<span> Find the molar mass of calcium acetate...158.165g.
So start with 255g Calcium acetate x 1 mole Calcium Acetate/158.165 g Calcium Acetate x 6 mol Hydrogen/1 mole Calcium Acetate x 6.022e23 atoms/mole.
So...(255 x 1 x 6 x 6.022e23)/(158.165 x 1 x 1) = 5.825e24 hydrogen atoms</span>
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There’s no picture if you meant to add a picture
A solution with a higher concentration of hydroxide ions than hydrogen ions is basic solution.
This solution formed by Base dissolved in water and release hydroxide ions.
The PH of the solution is greater than 7
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Hello.
The answer is: D it produces hyrogen ions in a solution.
This is correct because when Arrhenius acid it turns into hydrogen ions.substance as an acid if it produces hydrogen ions H(+) or hydronium ions in water. A substance is classified as a base if it produces hydroxide ions OH(-) in water.
have a nice day
