Each of the following substances is a liquid at –50°C. Place these liquids in order of increasing vapor pressure: dimethyl ether

Answer:

1) potential energy within the bond.

2) Linear

3) Electrons are transferred from K to Cl.

4) ClF

5) Oxygen

6) Electrolysis

7) Double displacement

Explanation:

When two atoms approach each other during bonding, the potential energy of the bond is reduced as the distance between the nuclei of the atoms diminishes.

BeH2 features two electron domains, indicating the central beryllium atom exhibits sp2 hybridization. According to the theory of valence shell electron pair repulsion, a molecule with two areas of electron density results in a linear structure.

KCl is an ionic compound, hence electrons are transferred from K (metal) to Cl (nonmetal).

ClF has partial charges due to its polar covalent bond. These partial charges arise from the molecule's dipole. LiF represents a pure ionic compound formed by the transfer of electrons from Li to F, resulting in full rather than partial charges.

When oxygen bonds with another oxygen atom, they create a homonuclear covalent bond. Given the equal electronegativity of both atoms, a pure covalent bond is established. Remember that polar covalent bonds form when there is a considerable electronegativity disparity between the bonding atoms.

During electrolysis, when direct current is utilized through certain salt solutions, gases may be produced and captured at the corresponding electrodes.

A double-replacement reaction occurs when cations and anions from two different ionic compounds react and exchange places, forming two new compounds on the product side. As an example, consider the reaction presented in question 7:

AgNO3 (s) + NaCl (s) → AgCl (s) + NaNO3 (s).

Answer: The percentage abundance for isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

  .....(1)

From the information provided:

Let the fractional abundance for isotope be 'x'

• For isotope:

Mass of isotope = 27.9769 amu

Percentage abundance of isotope = 92.22 %

Fractional abundance for isotope = 0.9222

• For isotope:

Mass of isotope = 28.9764 amu

Percentage abundance for isotope = 4.68%

Fractional abundance of isotope = 0.0468

• For isotope:

Mass of isotope = 29.9737 amu

Fractional abundance for isotope = x

• The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

To convert this fractional abundance into a percentage, multiply by 100:

This shows that the percentage abundance for isotope is 3.09 %.