Yes it can as it’s made of protons neutrons and electrons which is how we break the atoms down
Answer:
The empirical formula would be N₂Os * Page 2 Calculate the empirical formulaof a compound that is 94.1% oxygen, 5.9% hydrogen.
Answer:
2.67 × 10⁻²
Explanation:
Equation for the reaction is expressed as:
CaCrO₄(s) ⇄ Ca₂⁺(aq) + CrO₂⁻⁴(aq)
Given that:
Kc=7.1×10⁻⁴
Kc= ![[Ca^{2+}][CrO^{2-}_4]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5BCrO%5E%7B2-%7D_4%5D)
Kc= [x][x]
Kc= [x²]
7.1×10⁻⁴ = [x²]
x = 
x = 0.0267
x = 
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.
To learn more about the oxidation of cytochrome c, visit:
brainly.com/question/14473523
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Explanation:
These elements are rare because:
<u>Helium fuses into the carbon by the combination of three helium nuclei (Z = 2) and one carbon nucleus (Z = 6), therefore bypassing elements with Z= 3, 4 and 5 which are lithium, beryllium, and boron respectively. Therefore, the fusion processes in cores of the stars do not form these three elements. </u>
