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amid [387]
3 years ago
11

A more experienced colleague mentions to you that you should aim for a dissolved oxygen concentration around 1.0 mg/L at day 5 o

f the experiment. They show you data from a similar wastewater spill that happened last year, in which the BOD5 of the stream water was 60 mg/L. From your previous experiment, you found that the dissolved oxygen concentration at day 0 was 10 mg/L. Based on this information, what dilution factor, P, should you try in your reactor?
Chemistry
1 answer:
Sati [7]3 years ago
4 0

Answer:

3

Explanation:

Lt= Loe^(-kt)

Data:

Lo = 10 mg/mL

Assume k = 0.23/da

1. Calculate L5

L5 = 10e^(-5×0.23) = 10e^-1.15 = 10 × 0.317 = 3.17 mg/mL

2. Calculate the dilution factor

You expect to find L5 to be about 3

You want L5 to be about 1.

You should use a dilute your sample by a factor of 3.

P = 3

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Rank the following elements by effective nuclear charge, Zeff, for a valence electron. F LI Be B N
Stels [109]

Answer:

Rank in increasing order of effective nuclear charge:

  • Li < Be < B < N < F

Explanation:

This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.

<u>1) Effective nuclear charge definitions</u>

  • While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.

  • Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.

<u><em>2) Z eff for a F valence electron:</em></u>

  • F's atomic number: Z = 9
  • Total number of electrons: 9 (same numer of protons)
  • Period: 17 (search in the periodic table or do the electron configuration)
  • Number of valence electrons:  7 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
  • Zeff = Z - S = 9 - 2 = 7

<u><em>3) Z eff for a Li valence eletron:</em></u>

  • Li's atomic number: Z = 3
  • Total number of electrons: 3 (same number of protons)
  • Period: 1 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 1 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
  • Z eff = Z - S = 3 - 2 = 1.

<em>4) Z eff for a Be valence eletron:</em>

  • Be's atomic number: Z = 4
  • Total number of electrons: 4 (same number of protons)
  • Period: 2 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 2 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
  • Z eff = Z - S = 4 - 2 = 2

<u><em>5) Z eff for a B valence eletron:</em></u>

  • B's atomic number: Z = 5
  • Total number of electrons: 5 (same number of protons)
  • Period: 13 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 3 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
  • Z eff = Z - S = 5 - 2 = 3

<u><em>6) Z eff for a N valence eletron:</em></u>

  • N's atomic number: Z = 7
  • Total number of electrons: 7 (same number of protons)
  • Period: 15 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 5 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
  • Z eff = Z - S = 7 - 2 = 5

<u><em>7) Summary (order):</em></u>

  Atom          Zeff for a valence electron

  • F                   7
  • Li                   1
  • Be                 2
  • B                   3
  • N                   5

  • <u>Conclusion</u>: the order is Li < Be < B < N < F
6 0
3 years ago
What is the nuclear binding energy of an atom that has a mass defect of 5.0446 x 10-29 kg?
lutik1710 [3]
I think it is 4.54 x 10-12
3 0
2 years ago
Read 2 more answers
Help me. please help me.
igor_vitrenko [27]
B- 90 grams bc 45 mL times 2 equals 90
3 0
3 years ago
Why should scientists control most possible variables in their experiments?
irina [24]
Scientists should control most possible variables in experiments to get the most valid and correct data. If many variables are included in experiments it is more difficult to interpret what is causing a different outcome.
8 0
3 years ago
The smallest unit of an element that can exist either alone or in combination with other such particles of the same or different
Mekhanik [1.2K]

Answer

im not quite sure but I think the answer is <em>D atom</em><em> </em>

Explaination

5 0
2 years ago
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