A whiteout occurs when strong updrafts and downdrafts combine
If you have an aqueous solution that contains 1.5 moles of HCl, the number of moles of ions in the solution is 3.0 moles.
<h2>Further Explanation
</h2><h3>Strong acids </h3>
- Strong acids are types of acids that undergo complete dissociation to form ions when dissolved in water.
- Examples of such acids are, HCl, H2SO4 and HNO3
- Dissociation of HCl
HCl + H₂O ⇔ H₃O⁺ + OH⁻
<h3>Weak acids </h3>
- Weak acids are types of acids that undergo incomplete dissociation to form ions when dissolved in water.
- Examples of such acids are acetic acids and formic acids.
- Dissociation of acetic acid
H₃COOH ⇔ CH₃COO⁻ + H⁺; CH₃COO⁻ is a conjugate base of acetic acid.
<h3>In this case;</h3>
- HCl which is a strong acid that ionizes completely according to the equation;
HCl + H₂O ⇔ H₃O⁺ + OH⁻
- From the equation, 1 mole of HCl produces 1 mole of H₃O⁺ ions and 1 mole of OH⁻ ions.
Therefore;
1.5 moles of HCl will produce;
= 1.5 moles of H₃O⁺ ions and 1.5 moles of OH⁻ ions.
This gives a total number ions of;
= 1.5 + 1.5
= 3 moles of ions
Keywords: Strong acid, weak acid, ions, ionization
<h3>Learn more about: </h3>
Level: High school
Subject: Chemistry
Topic: Salts, Acids and Bases
Answer:
The correct answer is 146 g/mol
Explanation:
<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:
ΔTf = Kf x m
Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:
ΔTf = 1.02ºC
Kf = 5.12ºC/m
From this, we can calculate the molality:
m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m
The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:
0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute
There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:
molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol
<em>Therefore, the molar mass of the compound is 146 g/mol </em>
