Answer:
When our bodies are dry and wind blows by, we lose some energy to the air molecules. When are bodies are wet, we have a substance on our skin that likes to absorb heat. So when wind blows by, we lose a LOT of energy to the air molecules. When the body loses heat energy, our body temperature drops.
Explanation:
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We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH
We know that NaOH dissociates by the following reaction:
NaOH → Na⁺ + OH⁻
Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions
Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻
Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴
<u>pOH of the solution:</u>
pOH = -log[OH⁻] = -log(3 * 10⁻⁴)
pOH = -0.477 + 4
pOH = 3.523
<u>pH of the solution:</u>
We know that the sum of pH and pOH of a solution is 14
pH + pOH = 14
pH + 3.523 = 14 [subtracting 3.523 from both sides]
pH = 10.477
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
