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cupoosta [38]
3 years ago
6

Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction

potential for copper(II) is 0.34 V and the standard reduction potential for silver(I) is 0.80 V what is the cell potential for this cell, in volts
Physics
1 answer:
lutik1710 [3]3 years ago
8 0

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

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A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
3 years ago
What would happen to a loop in a metal tube when it is heated
Galina-37 [17]
I think it most likely burn since th metal tube is going to transfer so much heat
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2 years ago
A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling
KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

     \frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} ...... (2)

Now, equating both equations (1) and (2) as follows.

 Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} = Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}        

        \gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}

                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

                    = \frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})

                    = 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0
2 years ago
Which THREE forms of light are invisible light?
PtichkaEL [24]

Answer:

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3 0
3 years ago
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A net force of 3000 N is accelerating a 1200 kg elevator upward. If the elevator starts from rest, how long will it take to trav
alexdok [17]

F = net force acting on the elevator in upward direction = 3000 N

m = mass of the elevator = 1200 kg

a = acceleration of the elevator = ?

Acceleration of the elevator is given as

a = F/m

a = 3000/1200

a = 2.5 m/s²

v₀ = initial velocity of the elevator = 0 m/s

Y = displacement of the elevator = 15 m

t = time taken

Using the kinematics equation

Y = v₀ t + (0.5) a t²

15 = (0) t + (0.5) (2.5) t²

t = 3.5 sec

5 0
3 years ago
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