How did the instrument in the picture help to disprove part of Dalton's atomic model?

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

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

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}$

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142$

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen = $d_{O_2}=\frac{1 m}{t}$

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane= $d_{CH_4}=\frac{y }{t}$

$\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142$

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0

3 years ago

2) Which liberates the most energy?

evablogger [386]

Answer: $F(g)+e^-\rightarrow F^-(g)$

Explanation:

Electron gain enthalpy is defined as energy released on addition of electron to an isolated gaseous atom.

The amount of energy released will be maximum when the tendency to attract electrons is maximum. As flourine has atomic number of 9 and has electronic configuration of 2,7. It can readily gain 1 electron to attain stable noble gas configuration and hence liberates maximum energy.

$F(g)+e^-\rightarrow F^-(g)$

4 0

3 years ago

What is the molarity of 225 grams of Cu(NO2)2 in a total volume of 2.59 L?

Kay [80]

Answer:

The molarity is 0.56 $\frac{moles}{L}$

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.

In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by:

$Molarity (M)=\frac{number of moles of solute}{Volume}$

Molarity is expressed in units ( $\frac{moles}{liter}$ ).

Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:

Cu: 63.54 g/mol
N: 14 g/mol
O: 16 g/mol

the molar mass of Cu(NO₂)₂ is:

Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol

Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?

$moles=\frac{225 g*1 mole}{155.54 g}$

moles= 1.45

So you know:

number of moles of solute= 1.45 moles
volume=2.59 L

Replacing in the definition of molarity:

$Molarity=\frac{1.45 moles}{2.59 L}$

Molarity= 0.56 $\frac{moles}{L}$

The molarity is 0.56 $\frac{moles}{L}$

5 0

3 years ago

What volume of a 6.67 m nacl solution contains 3.12 mol nacl??

natima [27]

The volume of a 6.67 m nacl solution that contains 3.12 mol nacl is 0.468.

You just need to follow this formula:

C = n / V

So, it will be

V = n / C

= 3.12 / 6.67 moles / Liter

= 0.468 L

6 0

3 years ago

Read 2 more answers

Select all statements that are correct:____

topjm [15]

Answer:

A, C and D are correct.

Explanation:

Hello.

In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:

$P_{vap}^{solution}=x_{solvent}P_{solvent}$

Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.

Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.

Best regards.

4 0

3 years ago