Answer: The number 4 indicates 4 electrons.
Explanation: We are given an electronic configuration, which is:

Here,
- The letters denote the sub-shells of an element.
- The numbers written before the letters which are 1, 2 and 2 represents the Principle Quantum Number and these represents the energy level of the sub-shells.
- The number which are written in the superscripts which are 2, 2 and 4 denotes the electrons which are present in the sub-shell.
Hence, 4 indicates 4 electrons present in 2p sub-shell.
Explanation:
The given data is as follows.
Volume of lake =
= 
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake = 
=
mg
=
kg
Flow rate of river is 50 
Volume of water in 1 day = 
=
liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are
or 
Flow rate of sewage = 
Volume of sewage water in 1 day =
liter
Concentration of sewage = 300 mg/L
Total amount of pollutants =
or 
Therefore, total concentration of lake after 1 day = 
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence,
= 
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
moles CO₂ = 5.57.10⁻⁴
<h3>Further explanation
</h3>
A mole is a number of particles(atoms, molecules, ions) in a substance
Can be formulated :

0.7% percent change with 3.5g of plant matter
mass :

moles :

Answer:
<h2>Oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.</h2>
<h3>Valence electrons are the electrons in the outermost shell, or energy level, of an atom. </h3>
<h3>Configuration of oxygen's valence electrons as 2s²2p⁴.</h3>
Explanation:
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Answer:
Balanced reaction:
3 H2 (g) + N2 (g) → 2 NH3 (g)
Use stoichiometry to convert g of H2 to g of NH3. The process would be:
g H2 → mol H2 → mol NH3 → g NH3
12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3
Explanation: See above
Hope this helps, friend.