What prevents the pressure from increasing as a cloud contracts due to its gravity?

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

2 years ago

ANSWER QUICK 30 POINTS

Lelu [443]

What do you need help on

4 0

3 years ago

In an experiment what are all the parts of an experiment that remain unchanged are called<br>

stira [4]

The answer for this question is Control Variable because it doesn’t change throughout the experiment.

3 0

3 years ago

A. make an argument for halting the advances of technology? b. make an argument for continuing advances in technology

Naddik [55]

<span>Halting advances in tech could be useful because the automation in many tasks has rendered the human touch obsolete in many lower-level positions. They can be done by machines without the need for stoppage time for breaks. Continuing technology can be useful because it can increase output and make it easier for people to get what they're wanting or needing without spending extravagant amounts of resources.</span>

6 0

3 years ago

Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro

marishachu [46]

Answer:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N , θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y= -46.1 *sin28.8° = -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N -22.2 N = + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey= - 8.4 N

Magnitude of the equilibrant (Fe)

$F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2} }$

$F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2} }$

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

$\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )$

$\beta = tan^{-1} (\frac-8.4 }{26.92 } )$

β = -18.34°

β = 18.34° , clockwise from the positive x axis

8 0

2 years ago