0.116 V is the e value for the oxidation of cytochrome c by the cue redox center in complex iv when the ratio of cyst c (fe3 ) /cyst c (fe2 ) is 20 and the ratio of cue (cu2 )/cue (cu ) is 3.
<h3>
Explain the process of oxidation of cytochrome c.</h3>
When cytochrome c is oxidized by mitochondrial cytochrome oxidase (COX), it attaches to Apaf-1 to produce the apoptozole, which activates pro-caspase-9 and causes cell death. Cyst can be created from cytosolic cytochrome c. In the IMS, oxidized cytochrome c can scavenge superoxide without converting it into H2O2, a process that happens naturally but is accelerated by SOD. The benefit of scavenging superoxide independently of H2O2 synthesis is reducing the possibility of hydroxyl radical generation via the Fenton reaction.
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<span>1. Tap water has a small concentration of H+ & OH- ions as well as water molecules, hence there would be permanent dipole-permanent dipole (p.d.-p.d.) forces of attraction between the water molecules (aka H-bonds) as well as ionic bonds between the H+ & OH- ions.
2. Distilled water does not have H+ & OH- ions, hence only H-bonds exist between the water molecules.
3. There are covalent bonds between the individual sugar molecules.
4. There are ionic bonds between the Na+ & Cl- ions in NaCl.
5. There are p.d.-p.d. forces of attraction between the Na+ ions and the O2- partial ions of the water molecules as well as between the Cl- ions and the H+ partial ions of the water molecules. There are also H-bonds between the individual water molecules and ionic bonds between the Na+ & Cl- ions (although these are in much lower abundance than in unsolvated solid NaCl).
6. There are i.d.-i.d. as well as p.d.-p.d. forces of attraction between the sugar molecules and the water molecules. There are also H-bonds between the individual water molecules and covalent bonds within the sugar molecules.</span>
Answer is: D. Cl (chlorine).
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Barium, potassium and arsenic are metals (easily lost valence electrons), chlorine is nonmetal (easily gain electrons).
Alkaline metals (in this example, potassium) have lowest ionizations energy and easy remove valence electrons (one electron), earth alkaline metals (in this example, barium) have higher ionization energy than alkaline metals, because they have two valence electrons.
Nonmetals (in this example chlorine) are far right in the main group and they have highest ionization energy, because they have many valence electrons.
