When sugar is burned, water vapor and carbon dioxide are produced.
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Answer:
29Si has a natural abundance of 4.68%.
30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.
Explanation:
The atomic mass of silicon is given by:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
Where:
Si: atomic mass of silicon (28.086)
Si²⁸: relative atomic mass of 28Si (27.97693)
A₁: natural abundance of 28Si (92.23%)
Si²⁹: relative atomic mass of 29Si (28.97649)
A₂: natural abundance of 29Si
Si³⁰: relative atomic mass of 30Si
A₃: natural abundance of 30Si
We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.
We have to set up a system of three equations in three unknowns:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
A₃=0.6592×A₂
A₁+A₂+A₃=1
First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:
A₁+A₂+0.6592×A₂=1
A₁+1.6592×A₂=1
1.6592×A₂=1-A₁
A₂==0.0468
Then, we find the value of A₃:
A₃=0.6592×A₂
A₃=0.6592×0.0468=0.0309
Finally, we find the value of Si³⁰ in the first equation:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309
28.086=27.15922+Si³⁰×0.0309
28.086-27.15922=Si³⁰×0.0309
=Si³⁰
Si³⁰=29.99288
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
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