False reading may be obtained if there is a run over from adjacent reagent area in excessively wetted strips
Phenolphthalein only works efficiently from 4 to 10 ph
Litmus and phenolphthalein and methyl orange indicators can only allow you to tell if solution is alkaline,neutral or acidic
The two indicators with less limitations are methyl orange and thymol blue
Answer:
The answer is C. Gas particles have no attractive forces between them.
Explanation:
Answer:
a) 3.98 x 10^-10
Explanation:
Hello,
In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Hence, solving for the concentration of hydronium:
![[H^+]=10^{-pH}=10^{-9.40}\\](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-9.40%7D%5C%5C)
![[H^+]=3.98x10^{-10}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D3.98x10%5E%7B-10%7DM)
Therefore, answer is a) 3.98 x 10^-10
Best regards.
Answer:
A. The top layer will be diethyl ether, and the top layer will be yellow.
Explanation:
The purpose of the addition of the saturated aqueous solution of polar solvents like sodium chloride in the liquid-liquid extraction techniques is to remove as well as separate any kind of water which may be dissolved in the ether. Water and sodium chloride are both polar and thus, they forms the bottom layer and only ether forms the top layer. The compound being organic and is colored is in the top layer with the ether.
Hence, answer - A. The top layer will be diethyl ether, and the top layer will be yellow.
Answer:
Explanation:
Given that:
Half life = 30 min
Where, k is rate constant
So,
The rate constant, k = 0.0231 min⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 0.0231 min⁻¹
Initial concentration = 7.50 mg
Final concentration = 0.25 mg
Time = ?
Applying in the above equation, we get that:-
