Complete Question:
Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺
Answer:
pNa = 0.307
pCl = 0.093
pNH₄ = 0.503
Explanation:
The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.
Both substances are salts that solubilize completely, thus, by the solution reactions:
NaCl → Na⁺ + Cl⁻
NH₄Cl → NH₄⁺ + Cl⁻
So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.
[Na⁺] = 0.493 M
[Cl⁻] = 0.493 + 0.314 = 0.807 M
[NH₄⁺] = 0.314 M
The p-values are:
pNa = -log[Na⁺] = -log(0.493) = 0.307
pCl = -log[Cl⁻] = -log(0.807) = 0.093
pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503
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Solution :
From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .
Moles of KBr in 102 g of potassium bromide.
n = 102/119.002
n = 0.86 mole.
So, number of miles of KCl produced are also 0.86 mole.
Mass of KCl produced :
Hence, this is the required solution.
Answer:
The first option: Strontium Fluorate.
Explanation:
because Fluorine and oxygen combines to make fluorate, Strontium stays the same.
p.s: i need help in geo and there's an exam tomorrow.
