Answer:
The ratio [A-]/[HA] increase when the pH increase and the ratio decrease when the pH decrease.
Explanation:
Every weak acid or base is at equilibrium with its conjugate base or acid respectively when it is dissolved in water.
⇄ 
This equilibrium depends on the molecule and it acidic constant (Ka). The Henderson-Hasselbalch equation,
![pH = pKa + Log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20Log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
shows the dependency between the pH of the solution, the pKa and the concentration of the species. If the pH decreases the concentration of protons will increase and the ratio between A- and AH will decrease. Instead, if the pH increases the concentration of protons will decreases and the ratio between A- and AH will increase.
Answer:
C
Explanation:
Temperature is directly related to kinetic energy (KE). As we raise temperature, we are raising KE, as well. Particles with more KE move more quickly and with more force.
This means that these particles are more likely to collide with each other and react to allow the chemical reaction to follow through. In turn, if the chemical reaction is more likely to go to completion, the reaction rate increases, eliminating A and B.
The concentration of the solute is not affected by the temperature; in other words, temperature will not increase or decrease the amount of solute in the solution, so eliminate D.
Thus the answer is C.
Hope this helps!
<span>To work out the volume of something from its density, use the compound measures triangle: mass over density and volume. To find volume that the beaker holds, divide the mass by the density. V = (388.15 - 39.09)/1. V = 349.06g/cm3. To find the weight of the beaker and the contents, first work out the weight (mass) of the mercury, with this formula: mass = d x v. M = 13.5 x 349.06. M = 4712.31. Then add on the weight of the beaker (39.09g). The total weight is 4751.40g.</span>
Answer:
0.5667 M ≅ 0.57 M.
Explanation:
It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.
It can be expressed as:
(MV) before dilution = (MV) after dilution.
M before dilution = 1.5 M, V before dilution = 340 mL.
M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.
∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.
<span>M(NO3)2 ==> [M2+] + 2 [NO3-]
0.202 M ==> 0.202 M
M(OH)2 ==> [M2+] + 2[OH-]
5.05*10^-18 ===> s + [2s]^2
5.05*10^-18 ===> 0.202 + [2s]^2
5.05*10^-18 = 0.202 * 4s^2
4s^2 = 25*10^-18
s^2 = 6.25*10^-18
s = 2.5*10^-9
So, the solubility is 2.5*10^-9</span>
