Answer:
Following are the solution to the given question:
Explanation:
Due to the strong attachment among ions, it's indeed hard to break the covalent compounds in a combination. The ionic compounds, as they have a great melting point, become solid at room temperature, and may we say that the stronger between both the ions, as it is on a high point of fusion, is quite desirable.
Water molecules are polar and form hydrogen bonds.
Answer:
The concentration of H⁺ ions is 0.0165 M.
Explanation:
Let's consider the dissociation of H₂SO₄. In the first step, H₂SO₄ acts as a strong acid, completely dissociating into HSO₄⁻ and H⁺. Therefore, the concentrations of these ions will be the <em>same</em> that the initial concentration of the acid.
H₂SO₄ ⇒ HSO₄⁻ + H⁺
Initial 0.010M 0 0
Final 0 0.010M 0.010M
Now, HSO₄⁻ is a weak acid that will dissociate partially to form H⁺ and SO₄²⁻.
HSO₄⁻ ⇄ H⁺ + SO₄²⁻
To find out the concentration of H⁺ from HSO₄⁻ we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and complete each row with the concentration or change in concentration.
HSO₄⁻ ⇄ H⁺ + SO₄²⁻
I 0.010 0 0
C -x +x +x
E 0.010 -x x x
![Ka2=0.012=\frac{[H^{+}].[SO_{4}^{2} ]}{H_{2}SO_{4}} =\frac{x^{2} }{0.010-x}](https://tex.z-dn.net/?f=Ka2%3D0.012%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D.%5BSO_%7B4%7D%5E%7B2%7D%20%5D%7D%7BH_%7B2%7DSO_%7B4%7D%7D%20%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B0.010-x%7D)
This quadratic equation has 2 solutions: x₁ = -0.018 and x₂ = 0.00649. Since concentrations cannot be negative, we choose x₂. Then, [H⁺] coming from HSO₄⁻ is 0.00649 M.
The total concentration of H⁺ is:
[H⁺] = 0.010 M + 0.00649 M = 0.0165 M
Answer:
percentage by mass of Ba = 0.4583/0.8360 × 100 = 45.83
/0.8360 = 54.82%
Explanation:
The mass of Barium in the original compound is the same as the masss of barium in the precipitate BaSO4.
Ba²+
molar mass of BaSO4 = 137 + 32 + 16(4) = 137 + 32 + 64 = 233 g/ mol
since
137 grams of barium is in 233 g of BaSO4
? grams of barium is in 0.7794 g of BaSO4
cross multiply
grams of barium = 0.7794 × 137/233
grams of barium = 106.7778/233
grams of barium = 0.45827381974
grams of barium = 0.4583 grams
percentage by mass of Barium in the original compound =
mass of barium/ mass of the compound × 100
mass of barium = 0.4583 g
mass of compound = 0.8360 g
percentage by mass of Ba = 0.4583/0.8360 × 100 = 45.83
/0.8360 = 54.82%
NO₃⁻ is the polyatomic ion in this case.