How does electron sharing occur in forming covalent bonds?

1 answer:

7 0

Covelant bonds form when 2 atoms do not have full outer shells.
Both of the 2 atoms then join together and share their electrons in order to gain stability have have a full outer shell

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What is the mass of 6.2 mil of K2CO3

Oksi-84 [34.3K]

Answer:

6.2 moles of K2CO3 can be converted to 856.8741 grams.

6 0

3 years ago

This is the chemical formula for zinc bromate: . Calculate the mass percent of oxygen in zinc bromate. Round your answer to the

Paladinen [302]

Answer:

30%

Explanation:

<em>This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.</em>

Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂

M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)

M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol

M(Zn(BrO₃)₂) = 321.18 g/mol

Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂

There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.

6 × m(O) = 6 × 16.00 g = 96.00 g

Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂

%O = mO/mZn(BrO₃)₂ × 100%

%O = 96.00 g/321.18 g × 100% ≈ 30%

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3 years ago

Which of the following is unchanged at the end of the CNO cycle?

Musya8 [376]

Answer:

The correct option is: A. carbon-12

Explanation:

The CNO cycle, the abbreviation for the carbon-nitrogen-oxygen cycle, is a catalytic cycle by which the stars produce helium from elemental hydrogen, via a series of nuclear fusion reactions.

This cycle involves the fusion of four protons with carbon ( $_{6}^{12}\textrm{C}$ ), nitrogen isotope ( $_{7}^{13}\textrm{N}$ ), and oxygen isotope ( $_{8}^{15}\textrm{O}$ ), to give an alpha particle and two electron neutrinos and positrons.

The reaction involves the regeneration of carbon ( $_{6}^{12}\textrm{C}$ ) nucleus in the last step.

$_{6}^{12}\textrm{C}$ → $_{7}^{13}\textrm{N}$ → $_{6}^{13}\textrm{C}$ → $_{7}^{14}\textrm{N}$ → $_{8}^{15}\textrm{O}$ → $_{7}^{15}\textrm{N}$ → $_{6}^{12}\textrm{C}$