I am giving brainiest! All I need are the last two questions at the bottom the rest of this info is to help solve them. I hope I
1 answer:
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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 5,75
Have a nice day!
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:
So, the mole ratio of CH₃COOH to CH₃COONa is 5,75
Have a nice day!
2-methyl-3-hexyn-2-ol can be prepared from Acetylene by treating Acetylene with <span>NaNH</span>₂<span> followed by CH</span>₃<span>CH</span>₂<span>Br, then treating the intermediate with NaNH</span>₂<span> followed by acetone and then doing aqueous workup. NaNH</span>₂ acts as base and abstracts proton from acetylene as the terminal alkynes are acidic in nature. While acetone on reduction gives tertiary alcohol.
Answer:
[CO] = 7.61x10⁻³M
7.61x10⁻³x10³ = 7.61
Explanation:
For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:
We need to know the molar concentrations in the equilibrium. In the beginning, there is only COCl₂, and its concentration is the number of moles divided by the volume:
[COCl₂] = 7.73/10.0 = 0.773 M
So, the equilibrium will be:
COCl₂(g) ⇆ CO(g) + Cl₂(g)
0.773 0 0 <em>Initial</em>
-x +x +x <em> Reacts</em>
0.773-x x x <em>Equilibrium</em>
Supposing that x<<0.773, then:
7.5x10⁻⁵ = x²/0.773
x² = 5.7975x10⁻⁵
x = √5.7975x10⁻⁵
x = 7.61x10⁻³ M
The supposing is correct, so [CO] = 7.61x10⁻³ x 10³ = 7.61