The pH of the solution at 25 degree celsius of 1.3 × 10⁻⁶ moles of a sample of Sr(OH)₂ is 10.02.
<h3>How do we calculate pH?</h3>
The pH of any solution gives an idea about the acidic and basic nature of the solution and the equation of pH will be represented as:
pH + pOH = 14
Given that,
Moles of Sr(OH)₂ = 1.3 × 10⁻⁶ mol
Volume of solution = 25mL = 0.025L
The concentration of Sr(OH)₂ in terms of molarity = 1.3×10⁻⁶/0.025
= 5.2×10¯⁵M
Dissociation of Sr(OH)₂ takes place as:
Sr(OH)₂ → Sr²⁺ + 2OH⁻
From the stoichiometry of the reaction 1 mole of Sr(OH)₂ produces 2 moles of OH⁻.
Given that the base is a strong base and that it entirely dissociates into its ions, the hydroxide ion concentration is 5.2×10¯⁵×2 = 1.04×10¯⁴ M.
pOH = -log[OH⁻]
pOH = -log(1.04×10¯⁴)
pOH = 3.98
Now we put this value on the first equation we get,
pH = 14 - 3.98 = 10.02
Therefore, the value of pOH is 10.02.
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Answer:
Ionic or electrovalent bonds
Explanation:
Ionic or electrovalent bonds are interatomic or intramolecular bonds which are formed between two kinds of atoms having a large electronegativity difference usually 2.1.
Electronegativity is the property that combines the ability of an atom to gain or lose electrons. It is expressed as the tendency with which atoms of elements attracts valence electrons in a chemical bond.
In this bond type, a metal transfers its electrons to a more electronegative atom which is a non-metal.
Biotic are living organisms and abiotic are non living things such as rocks water soil
Answer:
I believe is A
Explanation:
Enthalpy change is the name given to the amount of heat evolved or absorbed in a reaction carried out at constant pressure.
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
