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Ksenya-84 [330]
3 years ago
8

During glycolysis, for every 6 moles of glucose that enter the pathway, how many moles of atp are used? how many moles atp are g

enerated? how many moles of nadh are used? how many moles of pyruvate are made?
Chemistry
1 answer:
Goryan [66]3 years ago
7 0
During the process of glycolysis 1 mole of glucose yields 2 pyruvic acid. In the process 2 ATPs molecules are used up and 4 other ATP molecules are produced by substrate level phosphorylation and 2 NADH are also produced. Therefore; for six moles of glucose; 12 ATP molecules will be used up, 24 ATP molecules will be generated, 12 moles of NADH will be used and 12 moles of pyruvate are made. 
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Help please! this is timed
Licemer1 [7]

Answer:

C, P, P, C, P

Explanation:

is it still the same thing but the physical property change or did the thing change too? that's what it's asking

4 0
3 years ago
What kind of chemical reaction does the chemical equation sodium + chlorine → sodium chloride represent?
tresset_1 [31]
This would represent a synthesis reaction 
7 0
3 years ago
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Which of the following are produced when a base is dissolved in water?
Irina-Kira [14]

Answer is: B. Hydroxide ions.

An Arrhenius base is a substance that dissociates in water to form hydroxide ions (OH⁻).  

For example sodium hydroxide: NaOH(aq) → Na⁺(aq) + OH⁻(aq).

Another example, balanced chemical reaction: Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).

According to the Arrhenius definition barium hydroxide is base.

Acids and bases when react (neutralisation) produce salt and water.

3 0
3 years ago
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The measure of the amount of dissolved salt in a liquid sample is called
Ann [662]

The answer is salinity, salinity is the saltiness or dissolved inorganic salt content of a body of water.

7 0
4 years ago
The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r
Vinil7 [7]

Answer:

Rate = k [OCl] [I]

Explanation:

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Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

5 0
3 years ago
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