Always. You never know what kind of chemical you’re dealing with and how powerful it is.
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
The lowest value of the henry's law for methane gas (CH₄) will be obtained with H₂O as the solvent and a temperature of 349 K.
The lowest value of the henry's law for methane gas (CH₄) will be obtained with H₂O as the solvent and a temperature of 349 K.
Henry's law: This law states that at a constant temperature, the amount of a gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas that in equilibrium with that liquid.
Mathematically it can be written as:

So, for the methane gas , lowest value of the henry's law obtained at 349 K and with H₂O as the solvent.
The scientific geographic name is usually a great election
Yes the ANSWER is Correct-
On first half life the mass will be 10 gram
On second half life the mass will be 5 gram
On third half life the mass will be 2.5 gram