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antoniya [11.8K]
3 years ago
5

How many atoms are there in 5.00 mol of sulphur (S)?

Chemistry
1 answer:
Mumz [18]3 years ago
6 0

Answer:

3.01 × 10²⁴ atoms S

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

5.00 mol S

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

Avogadro's Number

<u>Step 3: Convert</u>

<u />5.00 \ mol \ S(\frac{6.022 \cdot 10^{23} \ atoms \ S}{1 \ mol \ S} ) = 3.011 × 10²⁴ atoms S

<u />

<u>Step 4: Check</u>

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

3.011 × 10²⁴ atoms S ≈ 3.01 × 10²⁴ atoms S

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At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i
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Answer:

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

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Considering the ICE table for the equilibrium as:

\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}

Given:    

Equilibrium concentration of  O₂ = 0.130 mol

Volume = 1.50 L

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

Molarity=\frac{0.130}{1.50\ L}

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}  

K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}

K_c=0.0867

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