Given the following electronic configurations, identify how many valence electrons each atom has.
2 answers:
Answer : Valence electrons in atom A = 1, B = 6 and C = 2.
Explanation :
Valence electrons means the outermost electrons are the ones which involved in bonding. Electrons are arranged in energy levels or shells.
Valence electrons are found in the highest energy orbitals of an element.
- The electronic configuration of (Na) is
. The highest energy orbital is 3s orbital which contains 1 electron. So,
The number of valence electrons in (Na) = A = 1
Similarly,
- The electronic configuration of (O) is
. The highest energy orbital are 2s and 2p orbitals which contains (2 + 4) = 6 electrons. So,
The number of valence electrons in (O) = B = 6
- The electronic configuration of (Fe) is
. The highest energy orbital is 4s orbital which contains 2 electrons. So,
The number of valence electrons in (Fe) = C = 2
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Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
- V₂ = 295 mL
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L