Answer is: Li < Na < K.
Lithium (Li), sodium (Na) and potassium (K) are alkaline metals.
Alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons (one electron).
Potassium has lowest ionization energy and it is most reactive, because valcense electrons is farthest from nucleus.
Electron configuration of lithium atom: ₃Li 1s² 2s¹.
Electron configuration of sodium atom: ₁₁Na 1s² 2s² 2p⁶ 3s¹.
Electron configuration of potassium is: ₁₉K 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
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).
Answer:
It occurs during the continuous glyceraldehyde-3-phosphate phosphorilation.
Explanation:
Hello,
First of all, the process of extracting energy from glucose is called glycolysis. Then, the NADH oxidation happens during the glycolysis' sixth step when the glyceraldehyde-3-phosphate is phosphorilated by the addition of another phosphate group due to the electron carrier NAD+ which produces NADH. In fact, that is not the requested oxidation but it occurs when this process becomes continuous as long as the NADH is oxidized back to NAD+ to keep on the process.
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Answer: 60.7 g of will be formed.
Explanation:
To calculate the moles :
The balanced chemical reaction is
is the limiting reagent as it limits the formation of product and is the excess reagent.
According to stoichiometry :
6 moles of produce = 4 moles of
Thus 2.68 moles of will produce= of
Mass of
Thus 60.7 g of will be formed by reactiong 60 L of hydrogen gas with an excess of