Answer:
Explanation:
Hello,
In this case, given the acid, we can suppose a simple dissociation as:
Which occurs in aqueous phase, therefore, the law of mass action is written by:
That in terms of the change due to the reaction's extent we can write:
But we prefer to compute the Kb due to its exceptional weakness:
Next, the acid dissociation in the presence of the base we have:
Whose solution is which equals the concentration of hydroxyl in the solution, thus we compute the pOH:
Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:
Regards.
There are two kinds of mixtures which are homogeneous and the heterogeneous. Homogeneous mixtures are mixtures that maintain the same uniform appearance and composition throughout; whereas, heterogeneous mixtures are mixtures that contain different visible substances or phases. Here are examples of each mixture.
Homogeneous mixtures: Rainwater, Air and Dishwashing detergent (This kind of mixture only shows one phase of matter)
Heterogeneous mixtures: cereals with milk, ice in soda and mixed nuts. (Obviously, this kind of mixture contains different phases of matter either in liquid, solid or gas).
Answer:
pH = 6.5
Explanation:
Given data:
pH of substance = ?
[OH⁻] concentration = 3.2×10⁻⁸
Solution:
pOH = -log[OH⁻]
pOH = -log[3.2×10⁻⁸]
pOH = 7.5
we know that,
pH + pOH = 14
pH = 14- pOH
pH = 14 - 7.5
pH = 6.5
Answer:
Explanation:
The objective here is to draw the Lewis structure for the polyatomic trisulfide anion and to be sure all resonance structures that satisfy the octet rule are included.
The Lewis structure for Polyatomic trisulfide anion
The first step is to the layout the skeleton of the Polyatomic trisulfide anion
S S S
However, the next step is to make sure we fill in the bonding pairs of electrons on the central atom.
Then , we move over to filling the lone pairs electrons before we finally have the Lewis structure for Polyatomic trisulfide anion as shown in the image below.
Answer:
Explanation:
The chemical equation is
For simplicity, let's rewrite this as
1. Initial concentration of NH₃
2. Calculate [OH]⁻
We can use an ICE table to do the calculation.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.100 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.100 - x x x
Check for negligibility:
\
3. Solve for x
4. Calculate the pH