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Tatiana [17]
3 years ago
7

How many atoms are in 10.1 g Ne

Chemistry
1 answer:
LekaFEV [45]3 years ago
4 0

Answer:

3.01 × 10²³ atoms Ne

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

[Given] 10.1 g Ne

[Solve] atoms Ne

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

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

[PT] Molar Mass of Ne: 20.18 g/mol\

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 10.1 \ g \ Ne(\frac{1 \ mol \ Ne}{20.18 \ g \ Ne})(\frac{6.022 \cdot 10^{23} \ atoms \ Ne}{1 \ mol \ Ne})
  2. [DA] Divide/Multiply [Cancel out units]:                                                          \displaystyle 3.01398 \cdot 10^{23} \ atoms \ Ne

<u>Step 4: Check</u>

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

3.01398 × 10²³ atoms Ne ≈ 3.01 × 10²³ atoms Ne

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For the given chemical reaction:

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The equation for the enthalpy change of the above reaction is:

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Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

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(e) Another student investigated the rate of a different reaction
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0.07 g/s.

Explanation:

From the question given above, the following data were obtained:

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Finally, we shall determine the mean rate of the reaction. This can be obtained as illustrated below:

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Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
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