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Svetach [21]
1 year ago
3

What is the empirical formula for the compound C40H56

Chemistry
1 answer:
Stells [14]1 year ago
5 0

The empirical formula of the compound C₄₀H₅₆ is C₅H₇.

<h3>What is the empirical formula of a compound?</h3>

The empirical formula of a compound is the formula of a compound that shows the simplest ratio in which the elements in the compound are combined.

The molecular formulae of compounds are usually multiples of the empirical formulae of the compounds. However, for some compounds, the empirical formula and the molecular formula are the same.

The empirical formula of the compound C₄₀H₅₆ is obtained by dividing with the highest common factor of the ratios which 8.

Empirical formula = C₄₀H₅₆ / 8

Empirical formula = C₅H₇

Learn more about empirical formulas at: brainly.com/question/1439914

#SPJ1

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At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations
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This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, x, has taken place:

[HF]=[HF]_0-2x

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }

[HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M

Finally, the initial concentration of HF is calculated as follows:

[HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M

Learn more:

  • brainly.com/question/13043707
  • brainly.com/question/16645766
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