A becuase that is the one that is also b becuase a and c
The moving of molecules from areas of high concentration to that of low concentration to gain energy is best described as passive transport
<h3>What is passive transport?</h3>
Passive transport is a type of membrane transport in which chemicals are moved across cell membranes without using energy. Unlike active transport, which uses cellular energy, passive transport uses the second law of thermodynamics to cause the movement of substances across cell membranes.
<h3>Why is passive transport important?</h3>
Passive transport processes are critical to homeostasis. They maintain proper conditions inside the cell and the organism as a whole by letting chemicals to pass into and out of the cell.
To know more about Passive transport visit:
brainly.com/question/13542102
#SPJ4
Boiling is the process by which a liquid turns into a vapor when it is heated to it’s boiling point. The change from a liquid phase to a gaseous phase occurs when the vapor pressure of the liquid is equal to the atmospheric pressure exerted on the liquid. Boiling is a physical change and molecules are not chemically altered during the process.
Hope that helped!
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.