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daser333 [38]
4 years ago
13

A certain rain cloud at an altitude of 1.80 km contains 3.20 107 kg of water vapor. How long would it take for a 2.90-kW pump to

raise the amount of water from Earth's surface to the cloud's position?
Physics
1 answer:
kvasek [131]4 years ago
7 0

Answer:

2255 days

Explanation:

height, h = 1.8 km = 1800 m

amount of water, m = 3.2 x 10^7 kg

Power, P = 2.9 kW = 2900 W

Let t be the time taken

Energy required to lift the water,

E = m g h

E = 3.2 x 10^7 x 9.8 x 1800 = 5.65 x 10^11 J

Power, P = Energy / time

t = E / P = (5.65 x 10^11) / 2900

t = 1.95 x 10^8 second

t = 2255 days

thus, the time taken is 2255 days.

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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
lutik1710 [3]

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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