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Arturiano [62]
3 years ago
11

BEFORE equilibrium has been reached in this container: (Circle a letter for each) 1. Movement of glucose across the membrane in

this container can best be described as A. Solution A to Solution B B. Solution B to Solution A C. both directions D. neither direction
Biology
1 answer:
qwelly [4]3 years ago
3 0

Hello. You forgot to put the image so that this question can be answered, but I will describe what the image shows.

The image shows two types of "cups" that have a type of connection between the two. In cup A there is 1 mole of glucose and 1 mole of fructose. In cup B there is 0.1 mol of glucose and 1.5 mol of fructose.

Answer:

A. Solution A to Solution B

Explanation:

Balance is achieved when the "cup" with the lowest concentration of glucose receives glucose from the "cup" with the highest concentration, to the point that the two glasses establish equal concentrations of glucose between them.

We know that cup B has a lower concentration of glucose, which indicates that the movement of this solute was from cup A towards cup B. With this we can conclude that the letter A is the correct answer.

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Answer:

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Explanation:

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Answer:

12 molecules of ATP can be made from one molecule of acetyl CoA.

Explanation:

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                        Now the reducing equivalent such as NADH and FADH2 enters electron transport chain and get oxidized to regenerate NAD+ and FAD along with ATP. 1 molecule of NADH produces 3 molecules of ATP whereas 1 molecule of FADH2 produces 2 molecules of ATP.

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As a result total 12 molecules of ATP is by the oxidation of 1 Acetyl CoA  by TCA cycle .

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3 years ago
The nucleus of "Lead-208", 208 82 Pb, has 82 protons within a sphere of radius 6.34×10-15 m. Each electric charge has a value of
jarptica [38.1K]

Answer:

2.94 × 10²⁰ N/C

Explanation:

Given that:  

The nucleus of "Lead-208 has 82 protons,

with a radius (r) 6.34×10-15 m, &

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∴ The formula for calculating an electrical field at the surface of the nucleus is:

 E=\frac{k*q}{r^2}  

Substituting our values into the equation above, we have;

 E = \frac{8.98755*10^8*82(1.60218*10^{-19C)}}{(6.34*10^{-15}_m)^2}

E = 2.93870499×10²⁰ N/C

E ≅ 2.94 × 10²⁰ N/C

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3 years ago
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You should move this question to Mathematics instead. Or at least physics.
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