From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
Answer:
It will require<u> second round</u> of the cycle to release
Explanation:
<u>Reason behind the requirement of second round of the cycle to release </u> -:
The C4 carbon of succinyl CoA is acetyl from acetyl CoA. Succinyl CoA is converted to succinate, which is then converted to fumarate, fumarate, malate, and eventually oxaloacetate. 14C will be found in oxaloacetate at either C1 or C4. During the second round of the loop, each of these carbons will be converted to carbon dioxide.
The answer is Al.
If it is a main group element with 3 electrons in its Lewis dot structure, it must be in group 3A. If it is in the 3p orbital section, then it must be in period 3, since the p orbital is a valence orbital and the number that preceeds it is the principal quantum number. Therefore, your answer is the element in period 3 and group 3A, which is aluminum.