The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+. If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).
D! because when you record something you will probably analyze the results and those are two big KEY components of an Experiment! <span />
Please now mark this one Brainliest.. please!!!!! I could answer your questions too,,,if you rephrase it!!
Answer:
0.508 mole
Explanation:
NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.
The number of mole present in 78.2 g of CCl₄ can be obtained as follow:
Mass of CCl₄ = 78.2 g
Molar mass of CCl₄ = 12 + (35.5×4)
= 12 + 142
= 154 g/mol
Mole of CCl₄ =?
Mole = mass / molar mass
Mole of CCl₄ = 78.2 / 154
Mole of CCl₄ = 0.508 mole
Therefore, 0.508 mole is present in 78.2 g of CCl₄
Molarity can be defined as the number of moles of solute in 1 L solution.
Molarity of Na₂SO₄ solution - 0.200 M
this means there are 0.200 moles in 1 L solution
Molar mass of Na₂SO₄ - 142 g/mol
therefore mass of Na₂SO₄ in 1.00 L - 0.200 mol x 142 g/mol = 28.4 g
a mass of 28.4 g of Na₂SO₄ is present in 1.00 L
