<span>
It makes sense that an inner shell electron would be tougher to remove
than a valence electron because the inner shell electron is closer to
the positive nucleus of the atom. Seeing as an electron caries a
negative charge it would be too attracted to the positive core to leave
readily. Also, the inner shell electrons are constantly repelling
electrons outside of it's energy level (however the reason these
electrons outside innershell energy levels don't simply fly away is the
charge of the positive core overcomes the smaller charges of the
comparably negligible inner shell electrons, but that repulsion is still
there so keep that in mind) </span>
pH of buffer can be calculated as:
pH=pKa+log[salt]/[Acid]
As ka = 4.58 x 10-4
Concentration of [Salt] that is NO2(-1)=0.380M
Concentration of [Acid] that is HNO2=0.500M
So, pH= -log(4.58*10^-4)+log((0.380)/0.500))
=3.21
So pH of solution will be 3.21
Answer:
The correct answer is - may not be typical, and participant burden.
Explanation:
The 24-hour recall is nothing but a retrospective method of diet assessment. In this method, an individual is interviewed about his or her diet consumption during the last 24 hours.
The disadvantages or limitations of this method include the inability of a single day's intake to describe the typical diet, multiple recalls to intake, cost and administration time; participant burden, have to recall to reliably estimate usual intake.
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]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
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
