Answer:
Explanation:
Gilman reagent is also known as the Lithium organocuprates. They are used as nucleophiles in the organic synthesis.
Gilman reagent is used for C - C bond formation.
H H H
| | |
-----→ H - C -- C -- C -- C =
(1 bromo propane) | | | |
H H H H
(1 pentene)
Normally in C--C formation by Gilman reagent, the following reactions takes place :
R -- Br + R--R' + R' Cu + Li Br
In 1 pentene, i.e.
(from 1 bromo (from Gilman reagent)
propane)
So Gilman reagent is
Organo lithium compound is
The Maxwell-Boltzmann distributions is a description of the speeds of different gases at the same temperature.
<h3>What is Maxwell-Boltzmann distributions?</h3>
The Maxwell-Boltzmann distributions is a description of the speeds of different gases at the same temperature. Recall that at the same temperature, the relative speed of gas molecules is described by the Maxwell-Boltzmann distributions.
The question is incomplete hence we can not be able to show the exact relationship between the two graphs in order to arrive at a justification for the student's claim.
Learn more about Speed of gases: brainly.com/question/25713346
-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.
Explanation:
Given:
mass of Na2O2 = 25 grams
atomic mass of Na2O2 = 78 gram/mole
number of mole =
=
=0. 32 moles
The balanced equation for the reaction:
2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ
It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.
similarly 0.3 moles of Na2O2 on reaction would give:
=
x =
= -20.16 KJ
Thus, - 20.16 KJ of energy will be released.
12 atm is the new pressure of the air‑fuel mixture when A gaseous air‑fuel mixture in a sealed car engine cylinder has an initial volume of 600 ml at 1.0atm and final volume of 50 ml.
Explanation:
Data given:
The air fuel mixture is assumed to be having ideal behaviour
initial volume of gaseous air fuel mixture V1 = 600 ml
initial pressure of gaseous air fuel mixture P1= 1 atm
final volume when piston is removed, V2= 50 ml
final pressure of the gaseous air mixture, P2 = ?
Applying the Boyle's Law,
P1V1 = P2V2
rearranging the equation:
P2 =
putting the value in the equation,
P2 =
P2 = 12 atm
the pressure is increased to 12 atm when volume is reduced to 50 ml.