54g ag *(108mol ag/1 g ag) =5832mol ag
        
             
        
        
        
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH  +  H₂O  ⇄   CH₃COO⁻  +  H₃O⁺     Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
 
        
             
        
        
        
<u>Answer:</u> The volume of given amount of ethanol at this temperature is 159.44 mL
<u>Explanation:</u>
Specific gravity is given by the formula:

We are given:
Density of water = 0.997 g/mL
Specific gravity of ethanol = 0.787
Putting values in above equation, we get:

Density is defined as the ratio of mass and volume of a substance.
 ......(1)
 ......(1)
Given values:
Mass of ethanol = 125 g
Density of ethanol = 0.784 g/mL
Putting values in equation 1, we get:

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL
 
        
             
        
        
        
The mixture should be 
Heterogeneous
Explanation:
<u>D</u><u>e</u><u>f</u><u>i</u><u>n</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u> Diverse in character or content.