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
Answer:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Explanation:
Hello,
In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

The suitable equilibrium constant turns out:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Or in terms of the initial equilibrium constant:

Since the second reaction is a doubled version of the first one.
Best regards.
Answer:
~1.5 g/cm3 and it does NOT float in water.
Explanation:
If you look at the graph, Object A weighs ~6 grams and is ~4 cm3 in volume
Density = Mass/Volume
So 6 grams/4 cm3 = 1.5 g/cm3
Water has a density of 1 g/cm3 and because Object A density is higher than that of water, it sinks.
:)
the molar mass is how many grams it takes to make a full mole. The number in carbon 12 tells us that it takes 12 g to makes a mole, so one half of that would be <u><em>option d, 0.5</em></u>.