Answer:
C. because the 4s orbital is at a lower energy level
Answer:
5.0 x 10⁹ years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of K-40 = 1.251 × 10⁹ years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (K-40) ([A₀] = 100%).
[A] is the remaining concentration of (K-40) ([A] = 6.25%).
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.
∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.
<h3>
Answer:</h3>
150000 J
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Thermodynamics</u>
Specific Heat Formula: q = mcΔT
- <em>q</em> is heat (in J)
- <em>m</em> is mass (in g)
- <em>c</em> is specific heat (in J/g °C)
- ΔT is change in temperature (in °C or K)
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>m</em> = 225 g
[Given] <em>c</em> = 4.184 J/g °C
[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C
[Solve] <em>q</em>
<u>Step 2: Solve for </u><em><u>q</u></em>
- Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
- Multiply: q = (941.4 J/°C)(159.8 °C)
- Multiply: q = 150436 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
150436 J ≈ 150000 J
Topic: AP Chemistry
Unit: Thermodynamics
Book: Pearson AP Chemistry