Give the hybridization for the O in H3O+

kondor19780726 [428]

A. Ammonium = polyatomic ion
B. Oxygen = compound
C. Silver = element
D. Nitrate = polyatomic ion
E. Helium = element
F. Water = polyatomic ion
G. Nitrogen = element
H. Carbon dioxide = compound

3 0

3 years ago

The __ is the only vein in the body that carries oxygen-rich blood.

Vesnalui [34]

Answer:

pulmonary vein

Explanation:

The pulmonary vein is the only vein in the body that carries oxygen rich blood.

4 0

3 years ago

What is the electron configuration for a neutral atom of fluorine.

zavuch27 [327]

Answer and Explanation: The full ground-state electron configuration for a fluorine atom is 1s2, 2s2, 2p5. We determine this by first identifying the number of electrons in a fluorine atom. If the atom is neutral, it will have nine electrons to match the nine protons int he fluorine nucleus

6 0

3 years ago

Read 2 more answers

Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos

SCORPION-xisa [38]

Explanation:

More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a) Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, $PF_{3}$ is the most ionic compound and $PBr_{3}$ is the least ionic compound.

(b) Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, $BF_{3}$ is the most ionic compound and $NF_{3}$ is the least ionic compound.

(c) Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, $TeF_{4}$ is the most ionic compound and $BrF_{3}$ is the least ionic compound.

6 0

3 years ago

What happend to the pressure of a gas inside a container if the temperatures of the gas decreases

ludmilkaskok [199]

There is an increase in the number of collisions between particles and the walls of the container. b. There is an increase in the temperature of the gas. If the volume of a container of gas is reduced, what will happen to the pressure inside the container

4 0

3 years ago