I NEED HELP WITH #3 pleaseeee

a) One kmol of N2O4 dissociates at 25℃, 1 atm to form an equilibrium ideal gas mixture of N2O4 and NO2 in which the amount of N2

horsena [70]

Answer:

neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C

Explanation:

ni change eq.

N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C

NO2 0 0 + x x

∴ x = neq = Peq.V / R.T.....ideal gas mix

if P = 0.5 atm, T = 25°C; assuming: V = 1 L

⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))

⇒ x = neq = 0.0205 mol

⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol

3 0

3 years ago

A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and

Lunna [17]

The pH of the unknown solution is 3.07.

Explanation:

1.Find the cell potential as a function of pH

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0 lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

2. Find the Q value

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse) × 298.15

K × 2pH /2×96 485 C⋅mol

( inverse)

E cell= 0.0592 V × pH

3. Finding the pH value

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0

3 years ago

The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A

Mamont248 [21]

Answer:

The $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.

Explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for $B(g) = \frac{1}{2}A$

$K_c=[B]^2[A]$

= 0.045

The reverse will be $2B\leftrightharpoons A$

Then, $K_c = \frac{[A]}{[B]^2}$

= $\frac{1}{0.045}$

= $22m^{-1}$

The equilibrium constant for $B(g) = \frac{1}{2}A$ will be

$K_c = \sqrt{K_c}$

$=\sqrt{22}$

= 4.69

Therefore, $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.

5 0

3 years ago

Which electromagnetic radiation on the electromagnetic spectrum can be detected by the human eye?

qaws [65]

Answer:

D. Visible Light

Explanation:

D is the right answer because visible light is perceived by humans as color and the rest aren't visible at all. For example, if you broke a bone and have to have an X-ray done, you can't see the x-rays. When you are heating something in the microwave, you can't see the waves. And when you are listening to the radio in the car, you also can't see the waves.

4 0

2 years ago

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During cooling, the kinetic energy of the molecules falls. Why does this happen?

katrin [286]

During cooling, the kinetic energy of the molecules falls, because, when cooling a substace, the particles (molecules) slow down.

The kinetic energy is related to the speed, such that the lower speed the lower kinetic energy.

Particles can translate and vibrate, in the case of gases and liquids, and only vibrate (in the case of solids).

As a substance is cooled the particles get closer and the motion (translation and vibration), slows down. You can see by the equation of the kinetic energy (KE):

KE = [1/2]mass×(speed)² that as the speed is lower the KE will also be lower.

Additionally, when the cooling does not drive a change of phase (gas to liquid, liquid to solid, or solid to gas), it drives a decrease on temperature. In this case you should know that the temperature is a measure of the kinetic energy: the lower the temperature, the lower the kinetic energy.

3 0

3 years ago

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