It is considered to be a conductor
Answer:
Methanol would be used as a reagent in excess, since it is a very low-cost solvent. For product isolation, the first thing to do is remove the methanol through a distillation process. The residue produced can be dissolved in diethyl ether. Using a NaHCO₃ solution, extraction is performed. When it separates into two phases, the product will be in the ether and the reagent in the aqueous phase. The ether can also be removed by distillation, and at the end of this process you will have the product you want.
Explanation:
Rubisco is an important enzyme that helps in making lifeless carbon of carbon dioxide into organic molecules. Rubisco takes carbon dioxide and attaches it to ribulose bisphosphate, a
short sugar chain with five carbon atoms that has rubp as its shortcut. Rubisco then clips the
lengthened chain into to polyglycerate pices, which are pretty flexible molecules and are also used in the feeding of the plant. Most of it is used in the photosynthesis pathway, but some of it is used to make sucrose
(table sugar) to feed the rest of the plant, or stored away in the form
of starch for later use. Hence, rubisco is crucial in the storing of the energy that is created from photosynthesis.
Depends on the element it can by up to 3, 8, or maybe 16.
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
