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Verizon [17]
2 years ago
8

How many moles are in 6.5e23 atoms of Ne?

Chemistry
1 answer:
Savatey [412]2 years ago
7 0
<h3>Answer:</h3>

1.1 mol Ne

<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>

  • Moles
  • 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] 6.5 × 10²³ atoms Ne

[Solve] moles Ne

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 6.5 \cdot 10^{23} \ atoms \ Ne(\frac{1 \ mol \ Ne}{6.022 \cdot 10^{23} \ atoms \ Ne})
  2. [DA] Divide [Cancel out units]:                                                                       \displaystyle 1.07938 \ mol \ Ne

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.07938 mol Ne ≈ 1.1 mol Ne

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Explanation:

8 0
3 years ago
Hydrazine, n2h4, is used as a rocket fuel. in the reaction below, if 80.1 g of n2h4 and 92.0 g of n2o4 are allowed to react, how
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Answer is: excess of hydrazine is 16 grams.
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</span>m(N₂H₄) = 80,1 g.
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3 years ago
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 Use the coefficients of the reaction and sum the product entropies less the reactant entropies:

4*188.8 + 2*213.7 - 3*205.1 - 2* 126.8 = 313.7 J/mol*K

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3 years ago
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It is water-soluble. hope this helps


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