2. Combine lead (II) nitrate and potassium iodide solutions. Pb(NO3)2+ Kl →
1 answer:
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2-7-1
<h3>Further explanation </h3>Electrons can move the shell up or down by releasing energy or absorbing energy
Excited electrons show higher electron transfer to the shell by absorbing energy
So it can be concluded that there are 2 conditions:
Ground state is the state of electrons filling shell with the lowest energy levels.
Excited state is the state of electrons which occupies a higher energy level
The state of excited electrons can be seen from the presence of electrons which do not fill the skin completely but fill the skin afterward
2-7-1
From its 8 electron configuration, filling 3 shells, 2 electrons in the firs shell, 7 electrons in the second shell and 1 electron in the third shell
the electrons in the third shell should fill the electrons in the second shell first according to Aufbau rule (lower energy shells)
<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M
<u>Explanation:</u>
To calculate the concentration of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
Hence, the molarity of in the solution is 0.1 M.
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6