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ankoles [38]
8 months ago
14

After completion of the citric acid cycle, most of the usable energy from the original glucose molecule is in the form of ______

____.
A. acetyl CoA
B. ATP
C. CO2
D. FADH2
E. NADH
Chemistry
1 answer:
GaryK [48]8 months ago
5 0

After completion of the citric acid cycle, most of the usable energy from the original glucose molecule is in the form of B. ATP

The most basic kind of sugar, monosaccharides only have one kind of sugar molecule. The simplest sugar is glucose, which is also your body's primary energy source. In tests to determine blood sugar levels, sugar is measured. Fructose and galactose, two more monosaccharides, are converted into glucose via metabolism ( 1 , 2 ). For instance, ATP is necessary for both breathing and keeping your heart beating. As well as assisting in the synthesis of lipids and nerve impulses, ATP also facilitates the entry and exit of certain molecules from cells. Even some living things, including bioluminescent creatures like jellyfish and fireflies, use ATP to create light!

Learn more abut ATP here:

brainly.com/question/14637256

#SPJ4

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Ketones undergo a reduction when treated with sodium borohydride, NaBH4. The product of the above reaction has the following spe
Anna71 [15]

Answer:

The product is cyclohexanol

Explanation:

Firstly,

A ketone undergo a borohydride reduction reaction to form an alcohol as below,

R-CO-R'  ⇒ R-CO(OH)-R'

  1. IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹
  2. From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.
  3. Check with other spectroscopic properties,
  • 3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-<u>C</u>H₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-<u>C</u>H₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-<u>C</u>H(CH₂-)-OH).
  • 1H NMR confirms,

        1.56 δ (4H, triplet) - (-C<u>H</u>₂-CH₂-CH-OH) ; triplet as coupling with 2 H,

        1.78 δ (4H, multiplet)  - (-CH₂-C<u>H</u>₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH

         3.24 δ (1H, quintet); - (-CH₂-CH₂-C<u>H</u>(CH₂-)-OH), coupling with4 H of 2 group of CH₂

         3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-O<u>H</u>), hydrogen of alcohol group, not tend to coupling with other hydrogen

4 0
3 years ago
what is the molarity of a RbOH solution if 60.0 mL of the solution is neutralized by 52.8 mL of a 0.5M HCl solution (Hint: Ma x
Dahasolnce [82]
RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume 
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted 
Ma x Va = Mb x Vb
0.5 M x 52.8 mL = Mb x 60.0 mL 
Mb = 0.44 M 
molarity of base - 0.44 M 

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2 years ago
What is direct transmission?
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4 0
3 years ago
A battery changes chemical energy into _____.
Sophie [7]
It turns chemical into electricity 
5 0
3 years ago
Read 2 more answers
What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?
Solnce55 [7]
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+  + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
     = -㏒(4.81x10^-7) = 6.32

3 0
3 years ago
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