How many molecules of HCI are in 4.91 L of HCI acid at 25°C if the density equals 1.096 g/ml
To determine the number of HCl molecules in 4.91 L of HCl acid at 25°C, we can use the following steps:
Calculate the mass of the HCl acid in 4.91 L using its density.Convert the mass of HCl acid to the number of moles using its molar mass.Use Avogadro's number to convert the number of moles of HCl to the number of HCl molecules.Calculate the mass of the HCl acid in 4.91 L using its density:[tex]\qquad\sf {Density = \dfrac{mass}{volume}}[/tex]
[tex]\qquad\sf{mass = density \times volume}[/tex]
[tex]\qquad\sf{mass = 1.096 \: g/mL \times 4.91\: L = 5.38\: kg}[/tex]
Convert the mass of HCl acid to the number of moles using its molar mass. The molar mass of HCl is 36.46 g/mol.
[tex]\sf{moles = \dfrac{mass}{ molar\: mass} = \dfrac{5.38\: kg}{36.46\: g/mol} = 147.6\: mol}[/tex]
Use Avogadro's number to convert the number of moles of HCl to the number of HCl molecules. Avogadro's number is [tex]6.02 \times 10^23[/tex] molecules/mol.
[tex]\sf number\: of\: HCl\: molecules = moles \times Avogadro's\: number[/tex]
[tex]\begin{aligned}\sf number\: of\: HCl\: molecules& =\sf 147.6 \: mol \times 6.02 \times 10^23\: molecules/mol \\& =\sf 8.88 \times 10^25\: molecules\end{aligned}[/tex]
Therefore, there are [tex]8.88 \times 10^25[/tex] HCl molecules in 4.91 L of HCl acid at 25°C, assuming the density of the acid is 1.096 g/mL.
[tex]\rule{200pt}{5pt}[/tex]
Please help almost due?
Answer:
-lithium
-atomic number
-mass number
-protons
Explanation:
After addition of 20.00 mL of 0.500 M standard KOH solution to 10.00 mL of formic acid (HCOOH, Ka = 1.8 × 10-4), the equivalence point is reached. What is the molarity of the formic acid?
What is the pH at the equivalence point, based on the question above? Please make a suggestion for an appropriate indicator.
Answer: 3.79
Explanation: The balanced chemical equation for the reaction between formic acid (HCOOH) and KOH is:
HCOOH + KOH → HCOOK + H2O
We can use the stoichiometry of this reaction to calculate the number of moles of formic acid that reacted with the KOH:
moles of KOH = (20.00 mL)(0.500 mol/L) = 0.01000 moles
moles of HCOOH = moles of KOH
Therefore, the initial number of moles of formic acid is:
moles of HCOOH = (10.00 mL)(x mol/L) = 0.01000 moles
where x is the molarity of formic acid.
Solving for x, we get:
x = 1.00 M
Therefore, the molarity of the formic acid is 1.00 M.
At the equivalence point, all of the formic acid has reacted with the KOH, and the solution contains only the salt formed by the reaction, potassium formate (HCOOK). The pH at the equivalence point can be calculated using the equation for the salt hydrolysis constant:
Kb = Kw/Ka
where Kb is the base dissociation constant of the conjugate base (formate ion), Kw is the ion product constant for water (1.0 × 10^-14 at 25°C), and Ka is the acid dissociation constant of the acid (formic acid). Rearranging this equation, we get:
Kb/Ka = [OH^-][HCOO^-]/[HCOOH]
At the equivalence point, the concentration of the formate ion (HCOO^-) is equal to the concentration of the KOH added (0.01000 moles / 30.00 mL = 0.3333 M). We can assume that the concentration of the hydroxide ion (OH^-) is also equal to 0.3333 M, since KOH is a strong base and will dissociate completely. Substituting these values into the equation above, we get:
Kb/Ka = (0.3333)^2 / [HCOOH]
Solving for [HCOOH], we get:
[HCOOH] = (0.3333)^2 / (1.8 × 10^-4) = 6181.5 M
Taking the negative logarithm of this concentration, we get the pH at the equivalence point:
pH = -log[HCOOH] = -log(6181.5) = 3.79
Therefore, the pH at the equivalence point is 3.79.
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