Answer:
Ok so, b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...
i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced
ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.
iii. Calculate the standard potential (voltage) of the cell
Look up the reduction potential,
E
⁰
red
, for the reduction half-reaction in a table of reduction potentials
Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,
E
⁰
ox
=
-
E
⁰
red
.
iv. What kind of electrochemical cell is this? Explain your answer.
All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells
1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery
Explanation:
We can use the ideal gas equation:
PV = nRT
P = 202.6kPa = 202600 Pa (You have to
multiply by 1000)
n = 0.050 mole
R = 0.082 atm*l/(K*mol)
T = 400K
We will have to convert from Pa to atm or
viceversa.
101325 Pa________1 atm
202600 Pa________x = 2.00 atm
2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K
V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm
= 0.82 liters = 820 mililiters
Answer:
False
Explanation:
Cell division is a primary procedure by all cells. This process in the life cycle of cells ensures that new cells (daughter cells) are produced. The two major process are; meiosis and mitosis.
Normal cells would divide if the nutrients required are not present, since water is the main nutrient required. But in some cases, the division process may lead to an effect on one or both daughter cells. This effect is called mutation, thereby mutated cells may be produced.
Answer:
Se =[Ar] 3d¹⁰ 4s² 4p⁴
Explanation:
The noble gas notation is used for the shortest electronic configuration of other periodic table elements.
For example:
The atomic number of Argon is 18, and its electronic configuration is,
Ar₁₈ = 1s² 2s² 2p⁶ 3s² 3p⁶
The atomic number of selenium is 34, its electronic configuration is,
Se₃₄ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴
By using the noble gas notation, electronic configuration of selenium can be written is shortest form.
Se =[Ar] 3d¹⁰ 4s² 4p⁴
This electronic configuration is also called abbreviated electronic configuration.
14 since K has 1 valence but there’s two so 2 valence for k and oxygen has 6 but there’s two so 12
