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olga55 [171]
3 years ago
5

42 mole Al= how many atoms?

Chemistry
1 answer:
Masja [62]3 years ago
8 0
<h3>Answer:</h3>

2.5 × 10²³ atoms 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>

42 mol Al

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 42 \ mol \ Al(\frac{6.022 \cdot 10^{23} \ atoms \ Al}{1 \ mol \ Al})
  2. Multiply/Divide:                 \displaystyle 2.52924 \cdot 10^{25} \ atoms \ Al

<u>Step 4: Check</u>

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

2.52924 × 10²³ atoms Al ≈ 2.5 × 10²³ atoms Al

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2 years ago
What is the molar concentration of H2SO4 in a solution made by reacting 188.9 mL of 2.086 M H2SO4 with 269.3 mL of 0.4607 M NaOH
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Answer:

0.7246 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

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Volume = 188.9 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 188.9×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=2.086 \times {188.9\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.39405 moles

For NaOH :

Molarity = 0.4607 M

Volume = 269.3 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 269.3×10⁻³ L

Thus, moles of NaOH :

Moles=0.4607 \times {269.3\times 10^{-3}}\ moles

Moles of NaOH  = 0.1241 moles

According to the given reaction:

H_2SO_4_{(aq)}+2NaOH_{(aq)}\rightarrow Na_2SO_4_{(aq)}+2H_2O_{(aq)}

1 moles of H_2SO_4 react with 2 moles of NaOH to form 1 mole of sodium sulfate.

Thus,

2 moles of NaOH react with 1 mole of H_2SO_4

1 mole of NaOH react with 1/2 mole of H_2SO_4

0.1241 moles of NaOH react with (1/2)×0.1241 mole of H_2SO_4

Moles of H_2SO_4 that got reacted = 0.06205 moles

Unreacted moles = Total moles - Moles that got reacted = 0.39405 - 0.06205 moles = 0.332 moles

Total volume = 188.9×10⁻³ L + 269.3×10⁻³ L = 458.2×10⁻³ L

Concentration of H_2SO_4 :

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{H_2SO_4}=\frac{0.332}{458.2\times 10^{-3}}

<u>Concentration of H_2SO_4 = 0.7246 M</u>

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