Selfmade.ivyy hope this help
A because if you multiple it, you will be moving the decimal one time
Nucleus ,endoplasmic reticulum
Hello!
To solve this problem we are going to use the
Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is
4,76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
So, the mole ratio of CH₃COOH to CH₃COONa is
5,75Have a nice day!
Answer:
The correct answer to the question is
The standard heat of reaction for the reaction is
a. 216.8 kJ released per mole
Explanation:
The heat of reaction is given by [Heat of formation of products] - [Heat of formation of reactants]
In the question we have, heat of formation of the products Zn+2 (aq) = -152.4 kJ/mole and the heat of formation of the reactants = 64.4 kJ/mole
Therefore, the heat of formation of the reaction = (-152-64.4) kJ/mole or
-216.8 kJ/mole released