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jeka94
3 years ago
7

By what factor would the average kinetic energy of the particles by increased if the temperature of a substance was increased fr

om -49°C to 287°C?
Chemistry
1 answer:
natita [175]3 years ago
3 0
Average kinetic energy of a particle :
0.5 mv^2 = kT/2
so the kinetic energy = kT/2
assuming the same value of K
T1 = -49 + 273 = 224
T2 = 287 + 273 = 560

E2 / E1 = kT2 / 2 / kT1 / 2
E2 / E1 = T2 / T1
E2 / E1 = 560 / 224 = 2.5
so the average kinetic energy of the particle increases by 2.5
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3 years ago
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Alex17521 [72]
<h3>Answer:</h3>

91.2 g Mn

<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>

  • Reading a Periodic Table
  • 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] 1.00 × 10²⁴ atoms Mn

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

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})
  2. [DA] Multiply/Divide [Cancel out units]:                                                           \displaystyle 91.2321 \ g \ Mn

<u>Step 4: Check</u>

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

91.2321 g Mn ≈ 91.2 g Mn

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3 years ago
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3 years ago
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A sample of 211 g of iron (III) bromide is reacted with
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FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

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\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

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Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

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mol NaBr based on limiting reactant (FeBr₃) :

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