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What is the pin code for the lg vu cell phone?
Lx cu 920
What is the voltage of a galvanic cell with cu and mg 2.71?
The voltage of a galvanic cell can be calculated using the standard reduction potentials of the half-reactions involved. For a cell with copper (Cu) and magnesium (Mg), the standard reduction potential for Cu²⁺/Cu is +0.34 V, and for Mg²⁺/Mg, it is -2.37 V. The overall cell potential (E°cell) can be calculated as E°(cathode) - E°(anode), resulting in E°cell = 0.34 V - (-2.37 V) = 2.71 V. Therefore, the voltage of the galvanic cell with copper and magnesium is 2.71 V.
Weight of 1m by 1m by 1m rubber?
It depends on the density of rubber. Different kind of rubber have different density so different weights. weight = volume X Density in your case the volume is 1cuM. Natural rubber - 920 Kg/cu M Neoprene Rubber - 1230 kg/ cu M Silicone Rubber - 1150 kg/cu M EPDM Rubber - 860 kg/cu M Ramki- India
What gets oxidized In an electrolytic cell made with nickel and copper?
Cu(s)
Where can you buy Samsung touchscreen phones?
You can purchase Samsung touch screen phones from Samsung directly or you can purchase them from your cell phone carrier. Samsung is distributed by a bunch of retailers so you should check around to score a good deal.
Calculate the standard emf of a cell that uses the Mg Mg2 plus and Cu Cu2 plus half cell reactions at 25 degrees C?
Cu + Mg2 --------> Cu2 + Mg Cu --------------> Cu2 + 2e Mg2 + 2e --------> Mg Cu --------------> Cu2 + 2e (E = +0.35) Mg2 + 2e --------> Mg (E = -2.36V) +0.35 + (-2.36) = -2.01V --------------------------------------… Mg + Cu2 --------> Mg2 + Cu Mg --------------> Mg2 + 2e Cu2 + 2e --------> Cu Mg --------------> Mg2 + 2e (E = +2.36V) Cu2 + 2e --------> Mg (E = -0.35V) +2.36 + (-0.35) = +2.01V
What is the voltage of the electrochemical cell written as Cu's Cu2 aq Mg2 aq Mg's?
-2.71v
What is the overall voltage for the nonspontaneous redox reduction involving Mg and Cu and their ions?
The overall voltage for the nonspontaneous redox reaction involving magnesium (Mg) and copper (Cu) can be determined using standard reduction potentials. The reduction potential for Cu²⁺ to Cu is +0.34 V, while the oxidation potential for Mg to Mg²⁺ is -2.37 V. The overall cell potential (E°cell) is calculated by adding the reduction potential of the cathode (Cu) to the oxidation potential of the anode (Mg), resulting in E°cell = 0.34 V - 2.37 V = -2.03 V. Since the value is negative, the reaction is nonspontaneous under standard conditions.
What is the voltage of the electrochemical cell written as Cu(s) Cu2 (aq) Mg2 (aq) Mg (s)?
-2.71v
What is the total reduction potential of a cell which potassiumk is reduced and copper Cu is oxidized?
The total reduction potential of a cell where potassium is reduced and copper is oxidized can be calculated by finding the difference in the standard reduction potentials of the two half-reactions. The reduction potential for potassium reduction (K⁺ + e⁻ → K) is -2.92 V, and the oxidation potential for copper oxidation (Cu → Cu²⁺ + 2e⁻) is 0.34 V. Therefore, the total reduction potential of the cell is -2.92 V - 0.34 V = -3.26 V.
Why does hydrogen gas liberate at Cu electrode in Zn-Cu cell?
In a zinc-copper cell, zinc gives up electrons, forming the negative terminal of the cell, and the electrons flows as electric current through wires. When the electrons reach the other positive terminal, electrolysis of the electrolyte takes place at the positive terminal. Hydrogen ions and the cation of the electrolyte will be attracted to the positive Copper electrode. The hydrogen ions, being less reactive than the cation ions, will take up the electrons on the copper electrode, forming hydrogen gas.
What is the overall voltage for a redox reaction with the half reactions Mg s -- Mg2 plus plus 2e- and Cu2 plus -- Cu s?
The standard reduction potentials for Mg/Mg^2+ and Cu^2+/Cu are -2.37 V and +0.34 V, respectively. To determine the overall cell potential, you subtract the reduction potential of the anode (Mg/Mg^2+) from the reduction potential of the cathode (Cu^2+/Cu) since the anode is where oxidation occurs. Therefore, the overall cell potential would be 0.34 V - (-2.37 V) = 2.71 V.
