Answer:
Contributes to the membrane potential.
Explanation:
Sodium-potassium pump: In cellular physiology, a protein which is identified in many cells that helping in to maintain the higher concentration of potassium ions inside than that is in the surrounding medium and maintain the lower concentration of sodium ions inside than that of the surrounding medium.
This unbalanced charge transfer contributes in the separation of charge across the cell membrane. Sodium-potassium pump is known for important contributor to action potential which is produce by nerve cells.
Methane Volume : O.O
Methane Mass: 0.100g
Molar Mass of Methane : 16.04 g/mol
Answer:
Cell cycle.
Explanation:
A cell can be defined as the fundamental or basic functional, structural and smallest unit of life for all living organisms. Some living organisms are unicellular while others are multicellular in nature.
A unicellular organism refers to a living organism that possess a single-cell while a multicellular organism has many (multiple) cells.
Generally, cells have the ability to independently replicate themselves. These cells can be compared to the kind of structures found in a business or factory, where you have different workers performing different functions.
In a cell, the "workers" that perform various functions or tasks for the survival of the living organism are referred to as organelles and they include nucleus, cytoplasm, cell membrane, golgi apparatus, mitochondria, lysosomes, ribosomes, chromosomes, endoplasmic reticulum, vesicles, etc.
The regular sequence of growth and division that cells undergo is called the cell cycle. This cycle makes it possible for the cells found in living organisms to divide and produce new cells.
Basically, there are four (4) phases of the cell cycle and these are;
I. Prophase.
II. Metaphase.
III. Anaphase.
IV. Telophase.
Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
![E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)
Now we have to calculate the standard Gibbs free energy.
Formula used :
where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the standard Gibbs free energy is +91 kJ/mole
