Since there is one carbon with 4 Fluorines attached to it, and both compounds are no metals, we use the covalent method for naming,
Here we ignore the prefix for the first element if it is 1. Mono. Then pay attention to the second one, it would be tetra, because tetra means 4. Here there are 4 fluorines.
Drop ine and place ide
CF4 = carbon tetrafluoride.
Answer:
The one left in the hot sunlight.
Explanation:
The solubility of gases decreases when temperature increases. The gas in the soda pop (CO2) left in the sun will not stay dissolved as much as the on left in the refrigerator.
<span>1=H, 2=B, 3=F, 4=A,5=C,6=E, 7=D, 8=G
</span>9: 69Ga=60.12% and 71Ga=39.88%; total=69.797amu
10: 27 27.977 92.23; 28 28.976 4.67; 29 29.974 3.10; abundance =28.07 Silicon
I hope this helps!
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.
I believe that it would be Al1N1.