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
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Explanation:
The dichloromethane (DCM) has less density than water and also the polarity of water is much more than DCM. So the mixture of water and dichloromethane will always be a heterogeneous mixture. In the mixture dichloromethane will be always up of the water layer. The volume of the separatory funnel which contains the mixture of DCM and water must have to be more than the total volume of the liquids thus the volume of the funnel will be more than (50+50) = 100mL.
The caution have to consider during the separation are-
1. The separatory funnel have to shake well with lid and have to settle down for some times until the two liquid separated.
2. The lid should be open very slowly as the vapor pressure of DCM is more and it will float on the water.
3. After this the stopcock should be opened and slowly the water will come out first followed by DCM.
ANSWER
36.12 degrees fahrenheit
