<span>The effective nuclear charge of an atom = total electrons - inner electrons
For O, ENC = 8 - 2 = 6
For Li, ENC = 3 - 2 = 1
For C, ENC = 6 - 2 = 4
The electrons in O experience the greatest effective nuclear charge and that is why O is smaller than C (which is smaller than Li).</span>
Answer:
The protonated form is predominant when aspirin is absorbed more readily. The ratio of conjugate base to acid is 1 to 100.
Explanation:
Aspirin is more readily absorbed when it is protonated, that is when pH is lower than pKa (<em>more H⁺ available in the medium</em>). We can confirm this using Henderson-Hasselbalch equation for pH = 1.5:

When aspirin is absorbed more readily the ratio of conjugate base to acid is 1 to 100, being the acid the <em>predominant</em> form.
The answer is B. Commensalism.
<span>Commensalism is a relationship between two organisms in which only one of them has benefit, and the other one is not affected. In this example, the Great Burdock's plants spread their seeds using animals, so they benefit from this relationship. On the other hand, animals neither have benefits not are harmed from the relationship.</span>
<h3>
Answer:</h3>
1.2 × 10⁻⁸ mol Pb
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 7.2 × 10¹⁵ atoms Pb
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb
(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>.