True
when a substance is impure, it boils over a range of temperature rather than a specific temperature
Answer:
Explanation:
1)<u><em> Ionization equilibrium equation: given</em></u>
- H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>
<u>3) Stoichiometric mole ratio:</u>
As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:
- [H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M
- ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M
<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>
Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:
Then you can substitute the known values and solve for the unknown:
- 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M
- ⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M
As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.
When the particles of a substance (usually a liquid) is heated up, its particles absorb the energy provided thereby increasing their kinetic energy resulting to more movement of the individual particles.
Rusting iron is an example of a chemical change because iron reacts with air when moisture is present.
examples of chemical changes are:
• Molecules rearrange with other molecules to make new substance
• Can be production of flames
• Color change
• Bubbling/fizzing
• Temp. change
Examples of physical changes are:
• Melting
• Boiling
• Freezing
• Condensing
• Breaking
• Bending
• Dissolving
Answer:
V₂ = 946.72 mL
Explanation:
Given data;
Initial pressure = 0.926 atm
Initial volume = 457 mL
Temperature = constant = 29.5°C
Final pressure = 0.447 atm
Final volume = ?
Solution:
The given problem will be solved through the Boyle's law,
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
by putting values,
P₁V₁ = P₂V₂
0.926 atm × 457 mL = 0.447 atm × V₂
V₂ = 423.18 atm. mL/ 0.447 atm
V₂ = 946.72 mL
