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:
See explanation
Explanation:
According to Bronsted-Lowry, an acid is a proton donor while a base is a proton acceptor.
Hence, if we consider the reaction above, we will notice that for each base there is a conjugate acid and for each acid there is a conjugate base.
For the acid HNO3, its conjugate base is NO3^- while for the acid H3O^+, its conjugate base is H2O.
The number of significant figures in 369,132,000 is 6
D ................................
Answer:
Average atomic mass = 63.553 amu.
Explanation:
Given data:
Abundance of Y-63 = 69.17%
Abundance of Y-65 = 100 - 69.17 = 30.83%
Atomic mass of Y-63 = 62.940 amu
Atomic mass of Y-65 = 64.928 amu
Atomic mass of Y = ?
Solution:
Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass= (62.940×69.17)+(64.928×30.83) /100
Average atomic mass = 4353.560 + 2001.730 / 100
Average atomic mass = 6355.29 / 100
Average atomic mass = 63.553 amu.