Answer:
1.7 mL
Explanation:
<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
Step 1: Calculate [H⁺] in the dilute solution
We will use the following expresion.
pH = -log [H⁺]
[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M
Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.
Step 2: Calculate the volume of the concentrated HCl solution
We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂/C₁
V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL
Answer: The reaction of benzene with concentrated sulfuric acid at room temperature produces benzenesulfonic acid. The reaction forms a stable carbocation. SO3 being the electrophile
Explanation:
Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
Answer:
0.414 mole (3 sig. figs.)
Explanation:
Given grams, moles = mass/formula weight
moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)
≅ 0.414 mole (3 sig. figs.)
Divide that my the molar mass which is 23 so 1.4087 g
