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Sidana [21]
3 years ago
7

For each molecule of glucose that enters glycolysis, how many atp molecules are gained?

Chemistry
2 answers:
Semenov [28]3 years ago
4 0
D. 2 ATP will be gained.

The process of Glycolysis requires 2 ATP to be activated. After glycolysis, 4 new ATPs will be produced. However, as mentioned, 2 will be used, hence only 2.
kakasveta [241]3 years ago
4 0

Answer:

d. 2

Explanation:

The glycolysis is the first step in glucose degradation that generated energy for the cellular metabolism. The glycolysis has two phases: one that requires energy and other that releases energy. These phases are described in the attached image.

In the Energy-requiring phase the glucose is rearranged and are added 2 phosphate groups forming fructose-1,6-bisphosphate. This molecule is unstable, so it is divided in two phosphate-bearing three-carbon sugars. The two phosphates used in these steps come from ATP molecules, so it is needed two molecules of ATP in this first phase.

In the Energy-releasing phase each three-carbon sugar formed turn into pyruvate by a series of reactions. These reactions produce 2 ATP molecules for each three-carbon sugar. And since this phase occurs twice, in total, there are produced 4 ATP molecules.

The general reaction is:

Glucose + 2ATP + 2ADP + 2Pi + 2NAD+ ---> 2 pyruvate + 4 ATP + 2 NADH + H2O

Or

Glucose + 2ADP + 2Pi + 2NAD+ ---> 2 pyruvate + 2 ATP + 2 NADH + H2O

So, this process produces 2 ATP molecules.

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Answer:

Explanation:

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The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon
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Answer:

\boxed{\text{300 g}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:         32                          60

           CH₃OH + CO ⟶ CH₃COOH

m/g:        160

(a) Moles of CH₃OH

\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}

(b) Moles of CH₃COOH

\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}

(c) Mass of CH₃COOH

\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}

3 0
4 years ago
What vitamin is produced by a mammalian skin?
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3 years ago
How many moles of cf4 are there in 171g of cf4 ?
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sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the press
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Answer:

-81.5 degrees C or 191.5 K

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We want to use Charles' gas law: V/T = V/T

Our initial volume is 3.20 L, and our initial temperature is 125 degrees C, or 125 + 273 = 398 degrees Kelvin.

Our new Volume is 1.54 L, but we don't know what the temperature is. So, we use the equation:

3.20 L / 398 K = 1.54 L / T ⇒ Solving for T, we get: T = 191.5 K

If we want this in degrees Celsius, we subtract 273: 191.5 - 273 = -81.5 degrees C

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