If a liquid dissolves in a gas in what state will the solution be in? Explain

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago

Two protons are released from rest when they are 0.720 nm apart. For related problem-solving tips and strategies, you may want t

Gnesinka [82]

Answer:

a) Speed of the electrons at maximum speed = (1.384 × 10⁴) m/s

The maximum speed occurs at the point where all of the initial potential energy is converted into kinetic energy.

b) Maximum acceleration of the protons = (2.660 × 10¹⁷) m/s²

The maximum acceleration occurs at the minimum distance apart for the two protons.

Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

The potential energy between the two protons at the instant of release is given by

U = (kq₁q₂/r)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

U = (kq²/r) = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ (7.2 × 10⁻¹⁰) = (3.204 × 10⁻¹⁹) N/m or Joules

At the maximum speeds, the two protons will not possess any potential Energy, only kinetic energy.

The sum of kinetic and potential energies is always constant for the system

(Initial Kinetic Energy) + (Initial Potential Energy) = (Kinetic Energy at maximum speed) + (Potential Energy at maximum speed)

Initial Kinetic Energy of the system = 0 J (Since both protons were intially at rest)

Initial Potential Energy = (3.204 × 10⁻¹⁹) J

Kinetic Energy at maximum speed = Sum of the kinetic energies of the protons at this point = (½mv²) + (½mv²) = (mv²) J (Since theu are both protons, they have the same mass and the same speed at maximum speed)

Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v = speed of each of the protons at maximum speed = ?

v = √[(3.204 × 10⁻¹⁹) ÷ m]

v = √[(3.204 × 10⁻¹⁹) ÷ (1.673 × 10⁻²⁷)]

v = √(1.915 × 10⁸) = 13,838.8 m/s = (1.384 × 10⁴) m/s

b) Since the two protons repel each other and force of repulsion reduces as the dI stance between the protons increases, the maximum acceleration occurs at the minimum distance apart for the two protons.

Force of repulsion acting on each proton is given through Coulomb's law as

F = (kq₁q₂/r²)

And the force acting on each proton is obtainable using Newton's law that

F = ma

So, the acceleration of each proton at any time is obtainable through a relation of these 2 formulas.

ma = (kq₁q₂/r²)

a = (kq₁q₂/r²m)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ [(7.2 × 10⁻¹⁰)² × (1.673 × 10⁻²⁷)]

a = (2.660 × 10¹⁷) m/s²

Hope this Helps!!!

5 0

3 years ago

To have the highest magnification in a telescope, the focal length of the objective lens should be _________ and the focal lengt

Darina [25.2K]

Answer:

Large; small.

Explanation:

A telescope can be defined as an optical instrument or device which comprises of a curved mirror and lenses used for viewing distant objects i.e objects that are very far away such as stars and other planetary bodies. The first telescope was invented by Sir Isaac Newton.

To have the highest magnification in a telescope, the focal length of the objective lens should be large and the focal length of the eyepiece lens should be small.

This ultimately implies that, the eyepiece lens has a small focal length while the objective lens has a large focal length.

6 0

2 years ago

if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball

Ray Of Light [21]

There has to be 6 N for a reactant

5 0

3 years ago

Which of the following changes would decrease the coefficient of friction needed for this ride?

posledela

Friction occurs between two contacting surfaces. The coefficient of friction is very much dependent on the roughness of these surfaces. Some of the many ways in which the coefficient can be lessened or decreased are to lubricate the surface or make it shiny by eliminating the spikes which caused the roughness.

7 0

3 years ago