Answer:
2-methyl-2-pentanol is produced as a major product
Explanation:
- Double bond in 2-methyl-2-pentene is protonated first by consuming a proton from
. - A carbocation intermediate is formed after protonation of double bond. Most stable carbocation is formed as a major intermediate.
- Here a tertiary carbocation is formed as a major intermediate.
gives subsequent nucleophilic addition to the carbocation intermediate and produce 2-methyl-2-pentanol as a major product.- Reaction mechanism and major organic product has been shown below.
Internal combustion energy for TNT(Trinitro toluene) is found to be -170 kJ.
Internal energy= (- heat evolved by the system) + Work done by the system
ΔU = -Q+W
W = - PΔV
P= pressure
ΔV =difference in volume
Given,
V1 = 10 L
V2 = 90 L
∴ΔV = V2-V1 = 90-10 = 80 L
P = 1 atm
∴W = - ( 1× 80)
= -80 kJ
Heat evolved by the system = 90 kJ
∴Δ U= - 90 kJ - 80 kJ
= - 170 kJ
Thus, Internal combustion energy for TNT(Trinitro toluene) is found to be -170 kJ.
NOTE : here work done by the system is considered negative and also heat evolved by the system is taken -Ve.
Learn more about internal energy here..
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Answer : Option D) far left.
Explanation : If an unknown element displays many extremely strong metallic properties, where is it most likely to be located on the far left of the modern periodic table.
In the modern periodic table, the metallic character of the elements decreases across the periods in the periodic table from left to right. This occurs because the elemental atoms more readily accept electrons to fill their valence shell than losing them to remove the unfilled shell. Also, the metallic character increases down the groups in the periodic table. So, it is clear that the elements found on the far left are having high metallic properties in them.
Answer:
:O Fr? Bet let me go get them lol
Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
