answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
JulsSmile
26 days ago
12

Match each set of quantum numbers to the correct subshell description by typing in the correct number.

Chemistry
2 answers:
lorasvet [2.5K]26 days ago
8 0
Quantum numbers describe the location of the outermost or valence electron. There are five quantum numbers: principal (n), angular momentum (l), magnetic (ml), and spin (ms). Following Bohr's atomic theory, electrons are believed to orbit the nucleus within defined energy levels, beginning at level 1 which is nearest to the nucleus and increasing through levels 2, 3, 4, 5, 6, and 7. These levels correspond to the principal quantum number. Each orbital contains subshells, which follow the order of increasing complexity: s, p, d, and f. The angular momentum quantum numbers correspond to these subshells as follows: s=0, p=1, d=2, and f=3. Hence, for an electron in the 5p orbital, the quantum numbers are: 5, 1. The correct matches are as follows:

2p: n=2, l=1
3d: n=3, l=2
2s: n=2, l=0
4f: n=4, l=3
1s: n=1, l=0.
castortr0y [2.7K]26 days ago
6 0

2p: result is 4

3d: result is 2

2s: result is 1

4f: result is 5

1s: result is 3


You might be interested in
Four students are developing a model to illustrate covalent and ionic substances dissolving in
Anarel [2600]

Students dealing with ionic bonds comprehend better how to convey what the model should showcase.

Explanation:

  • Upon dissolving ionic compounds in water, the compounds separate into their constituent ions via a process called dissociation.
  • The ions become attracted to water molecules, which carry a polar charge.
  • If the pull between the ions and the water molecules is strong enough to disband the ionic bonds, the compound dissolves.
  • The ions disperse in the solution, each surrounded by water molecules to inhibit reattachment.
  • The ionic solution forms an electrolyte, allowing it to conduct electricity.
  • In contrast, while covalent compounds do dissolve in water, they separate into molecules, not individual atoms.
  • Water acts as a polar solvent, yet covalent compounds are generally nonpolar.
  • This implies that covalent compounds often do not dissolve in water and instead form a distinct layer on top of the water.
5 0
26 days ago
You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di
lorasvet [2515]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

500mM = 500mM \times (\frac{1M}{1000M})= 0.5M

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M

10 mL of this solution is further diluted to 250 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M

Convert μM:

2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M

5 mL of this solution is diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M

Convert μM:

0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M

10 mL of this solution is further diluted to 500 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

7 0
1 month ago
From the following list of elements, those that will always form ionic compounds in a 1:2 ratio with zinc.
eduard [2509]
A. iodine C. fluorine F. bromine Explanation: Ionic bonds primarily form between metals and non-metals, typically where there exists a significant difference in electronegativity between the constituent atoms. This situation results in one atom seeking to gain electrons while the other donates them. For zinc to form a compound in a 1:2 ratio, its combining power must align accordingly. The prevalent oxidation state of zinc is +2. The other combining atoms must also possess the capacity to accept two electrons. The halogens fit perfectly here, as they require only one electron to achieve stability and are highly electronegative. Hence, if two halogens combine with zinc, they will result in an ionic bond. The relevant halogens include fluorine, chlorine, bromine, iodine, and astatine which will yield compounds: ZnF₂, ZnBr₂, and ZnI₂.
3 0
11 days ago
Give an example of a rule of the natural world that a scientist can assume is always true.
Alekssandra [2711]
Laws of Nature should be differentiated from Scientific and Natural Laws. The Necessitarian Theory suggests that Laws of Nature are those principles which influence the natural phenomena in the universe, meaning the natural world adheres to them.
4 0
19 days ago
If you pump air into cycle tyre a slight warming effect is noticed at valve stem why
castortr0y [2731]
The slight warm feeling noticed at the valve stem when air is pumped into the tire is likely due to the kinetic energy generated by the friction from the pump and the resultant increase in gas pressure within the tire.
8 0
27 days ago
Other questions:
  • How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
    12·1 answer
  • Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter
    5·1 answer
  • A laboratory analysis of an unknown sample yields 74.0% carbon, 7.4% hydrogen, 8.6% nitrogen, and 10.0% oxygen. What is the empi
    14·1 answer
  • Marianne designs an experiment involving electrically charged objects. She wants to know which objects will be attracted to a ne
    7·2 answers
  • Which substance can be decomposed by chemical means?
    12·2 answers
  • An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroun
    7·1 answer
  • A sample of a compound contains 60.0 g C and 5.05 g H. Its molar mass is 78.12 g/mol. What is the compound’s molecular formula?
    11·2 answers
  • The graph shows a sample of gas when it is hot, cold, and at room temperature.
    15·1 answer
  • A substance with a density of 2.70 g/mL occupies a volume of 21.3 mL. What is the mass of the sample? Report your answer in unit
    11·1 answer
  • A sample of 88.0 g CO2 is held at 291 K in a 40.0 L container. What is the pressure this gas exerts on the container. Express yo
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!