PH scale is from 1 to 14 and indicates how acidic or basic a solution is. To find pH or pOH we need to know the H⁺ ion concentration or OH⁻ concentration.
pH can be calculated using the following equation;
pH = -log[H⁺]
the H⁺ concentration of the given acid is 1.0 x 10⁻⁴ M. substituting this we can find the pH
pH = -log[1x10⁻⁴]
pH = 4
answer is 1) 4
B. The unknown solution had the lower concentration.
Explanation:
Osmosis is a phenomenon in which the molecules of the solvent has a tendency to move through a membrane which is semipermeable from lower concentrated side to the higher concentration side, so that the concentrations on both sides of the membrane must be equal.
So the unknown solution may have lesser concentration than the isotonic solution so that molecules of that solution move from less concentrated side to the more concentrated side, so its level drops.
Answer:
0.007 M
Explanation:
pH is defined as the negative logarithm of the concentration of hydrogen ions.
Thus,
pH = - log [H⁺]
The expression of the pH of the calculation of weak acid is:-
Where, C is the concentration = ?
Given, pH = 3.45
So, for 
, 
C = 0.007 M
Answer:
I don't really get the options but it favoures the reactant side.
Explanation:
Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g
If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.
With the explanations I have made, I hope the question is now clear to you.
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
