Lead fluoride hope this helps
HCI is the limiting reactant, and 4.3 mol AICI3 can be formed.
Answer:
(a) sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d) sp³ sp²
H₃<u>C</u> - <u>C</u>H=O
Explanation:
Alkanes or the carbons with all the single bonds are sp³ hybridized.
Alkenes or the carbons with double bond(s) are sp² hybridized.
Alkynes or the carbons with triple bond are sp hybridized.
Considering:
(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.
Hence,
sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.
Hence,
sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is bonded by double bond to oxygen.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H=O
Answer:
c sounds like the most reasonable answer rember to always divide the pressure value by 101
Their identified by the number of protons they have in the nucleus which = their electrons.
