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
This can be, for example, halogensubstituted hydrocarbons.
CCl₄, C₂F₆.
Or halides halocarboxylic acids, and other compounds.
O
II
Cl₃C-Cl
According to Dalton's Atomic Theory, the <em>Law of Definite Proportion is applied when a compound is always made up by a fixed fraction of its individual elements.</em> This is manifested by the balancing of the reaction.
The reaction for this problem is:
H₂ + Cl₂ → 2 HCl
1 mol of H₂ is needed for every 1 mole of Cl₂. Assuming these are ideal gases, the moles is equal to the volume. So, if equal volumes of the reactants are available, they will produce twice the given volumes of HCl.
Answer:
heat increase, pressue loss, altitude gain,
Explanation:
