37.1×10−8M is the concentration of pb 2 ions in a solution prepared by adding 5. 00 g of lead(ii) iodide to 500. ml of 0. 150 m ki? [ k sp(pbi 2) =1. 4 × 10 –8] by common ion effect.
A phenomenon known as the "common ion effect" allows for the modification of a salt's molar solubility by adding another salt in which one ion dissociates in solution and maintains equilibrium with the undissociated salt.
The term "common-ion effect" describes the reduction in solubility of an ionic precipitate caused by the addition of a soluble molecule that shares an ion with the precipitate to the solution. Le principle for the equilibrium response of ionic association/dissociation leads to this behavior.
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The IUPAC name of the above mentioned compound is "2-Chloro-4,4-dimethylpentane"
<u>Exaplanation:</u>
- Since the above organic compound is an compound with only one saturated bond, It can be considered as a single bond compound, and hence we can conclude that as alkane.
- It also has 5 carbon atoms, so it is termed as pentane.
- From right to left we have to number the atoms, and 2 nd carbon atom contain Cl atom so it is termed as 2-Chloro and in the 4th position carbon atom contains 2 methyl groups, so it is termed as, 2-Chloro-4,4-dimethylpentane.
The substances Oxygen and Hydrogen are elements.
Answer: The laboratory value of potassium (3.0 mmol / L) is consistent with the client's symptoms of hypokalemia.
Explanation:
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Hypokalemia is a disorder in the body's electrolyte balance, when the decrease in blood potassium (K) ion levels is below 3.5 mmol / L. Potassium losses can occur through the digestive tract: such as vomiting and
diarrhea The most frequent symptoms of potassium loss include: tiredness, muscle weakness and cramping.
In conclusion, the laboratory value of potassium (3.0 mmol / L) is consistent with the client's symptoms of hypokalemia.
Answer:
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Explanation:
The unbalanced equation is
Al(s) + Ni²⁺(aq) ⟶ Ni(s) + Al³⁺(aq)
(i) Half-reactions
Al(s) ⟶ Al³⁺(aq) + 3e⁻
Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)
(ii) Balance charges
2 × [Al(s) ⟶ Al³⁺(aq) + 3e⁻]
3 × [Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)]
gives
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s)
(iii) Add equations
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
<u>3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s) </u>
2Al(s) +3Ni²⁺(aq) + <em>6e</em>⁻ ⟶ 2Al³⁺(aq) + 3Ni(s) + <em>6e⁻
</em>
Simplify (cancel electrons)
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
