Answer:
c when electrons drop down from a high energy level they emit a photon.
Explanation:
A is the answer
In an ozone molecule, the three atoms must be connected, so there must at least be a single bond between them. Place
dots in pairs around the oxygen atoms until each oxygen atom has eight valence electrons, starting with the atoms on the
outside and doing the central atom last if there are enough. Do not exceed the total number of valence electrons
identified in part A. Remember that the dashes between the oxygen atoms, which represent single bonds, each indicate
the presence of two valence electrons
<h3>
Answer:</h3>
1.43 × 10⁻²⁰ mol Li
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
8.63 × 10³ atoms Li
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after
is 0.00345 mol/L.
Explanation:

Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t = 
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after
is 0.00345 mol/L.
<span>Boyle's Law is k = PV so
Initial k = 13.0 L x 4.0 atm = 52 L atm
Final kf = 6.5 L x 8 atm = 52 L atm
The gas obeys Boyle's Law
The answer with two significant figures separated by a comma is k = 52, kf = 52.</span>