PH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g
Considering the definition of percentage by mass, the mass of ammonium sulfate that must be added to prepare 200.0 g of a 3.5% solution of ammonium sulfate in water is 7 g.
<h3>Percentage by mass</h3>
The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.
In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
<h3>Mass of ammonium sulfate</h3>
In this case, you know:
- mass of ammonium sulfate= ?
- mass of solution= 200 g
- percent by mass= 3.5 %
Replacing in the definition of percent by mass:
Solving:
3.5÷100=
0.035=
0.035×200 g= mass of ammonium sulfate
<u><em>7 g= mass of ammonium sulfate</em></u>
Finally, the mass of ammonium sulfate that must be added to prepare 200.0 g of a 3.5% solution of ammonium sulfate in water is 7 g.
Learn more about percent be mass:
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It’s the law of the outer
Answer:
Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase
Explanation:
The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion
I hope you undestand me
C₃H₈ + 5O₂ = 3 CO₂ + 4 H₂O
44.1 g ----------- 4* 18.02 g
? g --------------- 75 g
mass of C₃H₈ = 75 * 44.1 / 4 * 18.02
mass of C₃H₈ = 3307.5 / 72.08
= 45.886 g of propane
number of moles:
45.886 / 44.1 = 1.0404 moles
1 mole --------- 22.4 L ( at STP )
1.0404 moles ----- ?
v = 1.0404 * 22.4 / 1
v = 23.304 Lhope this helps!