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jenyasd209 [6]
3 years ago
11

The total number of electrons in one mole of fluorine molecules (F2) is

Chemistry
2 answers:
soldi70 [24.7K]3 years ago
6 0
<span>Fluorine's atomic number is 9, meaning it has 9 protons and 9 electrons in per atom. One mole is 6.0x10^23, so your answer is 4.</span>
galben [10]3 years ago
3 0
Has 9 protons and 9 electrons in per atom.........
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What distinguishes compounds from mixtures?
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Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai
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Answer : The concentration of HI and I_2 at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of H_2, I_2\text{ and }HI

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M

\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M

\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                            2HI(g)\rightleftharpoons H_2(g)+I_2(g)

Initial conc.      0.0163     0.00415      0.00276

At eqm.        (0.0163-2x) (0.00415+x)  (0.00276+x)

As we are given:

Concentration of H_2 at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of HI at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of I_2 at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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