Answer:
- <em>The solution that has the highest concentration of hydroxide ions is </em><u>d. pH = 12.59.</u>
Explanation:
You can solve this question using just some chemical facts:
- pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
- The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.
Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.
These mathematical relations are used to find the exact concentrations of hydroxide ions:
- pH + pOH = 14 ⇒ pOH = 14 - pH
- pOH = - log [OH⁻] ⇒
![[OH^-]=10^{-pOH}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D)
Then, you can follow these calculations:
Solution pH pOH [OH⁻]
a. 3.21 14 - 3.21 = 10.79 antilogarithm of 10.79 = 1.6 × 10⁻¹¹
b. 7.00 14 - 7.00 = 7.00 antilogarithm of 7.00 = 10⁻⁷
c. 7.93 14 - 7.93 = 6.07 antilogarithm of 6.07 = 8.5 × 10⁻⁷
d. 12.59 14 - 12.59 = 1.41 antilogarithm of 1.41 = 0.039
e. 9.82 14 - 9.82 = 4.18 antilogarithm of 4.18 = 6.6 × 10⁻⁵
From which you see that the highest concentration of hydroxide ions is for pH = 12.59.
Answer:
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Explanation:
The equation is given as;
2HCl(aq) + Zn(s) + H2(g) + ZnCl2(aq)
In writing an ionic equation, only the aqueous compounds dissociates into ions. This means HCl and ZnCl2 would dissociate to form ions.
This is given as;
2H+ + 2Cl- + Zn(s) --> H2(g) + Zn2+ + 2Cl-
The correct option is;
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Because the oily part of its molecule tends to lay up against solid surfaces, but without much in the way of areas of electric charge that would cause it to stick more firmly to that surface.
Answer:
(NH4)2Cr2O7
Explanation:
Hope this somehow helped.
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
