What is true of a covalent bond? (03.03)

Chemistry

2 answers:

garri49 [273]3 years ago

6 0

Answer:

I think B..

Explanation:

It is the sharing of electrons from one atom to another .

Brrunno [24]3 years ago

5 0

Answer:

The answer is b: It is the sharing of electrons by overlapping orbitals

Explanation:

The ionic bong involves the exchange of electrons from one atom to another to keep eight particles in their outermost orbit. In this way the atoms become stable. This happens between a metal, which losing electrons becomes a cation that now has a positive charge, and a non-metal, which accepting electrons becomes an anion that now has a negative charge. So the ionic bond is the electrostatic attraction between oppositely charged particle (cation and anion).

No chemical bond involves sharing neutrons between two nuclei. Chemical bonds occur between the electrons of atoms, which are found in the outer layer around the nuclei.

So, the correct answer is b: It is the sharing of electrons by overlapping orbitals. This happens between two non-metallic elements.

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Name 5 affects of weathering and erosion

nadezda [96]

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Explanation:

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3 years ago

A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$