Educated Guess Here!
Since Br-80 does not exist, maybe that means Br-79 or Br-81 have very unequal abundances. For example, Br-79 may have 75% abundance whereas Br-81 may have 25% abundance.
Answer:
And what is your question
Explanation:
Answer:
c. decarboxylation of an a-keto acid.
Explanation:
Decarboxylation refers to the removal of the carboxyl group from a carboxylic acid and thus releasing carbon dioxide. Decarboxylases are enzymes that speed up the removal of the carboxyl group from acids. These reactants could be amino acids, alpha-keto acids, and beta-keto acids. Biotin is known to catalyze the decarboxylation of malonyl CoA to acetyl CoA during fatty acid synthesis.
Malonyl CoA is converted to acetyl CoA after decarboxylation assisted by biotin also known as Vitamin H. Alpha keto acids are involved in fatty acids synthesis and Malonyl CoA is an alpha-keto acid because the keto group is located in the first carbon near the carboxylic acid group. Keto acids have both a carboxyl group and a ketone group.
Answer:
so they can find the aswer and not get hurt
Explanation:
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
