All categories

1 year ago

3. What is the total pressure of a mixture of nitrogen and hydrogen, in atm, if the

Chemistry

1 answer:

telo118 [61]1 year ago

8 0

Answer:

As per Dalton's law of Partial pressure,

The total pressure of a mixture of gases is equal to sum of partial pressure of individual components of the mixture.

So, P=P_H+P_N+P_WP=P

\implies 864\ mm=220\ mm+410\ mm+P_W\implies⟹864 mm=220 mm+410 mm+P

⟹ P_W=864-630=234P

=864−630=234 mm of Hg.

Explanation:

Hope it helps<3

You might be interested in

At a certain temperature, ethane has a vapor pressure of 304 torr and propane has a vapor pressure of 27 torr. What is the mole

SOVA2 [1]

Answer:

166 torr

Explanation:

Let’s call ethane Component 1 and propane Component 2.

According to Raoult’s Law,

$p_{1} = \chi_{1}p_{1}^{\circ}\\p_{2} = \chi_{2}p_{2}^{\circ}$

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Data:

p₁° = 304 torr

p₂° = 27 torr

n₁ = n₂

1. Calculate the mole fraction of each component

χ₁ = n₁/(n₁ + n₂)

χ₁ = n₁/n₁ + n₁)

χ₁ = n₁/(2n₁)

χ₁ = ½

χ₁ = 0.0.5

χ₂ = 1- χ₁ = 1- 0.5 = 0.5

2. Calculate the vapour pressure of the mixture

$p_{1} = 0.5 \times \text{304 torr} = \text{ 152 torr}\\p_{2} = 0.5 \times \text{27 torr} = \text{ 13.5 torr}\\p_{\text{tot}} = p_{1} + p_{2} = \text{152 torr + 13.5 torr} = \textbf{166 torr}$

$\text{The vapour pressure above the solution is $\boxed{\textbf{166 torr}}$}$

8 0

3 years ago

Explain why sodium chloride solution cannot be used as the electrolyte to produce

ira [324]

Explanation:

Because the solution only contains Na+ and Cl− ions, and water, and not the metal Na(s) . You cannot evaporate the water from the solution and hope to obtain anything but NaCl(s) .

Na(s) has an electron configuration of 1s^2 2s^2 2p^6 3s^1 , but Na+(aq) has an electron configuration of 1s^2 2s^2 2p^6 . That means they are not the same element, and thus, there is no straightforward way of extracting Na(s) from a NaCl (aq) solution

6 0

1 year ago

What can you tell me about the charge signs on the spheres

bazaltina [42]

<span>A similar thing occurred with the circles as did with the electroscope. When we initially brought the charged pole close to, the bar pulled in the circle since it was of polarization. At that point, once the bar touched the circle, the pole repulsed the circle. This is on the grounds that once the pole and circle touched, the electrons exchanged thus did the protons, consequently leaving the circle with a positive net charge. The nearer the bar is to the circle the more it repulsed, however, it didn't influence the charge of the circle once the circle was touched by the pole.</span>

6 0

2 years ago

Electrons orbit the nucleus like planets around the sun.<br>true or false?

gogolik [260]

Answer: true

Explanation: Electrons orbit the nucleus of an atom the way that planets revolve around the Sun. The electrons are like the planets in the solar system. The sun is like the nucleus in the solar system. The answer to the question is true.

7 0

2 years ago

Read 2 more answers

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0

2 years ago