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:
0.07172 L = 7.172 mL.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.
The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.
➡ ANSWER
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Answer:
0.120M is the concentration of the solution
Explanation:
<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>
<em />
Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.
To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:
<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>
2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3
<em>Molar concentration:</em>
0.0300 moles NaNO3 / 0.250L =
<h3>0.120M is the concentration of the solution</h3>