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alexandr402 [8]
3 years ago
11

The electronegativity values of fluorine, oxygen, and hydrogen are compared in the table. Comparison of Electronegativity Elemen

t Electronegativity Fluorine 4.0 Oxygen 3.5 Hydrogen 2.1 Which of the following statements is true about the strength of the hydrogen bonding in HF and H2O?
Chemistry
1 answer:
Dvinal [7]3 years ago
4 0

The hydrogen bonding in H₂O is stronger than that of HF

Explanation:

Hydrogen bonds are special dipole-dipole attraction in which electrostatic attraction is established between hydrogen atom of one molecule and the electronegative atom of a neighboring molecule.

  • The strength of hydrogen bonds depends on the how electronegative an atom is.
  • Electronegativity refers to the tendency of an atom to gain electrons.
  • The higher the value, the higher the tendency.
  • This why oxygen with a higher electronegativity will form a stronger hydrogen bond with hydrogen compared to fluorine.

Learn more:

hydrogen bond brainly.com/question/12408823

#learnwithBrainly

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Calculate the mass percent of 0.485 g of H, which reacts with oxygen to form 2.32 g H2O
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Answer:

Mass % of hydrogen = 20.9 %

Explanation:

Given data:

Mass of hydrogen = 0.485 g

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Mass percent of hydrogen = ?

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3 years ago
At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f
nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

  • If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
  • If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
  • If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

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