Ask question

Ask question

All categories

2 years ago

8

Freon-12, CF2Cl2, which has been widely used in air conditioning systems, is considered a threat to the ozone layer in the strat

osphere. Calculate the root-mean-square velocity of Freon-12 molecules in the lower stratosphere where the temperature is –65°C.

2 answers:

Vilka [71]2 years ago

5 0

Answer:

The root mean squared velocity for CF2Cl2 is $v_{rms}= 207.06 m/s$

Explanation:

From the question we are told that

The temperature is $T = -65 ^oC = -65+273 = 208K$

Root Mean Square velocity is mathematically represented as

$v _{rms} = \sqrt{\frac{3RT}{MW} }$

Where T is the temperature

MW is the molecular weight of gas

R is the gas constant with a value of $R = 8.314 JK^{-1} mol^{-1}$

For CF2Cl2 its molecular weight is 0.121 kg/mol

Substituting values

$v_{rms} = \sqrt{\frac{3 * 8.314 *208}{0.121} }$

$v_{rms}= 207.06 m/s$

otez555 [7]2 years ago

3 0

Answer:

The velocity of freon-12 is 207.0998 m/s

Explanation:

The molar mass of freon-12=120.91g/mol=0.1209kg/mol

The root-mean-square velocity of Freon-12 is:

$v=\sqrt{\frac{3RT}{m} }$

Where

R=8.31J/K mol

T=-65°C=208K

Replacing:

$v=\sqrt{\frac{3*8.31*208}{0.1209} } =207.0998m/s$

You might be interested in

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

Grace [21]

Answer : The value of $K_{eq}$ is, 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

3 0

2 years ago

Four balloons, each with a mass of 10.0 g, are inflated to a volume of 20.0 L, each with a different gas: helium, neon, carbon m

weeeeeb [17]

On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.

6 0

2 years ago

Read 2 more answers

The molecular weight of a gas is ________ g/mol if 6.7 g of the gas occupies 6.3 l at stp.

PIT_PIT [208]

At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol.
n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
6.7g/0.28125 mol = 23.82 g/mol

6 0

2 years ago

. Un cuerpo irregular se introduce dentro de una probeta que inicialmente tiene un volumen de 20 ml como muestra la figura Despu

MrMuchimi

Answer:

El volumen del cuerpo es el mismo al comienzo de la experiencia.

Explicación:

El volumen del cuerpo es el mismo al principio porque el volumen no cambia si la temperatura permanece igual. Si cambiamos la temperatura i. mi. Al aumentar la temperatura, las moléculas comienzan a expandirse y se produce un aumento de volumen mientras que cuando disminuimos la temperatura, las moléculas de esa sustancia comienzan a contraerse y el volumen de esa sustancia disminuye. Entonces concluimos que el volumen depende de la temperatura.

6 0

2 years ago

What vitamin a compounds bind with opsin to form rhodopsin??

irina [24]

The organic compound retinal binds with opsin and forms rhodopsin. Retinal is part of the molecule that is responsible for its color. This part is called chromophore. On the other hand, opsins are the proteins in photoreceptor cells. Retinal bounds with these opsins and forms rhodopsin: the basis of the human vision. Rhodopsin is also a protein.It is the pigment in the retinas of humans and animals.

7 0

2 years ago