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
Answer:
Ionic bonding occurs when atoms either gain or lose one or more valence electrons, resulting in the atom having either a negative or positive charge.
Through ionic bonding, an atom of each element will combine with the other to form a molecule, which is more stable since it now has a zero charge.
Explanation:
In the modern periodic table when elements are arranged according to their atomic number.
<span>0.38
You first calculate the total moles by dividing the grams by molecular weight:
45 g N2 / 28.02 g/mol = 1.6 mol N2
40 g Ar / 39.95 g/mol = 1.0 mol
Then you divide the moles of Ar by the total number of moles:
1.0 / (1.6 + 1.0) = 0.38 mol fraction</span>
<span>Ernest Rutherford was the scientist that developed a model of the atom that looked like a nucleus surrounded by electrons.
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