On complete conversion (100% yield) 9.75 g of ethyl butyrate will be produced. Below is the solution.....
<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
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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The molecules or atoms that are formed by gain or loss of one or more valence electrons are said to be ions.
When atom loss one or more valence electrons, results in formation of cation whereas when atom gain one or more valence electrons, then formation of anion occurs. Cations carry positive charge and anions carry negative charge.
In general, cations are smaller than the neutral atoms from which they are formed and anions are larger than the neutral atoms.
As cations are smaller than the related neutral atoms because the valence electrons are lost which are farthest away from the nucleus. After that, taking more electrons distant from the cation results in reduction of radius of the ion.
Thus, aluminium cation consist of few electrons which results in fewer occupied energy levels by the electrons further results in reduction of radius i.e. smaller size.
Hence, given statement is true i.e. aluminium atom is larger than the aluminium cation as cation has fewer occupied energy levels.
