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katovenus [111]
3 years ago
9

In the citric acid cycle (also known as the Krebs cycle), acetyl CoA is completely oxidized. From the following compounds involv

ed in cellular respiration, choose those that are the net inputs and net outputs of the citric acid cycle. Drag each compound to the appropriate bin. If a compound is not involved in the citric acid cycle, drag it to the "not input or output" bin. (Note that not all of the inputs and outputs of the citric acid cycle are included.)
Chemistry
1 answer:
REY [17]3 years ago
3 0

Answer:

We place ( O₂, CO₂, coenzyme A and acetyl CoA) into bin [not input or output];

We place (ADP, NAD⁺, Glucose) into [ Net Input] bin.;

And (ATP, NADH and Pyruvate) intothe [Net Output] bin.

Explanation:

Citric Acid Cycle is also known as Tricarboxylic Acid Cycle, Krebs Cycle and Szent-Györgyi-Krebs cycle.

It is a series of enzyme-catalyzed chemical reactions, which of central importance in all living cells that use oxygen as part of cellular respiration.

Here are few things to know about the CAC:

• The Krebs cycle uses the products of Glycolysis to produce 2 ATP molecules for each molecule of Glucose.

• Oxygen is required.

• The Krebs Cycle occurs in cellular power plant — Mitochondria of the cell.

• Each glucose molecule nets 2 pyruvic acids. The Krebs Cycle breaksdown 1 pyruvic acid one at å time.

• CAC begins and ends with Oxaloacetic Acid - a 4 carbon molecule.

• It is an 8 step cycle that begins when the Acetyl CoA that is a product of Glycolysis is 'picked up' by Oxalacetic Acid.

• Lipids, Proteins and Carbohydrates can all be metabolized in the Krebs Cycle.

• The NADH and FADH2 that are high energy co- enzymes (helper molecules) are by-products of the Krebs cycle and continue on to fuel the Electron Transport Chain.

The clues above helps us to know where to drag the compounds into;

We place these into bin of [Net Input]: ADP, NAD⁺, Glucose

We place these into the bin of [Net Output]: ATP, NADH and Pyruvate,

We place the following into [not input or output] bin: O₂, CO₂, coenzyme A and acetyl CoA

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Airida [17]

Answer:

They stay the same

Explanation:

4 0
3 years ago
25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is
vlabodo [156]

Answer:

a)   119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

\bigtriangleup T_f=K_fm, \  \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\

#We use the molality above to calculate the molar mass:

m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol

Hence, the molar mass of the compound is 119 g/mol

5 0
4 years ago
When bases dissolve in water they release what type of negative ion?
Leto [7]
A base generally releases a hydroxide ion (OH-) when dissolved in water. 

There are exceptions, such as ammonia NH3, which acts as a base but does not produce OH- ions. There are three definitions of acids and bases (Arrhenius, Bronsted-Lowry, and Lewis) and each one looks at acid/base characteristics differently. OH- donation is the Arrhenius definition.
4 0
4 years ago
In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].
Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose +  H^+\rightarrow  fructose+ glucose

The rate law of the reaction is given as:

R=k[H^+][sucrose]

[H^+]=0.01M

[sucrose]= 1.0 M

R=k[0.01M][1.0 M]..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

R'=[0.01 M][2.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

R'=[0.01 M][0.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when [H^+] is changed to 0.001 M = R'

R'=[0.0001 M][1.0 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}

R'=0.01\times R

If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and[H^+] both are changed to 0.1 M = R'

R'=[0.1M][0.1M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}

R'=1\times R

If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0
3 years ago
What is the solute when stirring salt in water until the salt disappears?
Umnica [9.8K]

Answer:

The solute is the substance being dissolved.

The solvent is the substance dissolving the solute.

Therefore, the salt is the solute and the water is the solvent.

Explanation:

The salt is the solute.

3 0
3 years ago
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