You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
Answer:
0.6378 M
Explanation:
Molarity is defined by Moles per liter.
Plugging the given information in, we get (14.968 moles)/(23.47 L) which comes out to be about 0.6378 M
Answer:
The true statements are:
The solution is acidic
The pH of the solution is 14.00 - 10.53.
![10^{-10.53}=[OH^-]](https://tex.z-dn.net/?f=10%5E%7B-10.53%7D%3D%5BOH%5E-%5D)
Explanation:
The pH of the solution is defined as negative logarithm of hydrogen ion concentration present in the solution .
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
- The pH value more 7 means that hydrogen ion concentration is less ,alkaline will be the solution.
- The pH value less 7 means that hydrogen ion concentration is more ,acidic will be the solution.
- The pH value equal to 7 indicates that the solution is neutral.
The pOH of the solution is defined as negative logarithm of hydroxide ion concentration present in the solution .
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
The pOH of the solution = 10.53
![10.53=-\log[OH^-]](https://tex.z-dn.net/?f=10.53%3D-%5Clog%5BOH%5E-%5D)
![10^{-10.53}=[OH^-]](https://tex.z-dn.net/?f=10%5E%7B-10.53%7D%3D%5BOH%5E-%5D)
The pH of the solution = ?


Here, the pH of the solution is less than 7 which means that solution acidic.
Answer:
option A is correct answer