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aniked [119]
3 years ago
8

What transition energy corresponds to an absorption line at 527 nm?​

Chemistry
2 answers:
Ede4ka [16]3 years ago
4 0

Answer:

E = 3.77×10⁻¹⁹ J

Explanation:

Given data:

Wavelength of absorption line = 527 nm (527×10⁻⁹m)

Energy of absorption line = ?

Solution:

Formula:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = speed of wave = 3×10⁸ m/s

by putting values,

E = 6.63×10⁻³⁴ Js ×  3×10⁸ m/s / 527×10⁻⁹m

E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m

E = 0.0377×10⁻¹⁷ J

E = 3.77×10⁻¹⁹ J

irina [24]3 years ago
3 0

Answer:

D

Explanation:

a p e x

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Answer:

Cl2.

Explanation:

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Note that the element Cl (chlorine) is missing from the left side.

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3 years ago
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How many moles in 2.21 x10E24 atoms of aluminum
lisov135 [29]
<h3>Answer:</h3>

3.67 mol Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<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>

2.21 × 10²⁴ atoms Al

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 2.21 \cdot 10^{24} \ atoms \ Al(\frac{1 \ mol \ Al}{6.022 \cdot 10^{23} \ atoms \ Al})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

3.66988 mol Al ≈ 3.67 mol Al

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How many moles are in a solution with a concentration of 5 M and a volume of 0.25 L?
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The reaction N O space plus thin space O subscript 3 space rightwards arrow space N O subscript 2 space plus thin space O subscr
nikklg [1K]

Answer:

4.3 × 10⁻⁵ M s⁻¹

Explanation:

Step 1: Given data

  • Rate constant (k):  2.20 × 10⁷ M⁻¹s⁻¹
  • Concentration of NO ([NO]): 3.3 × 10⁻⁶ M
  • Concentration of O₃ ([O₃]): 5.9 × 10⁻⁷ M
  • First order with respect to both NO and O₃

Step 2: Write the balanced reaction

NO + O₃ ⇒ NO₂ + O₂

Step 3: Calculate the reaction rate

The rate law is:

rate = k × [NO] × [O₃]

rate = 2.20 × 10⁷ M⁻¹s⁻¹ × 3.3 × 10⁻⁶ M × 5.9 × 10⁻⁷ M

rate = 4.3 × 10⁻⁵ M s⁻¹

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