Explanation:
The distribution of electrons of an atom or molecule in an atomic orbital is called electronic configuration.
1. The electronic configuration of Fluorine: [He] 2s²2p⁵.
2. The electronic configuration of Neon: [He]2s²2p⁶
3. The electronic configuration of Sulfur: 1s²2s²2p⁶3s²3p⁴
4. The electronic configuration of Calcium:[Ar] 4s²
5. The electronic configuration of Iron: [Ar] 3d⁶4s²
Hence, this is the required solution.
1. iron 2. gasoline 3. water 4. ice.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe
<h3>Explanation</h3>
Method One: Refer to electron transfers.
Oxidation states:
- Na: from 0 to +1; loses one electron.
- Fe: from +3 to 0; gains three electrons.
Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.
6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe
- There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
- There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Method Two: Atoms conserve.
Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume <em>one</em> as its coefficient.
? Na + <em>1</em> Fe₂O₃ → ? Na₂O + ? Fe
There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.
? Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.
<em>6</em> Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Answer:
The answer is Green
Explanation:
red wavelength: 650 frequency: 4.62
green wavenlength: 550 frequency: 5.45
Please tell me if it is correct.
This is what I found!!! I hope this helps!!!!