Answer:
A description of the experiment the chemicals were used for.
Explanation:
A chemical waste label is required to provide information about any hazardous waste present in the container. Some details are mandatory to mention of the chemical waste label to prevent any accident while doing the experiment, that includes:
- Chemical compound's name present in the container.
- Composition and physical state of the waste.
- Hazardous properties of the waste.
- The date of manufacturing.
- Amount of chemical compounds filled in the container.
A chemical waste label does not mention or describe the experiment for which the chemicals were used for, scientists, teacher or students should have knowledge of the chemical composition by reading their names and evaluate themselves on which chemical should be used for which experiment.
Hence, the correct answer is "A description of the experiment the chemicals were used for."
Answer:
pH = 2.18
Explanation:
Perchloric acid (HClO₄) is a strong acid. This means that in an aqueous solution it completely dissociates into H⁺ and ClO₄⁻ species.
First we <u>convert 249 mg HClO₄ into moles</u>, using its <em>molecular weight</em>:
- 249 mg HClO₄ ÷ 100.46 mg/mmol = 2.49 mmol HClO₄
<em>Because it is a strong acid</em>, 2.49 mmol HClO₄ = 2.49 mmol H⁺
We <u>calculate the molar concentration of H⁺</u>:
- 2.49 mmol H⁺ / 380 mL = 6.52x10⁻³ M
Finally we <u>calculate the pH of the solution</u>:
- pH = -log[H⁺] = -log(6.52x10⁻³)
Missing question: What is the vapor pressure of the solution at 25°<span>C?
n(NaCl) = 100 g </span>÷ 58,4 g/mol.
n(NaCl) = 1,71 mol.
NaCl → Na⁺ + Cl⁻, amount of ions are 2 · 1,71 mol = 3,42 mol.
n(CaCl₂) = 100 g ÷ 111 g/mol = 0,9 mol.
CaCl₂ → Ca²⁺ + 2Cl⁻, amount of ions 3 · 0,9 mol = 2,7 mol.
m(solution) = 1000 ml (1,00 L) · 1,15 g/ml = 1150 g.
m(H₂O) = 1150 g - 100 g - 100 g = 950 g.
n(H₂O) = 950 g ÷ 18 g/mol = 118,75 mol.
<span>water's mole fraction = 118,75 mol </span>÷ (118,75 mol + 2,7 mol + 3,42 mol).
water's mole fraction = 0,95.
p(solution) = 0,95 · 23 mmHg = 21,85 mmHg.
The two volumes are taken as additive ( in actuality they will not be) , then Final volume = 39.5+215 = 254.5mL % v/v ethanol = 39.5/254.5*100 = 15.5%l
