The solubility of nitrogen in water at 25 °C= 4.88 x 10⁻⁴ mol/L
<h3>Further explanation</h3>
Given
78% Nitrogen by volume
Required
The solubility of nitrogen in water
Solution
Henry's Law states that the solubility of a gas is proportional to its partial pressure
Can be formulated
S = kH. P.
S = gas solubility, mol / L
kH = Henry constant, mol / L.atm
P = partial gas pressure
In the standard 25 C state, the air pressure is considered to be 1 atm, so the partial pressure of N₂ -nitrogen becomes:
Vn / Vtot = Pn / Ptot
78/100 = Pn / 1
Pn = 0.78 atm
Henry constant for N₂ at 25 °c = 1600 atm/mol.L=6.25.10⁻⁴ mol/L.atm
The solubility :
Halogens (atoms with 7 valence electrons) and Hydrogen
or generally, atoms with their shells almost full
An atom's mass is determined by its protons and neutrons.
An atom's charge is determined by its number of protons minus it number of electrons.
Atoms become cations, or positively charged when they lose an electron, and since electrons have a negative charge, they become anions, or negatively charged.
Water is a universal solvent.
Carbohydrates (carbs) are used by the body for energy.
Steroids and triglycerides are lipids.
Proteins that catalyze chemical reactions are called enzymes.
Answer:
a) ammonium ion
b) amide ion
Explanation:
The order of decreasing bond angles of the three nitrogen species; ammonium ion, ammonia and amide ion is NH4+ >NH3> NH2-. Next we need to rationalize this order of decreasing bond angles from the valence shell electron pair repulsion (VSEPR) theory perspective.
First we must realize that all three nitrogen species contain a central sp3 hybridized carbon atom. This means that a tetrahedral geometry is ideally expected. Recall that the presence of lone pairs distorts molecular structures from the expected geometry based on VSEPR theory.
The amide ion contains two lone pairs of electrons. Remember that the presence of lone pairs causes greater repulsion than bond pairs on the outermost shell of the central atom. Hence, the amide ion has the least H-N-H bond angle of about 105°.
The ammonia molecule contains one lone pair, the repulsion caused by one lone pair is definitely bless than that caused by two lone pairs of electrons hence the bond angle of the H-N-H bond in ammonia is 107°.
The ammonium ion contains four bond pairs and no lone pair of electrons on the outermost nitrogen atom. Hence we expect a perfect tetrahedron with bond angle of 109°.
