Electron configuration is the electron distribution in the molecular and atomic orbital. An element with configuration ns²np¹ will be in the 3A group. Thus, option B is correct.
<h3>What is electronic configuration?</h3>
The electronic configuration has been the arrangement and distribution of the sub-atomic particle, an electron in the atomic shells.
The electronic configuration given is, ns²np¹. Here, there are three valence electrons in the outermost orbit. As it has been known that the number of the valence electron gives the number of the group.
Therefore, option B. 3A group or 13 group is the correct option.
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Your question is incomplete, but most probably your full question was, An element with the general electron configuration for its outermost electrons of ns2np1 would be in which group?
a. 2a
b. 3a
c. 4a
d. 5a
e. 8a
Answer:
I think the answer is 4) 41
Explanation:
APE= atomic number, proton and the electrons are the same number
MAN= mass = atomic number - neutrons
121 - 80 = 41
i haven't done this in a while so hope this helps :)
Answer:
The answer to your question is
1.- Volume = 3.4 ml
2.- Volume = 0.61 ml
3.- Mass = 2872.8 pounds
Explanation:
Problem 1
Volume = 18 ml
mass = 35.6 g
density = 10.5 g/ml
Process
1.- Calculate the volume of silver
Formula
solve for volume
Substitution
<u>volume = 3.4 ml</u>
2.- Problem 2
Total volume = ?
Volume = 18 + 3.4
Volume = 21.4 ml
Data
mass = 8.3 g
density = 13.6 g(ml
volume = ?
Formula
Solve for volume
Substitution
Result
<u>volume = 0.61 ml</u>
3.- Problem 3
Data
volume = 345 gal
density = 1 g/ml
mass = ?
Formula
Solve for mass
mass = density x volume
Covert gal to ml
1 gal --------------- 3785 ml
345 gal ------------- x
x = (345 x 3785) / 1
x = 1305825 ml
Substitution
mass = 1 x 1305825
mass = 1305825 g
Convert g to pounds
1 g ------------------- 0.0022 pounds
1305825 g ---------------- x
x = (1305825 x 0.0022)
<u> x = 2872.8 pounds</u>
the balanced equation for the formation of ammonia is
N₂ + 3H₂ ---> 2NH₃
molar ratio of N₂ to NH₃ is 1:2
mass of N₂ reacted is 8.0 g
therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol
according to the molar ratio,
1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant
therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃
therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g
a mass of 9.72 mol of NH₃ is formed
Answer: It is important to wet the filter paper in the Buchner funnel first with cold re crystallization solvent before the re crystallization mixture being filtered to minimize gaps around the edges of the filter paper which can prevent mechanical impurities from passing through. This gives better filtration where most impurities can be filtered. Furthermore, it provides good vacuum.
