<h3>
Answer:</h3>
A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent.
<h3>
Explanation:</h3>
- A solution is made by dissolving a solute in a solvent.
- For example dissolving a salt in a solvent such as water results to a solution.
- Solution may either be saturated or unsaturated.
- Unsaturated solution is a solution that can dissolve more solute upon addition because it has not reached saturation.
- A saturated solution on the other hand is a solution that has maximum solute and the concentration of solute is maximum and thus the solvent can not dissolve any more solute.
- Therefore, a saturated solutions contain maximum concentration of a solute dissolved in the solute.
Answer:
(iii) A has pH greater than 7 and B has pH less than 7
Explanation:
Phenolphthalein is a common indicator in acid base titrations. It turns pink in basic conditions and turns colorless in acidic conditions. Thus on addition of solution A it becomes pink so A should be basic having pH more than 7. On addition of B , it turn out to be colorless means that B is an acidic solution having pH less than 7.
The pH of a neutral aqueous solution at 37°C is 6.8.
<h3>What is Kw? </h3>
Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:
Kw = [OH⁻] . [H₃O⁺]
Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.
<h3>Calculation</h3>
Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:
Kw =[OH⁻] . [H₃O⁺] = x²
x² = 2.4 × 10⁻¹⁴ M²
x = 1.5919 × 10⁻⁷ M
Hence, the concentration of OH and H₃O⁺ (x) = [H₃O⁺] = [OH⁻] = 1.5919×10⁻⁷ M
pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)
pH = 6.8
Thus, we find that the pH of a neutral aqueous solution at 37 °c (which is the normal human body temperature) is 6.8.
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Answer:
The correct answer is 160.37 KJ/mol.
Explanation:
To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,
k = Ae^-Ea/RT
k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2
k2/k1 = e^-Ea/R (1/T2-1/T1)
ln(k2/k1) = Ea/R (1/T1-1/T2)
The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.
The temperatures T1 and T2 are 298 K and 425 K respectively.
Now by filling the values we get:
ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)
19.29 = Ea/R * 0.001
Ea = 160.37 KJ/mol
Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
