Answer:
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Explanation:
natural and social environment.
Answer: a) Anode: 
Cathode: 
b) Anode : Cr
Cathode : Au
c) 
d) 
Explanation: -
a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
At cathode: 
At anode: 
b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.
At anode which is a negative terminal, oxidation occurs which is loss of electrons.
Gold acts as cathode ad Chromium acts as anode.
c) Overall balanced equation:
At cathode:
(1)
At anode:
(2)
Adding (1) and (2)

d)
= -0.74 V
= 1.40 V

Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAu%5E%7B3%2B%7D%5D%7D%7B%5BCr%5E%7B3%2B%7D%5D%5E%7D)
where,
n = number of electrons in oxidation-reduction reaction = 3
= standard electrode potential = 2.14 V
![E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.14-%5Cfrac%7B0.0592%7D%7B3%7D%5Clog%20%5Cfrac%7B%5B1.0%7D%7B%5B1.0%5D%7D)

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.
Answer:
doppler radar, radiosondes, weather satellites, buoys
Answer: orbitals supernumerary
Explanation:
Answer:
0.001 mole of NaF.
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 100 mL
Molarity = 0.01 M
Mole of NaF =?
Next, we shall convert 100 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100 mL × 1 L / 1000 ml
100 mL = 0.1 L
Thus, 100 mL is equivalent to 0.1 L.
Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:
Volume of solution = 0.1 L
Molarity = 0.01 M
Mole of NaF =?
Molarity =mole /Volume
0.01 = mole of NaF / 0.1
Cross multiply
Mole of NaF = 0.01 × 0.1
Mole of NaF = 0.001 mole.
Thus, 0.001 mole of NaF is present in 100 mL of the solution.