Hello!
To solve this problem we'll use the
Henderson-Hasselbach equation, but first we need the vale for the pKa of Benzoic acid, which is pKa= -log(Ka)=
4,19Now, we apply the equation as follows:
![pH=pKa + log ( \frac{[C_6H_5COONa]}{[C_6H_5COOH]} )=4,19+log( \frac{0,15M}{0,25M} )=3,97](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BC_6H_5COONa%5D%7D%7B%5BC_6H_5COOH%5D%7D%20%29%3D4%2C19%2Blog%28%20%5Cfrac%7B0%2C15M%7D%7B0%2C25M%7D%20%29%3D3%2C97%20)
So, the pH of this solution of Sodium Benzoate and Benzoic Acid is
3,97Have a nice day!
Answer: Solids and liquids don't suddenly change their volumes.
But gases do...they expand to fill whatever container you
put them in.
Michael's mystery substance is in the gaseous phase.
Explanation:
Answer:
false
Explanation:
As we know that in sodium-potassium pump .
sodium potassium move 3Na+ outside the cells
and moving 2k+ inside the cells
so that we can say that given statement is false
Answer FALSE
Answer:
670.68°C
Explanation:
Given that:
volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g
specific heat (C) = 4.184 J/g˚C
Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C
The quantity of heat (Q) used to raise the temperature of a body is given by the equation:
Q = mCΔT
Substituting values:
Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J
Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.
-Q = mCΔT
-418.4 J = 5 g × 0.129 J/g˚C × ΔT
ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C
temperature change ΔT = final temperature - initial temperature
- 648 .68°C = 22°C - Initial Temperature
Initial Temperature = 22 + 648.68 = 670.68°C
