Answer:
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- <u><em>Because the x-intercet of the graph represents volume zero, which indicates the minimum possible temperature or absolute zero.</em></u>
Explanation:
Charle's Law for ideal gases states that, at constant pressure, the <em>temperature</em> and the <em>volume</em> of a sample of gas are protortional.

That means that the graph of the relationship between Temperature, in Kelivn, and Volume is a line, which passes through the origin.
When you work with Temperature in Celsius, and the temperature is placed on the x-axis, the line is shifted to the left 273.15ºC.
Meaning that the Volume at 273.15ºC is zero.
You cannot reach such low temperatures in an experiment, and also, volume zero is not real.
Nevertheless, you can draw the line of best fit and extend it until the x-axis (corresponding to a theoretical volume equal to zero), and read the corresponding temperature.
Subject to the experimental errors, and the fact that the real gases are not ideal, the temperature that you read on the x-axis is the minimum possible temperature (<em>absolute zero</em>) as the minimum possible volume is zero.
<span>electrons and space filled with wispy positive charge I would love if I could get brainliest :)
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Answer: With few exceptions, the mitotic process ensures that this is the case. Therefore, mitosis ensures that each successive cellular generation has the same genetic composition as the previous generation, as well as an identical chromosome set.
<span>In a chemical change, the molecular structure of a substance changes. I think this is right. I hope this helps.</span>
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.