Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
When edible oils are idle and stored for a long amount of time, they undergo oxidation due to the exposure to oxygen. This oxidation causes rancidity in oils.
Answ burh i dont even know
Explanation: i dont even know this
The products will be magnesium phosphate and potassium chloride. You then have to watch a solubility chart to see which one of these is not soluable. In this case it is magnesium phosphate.
Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.
Explanation:
According to the dilution law,

where,
= molarity of stock solution = 1M
= volume of stock solution = ?
= molarity of diluted solution = 0.075 M (1mM=0.001M)
= volume of diluted solution = 10 ml
Putting in the values we get:


Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.
Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration