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Determining the hybridization of each atom requires considering the number of bonds and lone pairs surrounding each atom.
1. Oxygen ([tex]\( O \)[/tex]):
- Typically, oxygen forms two bonds and has two lone pairs. This usually corresponds to [tex]\( sp^3 \)[/tex] hybridization.
2. Carbon-1 ([tex]\( C-1 \)[/tex]):
- If Carbon-1 forms three sigma bonds (single bonds), it usually adopts [tex]\( sp^2 \)[/tex] hybridization.
3. Carbon-2 ([tex]\( C-2 \)[/tex]):
- If Carbon-2 forms a triple bond (one sigma and two pi bonds) or two double bonds (two sigma and two pi bonds), it generally exhibits [tex]\( sp \)[/tex] hybridization.
4. Carbon-4 ([tex]\( C-4 \)[/tex]):
- If Carbon-4 forms three sigma bonds, it will usually have [tex]\( sp^2 \)[/tex] hybridization.
Given these guidelines:
- Oxygen ([tex]\( O \)[/tex]): [tex]\( sp^3 \)[/tex]
- Carbon-1 ([tex]\( C-1 \)[/tex]): [tex]\( sp^2 \)[/tex]
- Carbon-2 ([tex]\( C-2 \)[/tex]): [tex]\( sp \)[/tex]
- Carbon-4 ([tex]\( C-4 \)[/tex]): [tex]\( sp^2 \)[/tex]
Thus, the hybridizations are:
[tex]\( sp^3 \)[/tex] for Oxygen,
[tex]\( sp^2 \)[/tex] for [tex]\( C-1 \)[/tex],
[tex]\( sp \)[/tex] for [tex]\( C-2 \)[/tex], and
[tex]\( sp^2 \)[/tex] for [tex]\( C-4 \)[/tex].
The correct answer from the choices provided is:
[tex]\( sp^2 \quad sp^3 \quad sp^2 \quad sp^3 \)[/tex]
1. Oxygen ([tex]\( O \)[/tex]):
- Typically, oxygen forms two bonds and has two lone pairs. This usually corresponds to [tex]\( sp^3 \)[/tex] hybridization.
2. Carbon-1 ([tex]\( C-1 \)[/tex]):
- If Carbon-1 forms three sigma bonds (single bonds), it usually adopts [tex]\( sp^2 \)[/tex] hybridization.
3. Carbon-2 ([tex]\( C-2 \)[/tex]):
- If Carbon-2 forms a triple bond (one sigma and two pi bonds) or two double bonds (two sigma and two pi bonds), it generally exhibits [tex]\( sp \)[/tex] hybridization.
4. Carbon-4 ([tex]\( C-4 \)[/tex]):
- If Carbon-4 forms three sigma bonds, it will usually have [tex]\( sp^2 \)[/tex] hybridization.
Given these guidelines:
- Oxygen ([tex]\( O \)[/tex]): [tex]\( sp^3 \)[/tex]
- Carbon-1 ([tex]\( C-1 \)[/tex]): [tex]\( sp^2 \)[/tex]
- Carbon-2 ([tex]\( C-2 \)[/tex]): [tex]\( sp \)[/tex]
- Carbon-4 ([tex]\( C-4 \)[/tex]): [tex]\( sp^2 \)[/tex]
Thus, the hybridizations are:
[tex]\( sp^3 \)[/tex] for Oxygen,
[tex]\( sp^2 \)[/tex] for [tex]\( C-1 \)[/tex],
[tex]\( sp \)[/tex] for [tex]\( C-2 \)[/tex], and
[tex]\( sp^2 \)[/tex] for [tex]\( C-4 \)[/tex].
The correct answer from the choices provided is:
[tex]\( sp^2 \quad sp^3 \quad sp^2 \quad sp^3 \)[/tex]
Therefore, there are approximately \( 2.037 \) moles of aluminum in \( 54.96 \) grams of aluminum.Answer:
Explanation:
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