Answer:
The physical properties of a solution are different from those of the pure solvent. ... Colligative properties are those physical properties of solutions of nonvolatile solutes that depend only on the number of particles present in a given amount of solution, not on the nature of those particles.
Answer: They can be separated by physical processes.
Explanation: A mixture is made up of two or more substances that are not chemically combined and can be easily separated into its constituents by physical methods.
Answer:
11.39
Explanation:
Given that:
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
Answer:
Plants consume carbon through transpiration
Explanation:
In transpiration, plants lose water vapor through the stomata in their leaves. No carbon is involved in transpiration, which has an outbound direction. Nothing can be consumed through the stomata when vapor is going out of the plant. It´s like trying to get in through the exit.
