<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>
The correct option is B.
Coal dust refers to the powered form of coal. Because of the high surface area of coal dust it is highly prone to dust explosion, which involves rapid combustion of fine particles that are suspended in the air; this usually occur in an enclosed place. Coal dust in an enclosed place is more explosive than coal dust that is blown outdoor in an open space because the coal dust in an enclosed place is more concentrated due to restricted space, thus it is more liable to explosion.
Answer:
2
Explanation:
Data:
[H⁺] = 0.01 mol·L⁻¹
Calculation:
pH = -log[H₃O⁺] = -log(0.01) = -log(1) - log(10⁻²) = -0 - (-2) = 0 + 2 = 2
Answer:
Cl₂O₇
Explanation:
For the reaction:
ClₓOₙ + H₂ → HCl + H₂O
Moles of HCl and moles of H₂O are:
HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl
H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O
As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:
6.39x10⁻³ mol Cl
2.236x10⁻² mol O
Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:
2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>
As empirical formula must be given in natural numbers, the empirical formula is:
<em>Cl₂O₇</em>
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Answer:
50 g of S are needed
Explanation:
To star this, we begin from the reaction:
S(s) + O₂ (g) → SO₂ (g)
If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.
According to stoichiometry, we can determine the moles of sulfur dioxide produced.
100 g. 1mol / 64.06g = 1.56 moles
This 1.56 moles were orginated by the same amount of S, according to stoichiometry.
Let's convert the moles to mass
1.56 mol . 32.06g / mol = 50 g
