Answer:
5.555.
Explanation:
∵ pH = - log[H⁺],
<em>[H⁺] for weak acids = √Ka.C.</em>
Ka for CH₃COOH = 1.74 x 10⁻⁵, C = 0.16 M.
∴ [H⁺] = √Ka.C = √(1.74 x 10⁻⁵)(0.16 M) = 2.784 x 10⁻⁶ M.
∴ pH = - log[H⁺] = - log(2.784 x 10⁻⁶ M) = 5.555.
Answer:
Al (s) | Al³⁺ (aq) || Ni²⁺ (aq) | Ni (s)
Explanation:
It seems that you meant "A voltic cell".
Here, Al (s) represents the solid anode while Al³⁺ is in aqueous phase, these two forms are separated by a solid line. || is the salt bridge while Ni²⁺ is in aqueous phase and is reduced on the nickel cathode that is Ni (s).
Explanation:
In a Lewis dot diagram, the electrons are arranged so that, when full, two electrons placed beside each other are on four sides of the element symbol.
An electron is 3-dimensional, so you can start on any side on a 2D diagram.
Continue this pattern depending on how many electrons you have:
When adding an electron after the first, it is placed on the next side going either counterclockwise or clockwise. Which ever direction you choose, be consistent.
After the fourth electron, so that one electron is on each side, place a dot beside your first electron and continue the pattern.
Explanation:
We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.
Next we will calculate how many moles of 
are present in 85.00 mL of 1.500 M sulfuric acid.
As, Molarity = 
1.500 M = 
n = 0.1275 mol
Now set up and solve a stoichiometric conversion from moles of to grams of 
. As, the molar mass of is 84.01 g/mol.
= 21.42 g
So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.
