All categories

3 years ago

According to Graham’s law, the rate of effusion of a gas is inversely proportional to

Chemistry

1 answer:

VLD [36.1K]3 years ago

5 0

C. the square root of the mass of the particles.

<h3>Further explanation </h3>

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or

the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

$\rm M_1\times r_1^2=M_2\times r_2^2$

From this equation shows that the greater the mass of the gas, the smaller the effusion rate of the gas and vice versa, the smaller the mass of the gas, the greater the effusion velocity.

So if both gases are at the same temperature and pressure, the above formula can apply

You might be interested in

Consider the reaction: 2 SO2(g)+O2(g)→2 SO3(g) If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mm

Tanzania [10]

Answer:

a. Oxygen is the limiting reagent. $n_{SO_3}^{Theoretical}=8.096x10^{-4}mol SO_3$

b. $Y=58.9$ %

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

$n_{SO_2}=\frac{PV}{RT}=\frac{0.0658atm*0.2855L}{0.082\frac{atm*L}{mol*K}*315K} =7.273x10^{-3}molSO_2 \\n_{O_2}=\frac{PV}{RT}=\frac{0.0658atm*0.1589L}{0.082\frac{atm*L}{mol*K}*315K} =4.048x10^{-4}molO_2$

Afterwards, by considering the properly balanced chemical reaction:

$2SO_2(g)+O_2(g)-->2SO_3$

We compute the oxygen's moles that completely reacts with the previously computed $7.273x10^{-3}$ moles of $SO_2$ as follows:

$7.273x10^{-3}molSO_2*\frac{1molO_2}{2molSO_2} =3.6365x10^{-3}molO_2$

That result let us know that the oxygen is the limiting reagent since just $4.048x10^{-4}$ moles are available in comparison with the $3.6365x10^{-3}$ moles that completely would react with $7.273x10^{-3}$ moles of $SO_2$ .

Now, to compute the theoretical yield of sulfur trioxide, we apply the following stoichiometric relationship:

$n_{SO_3}^{Theoretical}=4.048x10^{-4}molO_2*\frac{2molSO_3}{1molO_2} =8.096x10^{-4}mol SO_3$

b. Percent yield.

At first, we must compute the collected (real) moles of sulfur trioxide:

$n_{SO_3}^{real}=\frac{PV}{RT}=\frac{0.0658atm*0.1872L}{0.082\frac{atm*L}{mol*K}*315K} =4.769x10^{-4}molSO_3$

Finally, we compute the percent yield:

$Y=\frac{n_{SO_3}^{real}}{n_{SO_3}^{Theoretical}} *100$ %

$Y=\frac{4.769x10^{-4}mol SO_3}{8.096x10^{-4}mol SO_3} *100$ %

$Y=58.9$ %

Best regards.

7 0

3 years ago

Read 2 more answers

What is the empirical formula for a compound if a sample contains 1.0 g of S and 1.5 g of O?

sammy [17]

I think it’s SO3 I’m not quite sure though

6 0

2 years ago

Does a relationship exist between a planets mass and becoming a ring system?

joja [24]

Answer:

A planet is believed to form when the mutual gravitational attraction of debris orbiting a star accumulates into a growing mass.

Explanation:

Planetary rings, then, consist of millions of separate small rock and ice particles, each maintaining their own orbit around the host planet.

6 0

3 years ago

How many formula units are in 26.5 g MgCl2?

kaheart [24]

Answer:

95.211 g/mol

Explanation:

For example magensium chloride has the formula unit MgCl2 , i..e., for every magnesium cation there are two chloride anions in the crystal. Or, we would say that the formula weight of MgCl2 = 95.211 g/mol.

7 0

3 years ago

Which element was used by the ancient egyptians, in the form of borax, to make mummies?

Kamila [148]

Answer:

boron compound - known as Borax

Explanation:

5 0

3 years ago