Answer:
Hydrogen concentration = 7.94×10^-3 M
Explanation:
from potenz Hydrogen ( pH ) definition
pH = -log[H+]
2.1 = -log[H+]
2.1/-log = -log[H+]/-log
10^-2.1 = [H+]
[H+] = 7.94×10^-3M
See the sketch attached.
<h3>Explanation</h3>
The Lewis structure of a molecule describes
- the number of bonds it has,
- the source of electrons in each bond, and
- the position of any lone pairs of electrons.
Atoms are most stable when they have eight or no electrons in their valence shell (or two, in case of hydrogen.)
- Each oxygen atom contains six valence electrons. It demands <em>two</em> extra electrons to be chemically stable.
- Each sulfur atom contains six valence electrons. It demands <em>two </em> extra electrons to be chemically stable.
- Each hydrogen atom demands <em>one</em> extra electron to be stable.
H₂O contains two hydrogen atoms and one oxygen atom. It would take an extra 2 + 2 × 1 = 4 electrons for all its three atoms are stable. Atoms in an H₂O would achieve that need by sharing electrons. It would form a total of 4 / 2 = 2 O-H bonds.
Each O-H bond contains one electron from oxygen and one from hydrogen. Hydrogen has no electron left. Oxygen has six electrons. Two of them have went to the two O-H bonds. The remaining four become 4 / 2 = 2 lone pairs. The lone pairs repel the O-H bonds. By convention, they are placed on top of the two H atoms.
Similarly, atoms in a SO₂ molecule demands an extra 2 × 2 + 2 = 6 electrons for its three atoms to become chemically stable. It would form 6 / 2 = 3 chemical bonds. Loops are unlikely in molecules without carbon. As a result, one of the two O atoms would form two bonds with the S atom while the other form only one.
Atoms are unstable with an odd number of valence electrons. The S atom in SO₂ would have become unstable if it contribute one electron to each of the three bond. It would end up with 3 × 2 + 3 = 9 valence electrons. One possible solution is that it contributes two electrons in one particular bond. One of the three bonds would be a coordinate covalent bond, with both electrons in that bond from the S atom. In some textbooks this type of bonds are also known as dative bonds.
Dots and crosses denotes the origin of electrons in a bond. Use the same symbol for electrons from the same atom. Electrons from the oxygen atoms O are shown in blue in the sketch. They don't have to be colored.
In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,
Answer:
9.3
Explanation:
This is long and complicated so get ready
We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)
Let the conjugate base be A- and Carbonic acid be HA
A- + H20 ⇄ HA + OH-
To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.
to get moles of acid we take the concentration and multiply by liters to cancel
.2653 x .150 = .039795 mol HA
Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters
We plug it into the equation and found: .181 L
Now use moles and combined volums to fins concentrarion which is .120 M
Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-
Ka is (10^-3.60) = 2.4E-4
Kb x Ka is 10^-14
Kb = 3.98E-11
Now we know Kb = [HA] [OH] / [A-]
Solve for this through algebra by using x for the values you dont know
youll find x^2 = 3.3E-10
X = 1.8 E -5
this is the OH- concentration
-log [oh] = pOH
pOH = 4.73
We know 14-pOH = ph so pH= 9.3