Answer:
yes the ones u chose are correct.
Explanation:
One way of knowing that oxygen was the gas removed from the volume of air and not another is to know what the volume of air is made of first. When the composition of the volume of air is already identified, then next would be the process of separating these elements from each other and as to which is to be separated first. This would usually lead to knowing their masses, their boiling and freezing points, the temperatures at which they condense, and so on. This is to identify their differences to each other and use those differences to successfully separate those elements to each other.
From the information given, the total volume of rubbing alcohol is 88.2 ml
68.6 % of this volume is isopropanol.
We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %
The volume of isopropanol is
68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml
The volume of isopropanol is 60.5 ml.
Volume of water will be 88.20 - 60.5 = 27.7 ml
(27.7 / 88.2 × 100 = 31.4% )
Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water solution.
Answer:
PH= 6.767 (answer is the A option)
Explanation:
first we need to correct the value in Kw at this temperature is 2.92*10^-14
so, in this case we have that:
Kw=2.92*10^-14 M²
[ H3O^+] [ H3O^+]
![[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%5BOH%5E%7B-%7D%20%20%5D%20%3D%20Kw%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D%20%20%20%5C%5C%5C%5C)
at 40ºC
![[H_{3}O^{+} ] = [OH^{-} ]](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%5BOH%5E%7B-%7D%20%20%5D)
![[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%5E%7B2%7D%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D)
![[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%282.92%2A10%5E%7B-14%7D%29%5E%7B1%2F2%7D%20%3D%201.71%2A10%5E%7B-7%7D%20M)
![PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767](https://tex.z-dn.net/?f=PH%3D%20-log10%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20-log10%281.71%2A10%5E%7B-7%7D%20%29%20%3D%206.767)
