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iris [78.8K]
4 years ago
5

In oxidative phosphorylation, electrons are passed from one electron carrier to another. The energy released is used to ________

__. a. generate large amounts of NADH and FADH2 b. synthesize carbon dioxide c. pump protons (H⁺) across the mitochondrial membrane d. form ATP during glycolysis
Chemistry
1 answer:
Zepler [3.9K]4 years ago
7 0

Answer:

d. Form ATP during glycolysis.

Explanation:

During oxidative phosphorylation, an oxidation-reduction reactions will happen and the electrons transferred and energy released will be used in the conversion of adenosine diphosphate (ADP) into adenosine triphosphate (ATP).

This happens as a step of glycolysis, wich is the process in which the organism breaks gluscose molecules to obtain energy.

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On a graph , which type of line shows a direct proportion?
svetoff [14.1K]
C - straight line
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3 years ago
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Calculate the χacetone and χcyclohexane in the vapor above the solution. p°acetone = 229.5 torr and p°cyclohexane = 97.6 torr.
Ulleksa [173]
First calculate the mole fraction of each substance:
Acetone: 2.88 mol ÷ (2.88 mol + 1.45 mol) = 0.665
Cyclohexane: 1.45 ÷ (2.88 mol + 1.45 mol) = 0.335
Raoult's Law: P(total) = P(acetone) · χ(acetone)  + P(cyclohexane) · χ(cyclohexane).
P(total) = 229.5 torr · 0.665 + 97.6 torr · 0.335
P(total) = 185.3 torr
χ for acetone: 229.5 torr · 0.665 ÷ 185.3 torr = 0.823 
χ for cyclohexane:  97.6 torr · 0.335 ÷ 185.3 torr = 0.177
7 0
3 years ago
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Platinum crystallizes with the face-centered cubic unit cell. The radius of a platinum atom is 139 pm. Calculate the edge length
horrorfan [7]

Answer:

Length = 393pm, Density = 21.3 g/cm^3.

Explanation:

From the question above, we have the following parameters or data which is going to aid in solving the above Question.

=> The radius of a platinum atom = 139 pm.

Therefore, the length can be calculated by making use of the formula given below:

Length = 2 √( 2r) = 2 × √ (2 × 139 × 10^-12m ) = 393 × 10^-10 m = 393pm.

The density can be calculated by making use of the chemical formula given below:

Density = mass ÷ volume = (195.064/ 6.02 × 10^23) ÷ (3.93 × 10^-10/ 10^-2) = 21.3 g/cm^3.

3 0
3 years ago
Element Y has two natural isotopes Y-63 (62.940 amu) and Y-65 (64.928 amu). Calculate the atomic mass of element Y, given the ab
djverab [1.8K]

Answer:

Average atomic mass = 63.553 amu.

Explanation:

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Abundance of Y-65 = 100 - 69.17 = 30.83%

Atomic mass of Y-63 = 62.940 amu

Atomic mass of Y-65 = 64.928 amu

Atomic mass of Y = ?

Solution:

Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass= (62.940×69.17)+(64.928×30.83) /100

Average atomic mass =  4353.560 + 2001.730 / 100

Average atomic mass = 6355.29 / 100

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