The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.
Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
Hey there!
For SN1 mechanism; the activation barrier is the C-I bond energy which is broken in the first step of the reaction.
The activation barrier is : 56 kcal/ mol = 5.6 kcal/ mole ( nearest 0.1)
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:
In order to find the molecular formula from an empirical formula you must find the ratio of their molecular masses.
We know that the molecular mass of the molecule is 70
gmol-1
. We can calculate the molar mass of
CH2
from the periodic table:
C=12.01
gmol−1
H=1.01
gmol−1
CH2 =14.03
gmol−1
Hence we can find the ratio:
14.03
70
≈
0.2
Answer:
C. Double Replacement
Explanation:
A. Wrong because SR uses a compound and element. This equation is a compound and compound.
B. Wrong because it does not have O2 in the formula. All combustion reactions must have O2.
C. Correct because it is a compound reacting with a compound.
D. Wrong because the reactants did not form a single product. ex. (x + y > xy)
E. Wrong because the equation did not start with a single compound and break down. ex. (xy > x + y)
