Answer:
a.option is the correct answer
Answer:
B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-
Explanation:
In the hydroboration of alkenes, an alkene is hydrated to form an alcohol with anti-Markovnikov orientation.
the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.
Answer:
d) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity increases because the arsenic atom loses the electron.
Explanation:
This is an example of a n-type semiconductor. The additional electron introduced to the 'grid' of silicon atoms causes an increase in the conductivity of the silicon. This additional electron is introduced as arsenic loses its extra electron.
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
