Answer: Compounds that contain a carbon-carbon double bond are known as _____alkenes_______ . Alkenes are electron rich; therefore, simple alkenes do not react with _____nuceophiles_______ or bases, but will react with _____electrophiles_______ or acids. In the IUPAC system, an alkene is identified by the suffix _____-ene_______ . Addition reactions in which two parts of a reagent are added to the same side of a double bond are known as _____syn_______ additions. Addition reactions in which two parts of a reagent are added from opposite sides of a double bond are called _____anti_______ additions. Halogenation and halohydrin formation occur with _____anti_______ addition. Hydroboration occurs with _____syn_______ addition.
Explanation:
The right terms have been filled into the statement.
Electrophiles are election rich so they react with nucleophiles. And vice versa.
Addition reactions in which two parts of a reagent are added to the same side of a double bond are known as _syn reaction and that involving opposite is anti reaction.
The Halohydrin formation reaction involves breaking a pi bond and creating a halohydrin in its place. Halo = halogen and Hydrin = OH. This reaction takes place in water and yields an anti-addition reaction which follows Markovnikov's rule.
Answer:
<h3>0.2498mol
</h3>
Explanation:
7gN2 x 
=0.2498mol N2
Nitrogen gas has the formula so therefore that means you would have to multiply the mass in the molar by 2. To solve for the number of moles you need to cancel out the grams, you do this by using the molar mass of nitrgoen gas. You get the value on in the denominator from the periodic table (atomic mass of element). The grams will cancel out, leaving you with the number of moles when you divide 7/2(14.01).
The formula to be used for this problem is as follows:
E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J
0.696×10⁻¹⁸ = (6.62607004×10⁻³⁴ m²·kg/s)(3×10⁸ m/s)/λ
Solving for λ,
λ = 2.656×10⁻⁷ m or <em>0.022656 nm</em>
