Answer:
The pH of the solution is 8.
Explanation:
To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:
1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:
pOH = –Log [OH-]
pOH = 6
6 = –Log [OH-]
–6 = Log [OH-]
[OH-] = Antilog (–6)
[OH-] = 1x10^–6 mol/L
2. The pH of the solution can be obtained as follow:
pH + pOH = 14
pOH = 6
pH + 6 = 14
pH = 14 – 6
pH = 8.
From the calculations made above,
[OH-] = 1x10^–6 mol/L
pH = 8.
Therefore, the correct answer is:
The pH of the solution is 8
<span>Answer:
.01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.
so .01/.015 to .005/.015 ~ 67% D to 33% L.
And thus, the enantiomer excess will be 34%.</span>
To determine the equilibrium concentration of hydronium ions in the solution, we use the given value of the percent ionized. Percent ionized is the percent of the ions that is dissociated into the solution. It is equal to the concentration of an ionized species over the initial concentration of the compound multiplied by 100 percent. For this case, the dissociation of the weak acid has a 1 is to 1 ratio to the ionized species such that the concentration of the CH3COO- and H+ ions at equilibrium would be equal. We calculate as follows:
5.2% = 5.2 M H3O+ / 100 M CH3COOH
5.2 M H3O+ / 100 M CH3COOH = [H3O+] / 0.048 M CH3COOH
[H3O+] = 0.2496 M
Answer:
to which cations from the salt bridge migrate
Explanation:
A voltaic cell is an electrochemical cell that uses spontaneous redox reactions to generate electricity. It's composed of a cathode, an anode, and a salt bridge.
In cathode, the substance is gaining electrons, so it's reducing, in the anode, the substance is losing electrons, so it's oxidating. The flow of electrons is from the anode to the cathode.
The salt bridge is a bond between the cathode and the anode. When the redox reaction takes place, the substances produce its ions, so the solution is no more neutral. The salt bridge allows the solutions to become neutral and the redox reaction continues.
So, the cathode produces anions, which goes to the anode, and the anode produces cations, which goes to the cathode. Then, the cathode n a voltaic cell is the electrode to which cations from salt bridge migrate and where the reduction takes place.
Molar mass CaCl₂ = 110.98 g/mol
Number of moles:
1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g
n = 85.3 / 110.98
n = 0.7686 moles of CaCl₂
Volume = ?
M = n / V
0.788 = 0.7686 / V
V = 0.7686 / 0.788
V = 0.975 L
hope this helps!
