Electrophiles are reagents attracted to electrons.
Electrophiles tend to be electron-deficient and carry partial positive charges. They are attracted to species with lone pairs of electrons. For example, protons 
have no electrons and tend to share ones with other species, hence behaving as electrophiles in aqueous reactions. In the reaction between 
and ammonia 
, protons would be attracted to lone electron pairs on nitrogen atoms in ammonia molecules, which carry partial positive charges.
The Lewis Acid-base theory define Acids as species that accept electron pairs in a particular acid-base reaction. Electrophiles, by definition, tend to accept electrons. Lewis acids thus behaves as electrophiles in acid-base reactions. In the previous example, demonstrates acidic behavior and can be inferred as an electrophile.
<span>N2 - Element; Pure substance O2 - Element; Pure substance N2O - Compound; Pure Substance Air - Homogeneous; Solution</span>
Answer:
Explanation:
From the given information:
Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.
Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.
If the molecular weight of camphor = 152.24 g/mol
and it mass = 200 mg
The its no of moles = 200 mg/ 152.24 g/mol
= 1.3137 mmol
Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol
= 6.831 mmol
since the molar mass of NaBH4 = 37.83 g/mol
Then, using the same formula:
No of moles = mass/molar mass
mass = No of moles × molar mass
mass = 6.831 mmol × 37.83 g/mol
mass of NaBH4 used = 258.42 mg
ANSWER:
What is the measured component of the orbital magnetic dipole moment of an electron with the values
(a) ml=3
(b )
ml= −4
a) -278 x 
J/T
b) 3.71 x J/T
STEP-BY-STEP EXPLANATION:
a) ml= 3
Цorb,z = ml Цв = - (3) * (9.27e - 24) = -278 x J/T
b) ml= 3
Цorb,z = ml Цв = - (-4) * (9.27e - 24) = 3.71 x J/T
It depends, for example, it is quite important to know the Kelvin scale (i.e 0 degrees Celsius is 273 K and -273 degrees Celsius is 0 K ) when dealing gases. But I don't know other situations where you would need to know other temperature scales.
Hope this helps and also if you are using Fahrenheit 1 Fahrenheit is -17.22 degrees Celsius
