The Standard state free energy for NO is constant with a value

$\Delta G^o _{NO} = 86.55 kJ/mol$

The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

The Standard state free energy for $NOCl$ is constant with a value

$\Delta G^o _{NOCl} =66.1kJ/mol$

Now substituting this into the equation

$\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6$

$= -43 kJ/mol$

The pressure constant is evaluated as

$Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }$

Substituting values

$Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}$

$= 0.0765$

The free energy for this reaction is evaluated as

$\Delta G = \Delta G^o + RT ln Q$

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

T is temperature in K with a given value of $T = 25+273 = 298 K$

Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

$= -43-6.36$

$\Delta G = -49.36 kJ$

4 0

3 years ago

What is the mass, in grams, of one mole of any substance known as?

KATRIN_1 [288]

Answer:

I think it's answer would be A

6 0

3 years ago