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3 years ago

Please help me:)

Physics

1 answer:

Alenkinab [10]3 years ago

8 0

Answer:

The correct option is;

Sphere I is positively charged and sphere II is negatively charged

Explanation:

The charging of the spheres by induction is achieved by introducing a charge to the metal spheres that are insulated from the ground to prevent loss of charge by placing them on insulating stand

The two spheres are brought into contact by the connection of a conducting wire between the spheres I and II

The presence of the positively charged sphere III draws attracts electrons towards sphere II while the net positive charge moves towards sphere I

While the spheres I and II are still polarized, the conducting wire is removed while the presence of sphere III continues to keep sphere II negative compared to sphere I

After removing the connecting wire, sphere III is removed leaving the excess negative charge on sphere II and the excess positive charge on sphere I

The net charges then evenly redistribute themselves on each sphere creating two oppositely charged spheres.

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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

Which type radiation can be observed well from Earth's surface?

umka2103 [35]

Answer:

Eletromagnetic radiation which is also known as visible light.

Explanation:

4 0

3 years ago

Read 2 more answers

How does the time at which you see things happen at a baseball game compare to the time when you hear things happen? Explain you

hoa [83]

Answer:

Because of the speed of the sound.

Explanation:

The first thing that happens in such cases is to take into account the speed of the sound. First, we see that the player hits the ball with the bat, if we are in the stands far enough we will hear the sound of the batting time later, this time depends on the speed of the sound which is equal to 345 [m/s].

Another visible and practical example is a fireworks display, where people nearby immediately hear the explosion. while those at a great distance will be able to see first the explosion followed by the sound.

With the following equation, we can calculate how long it takes to hear a hit or explosion

t = x / v

where:

x = distance [m]

v = sound velocity = 345 [m/s]

t = time [s]

7 0

3 years ago

The density of lead is 30.2g/cm^3.what is the value in kilograms per meter cube?

V125BC [204]

Answer:
11300 kgm3

Hope this helps

3 0

2 years ago

Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam

Hatshy [7]

Think of it like a graph. You start at the origin which is (0,0). go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6). 1 to the east, ((4,6). Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite.

Hope this helps!

6 0

3 years ago