The numerical value for the Keq of the N2O4 ⇔ 2NO2 is 10 M
calculation
Keq = ( concentration NO)^2/ ( concentration of N2O4)
Keq= (0.5 )^2/0.025 = 10 M
The power of NO2 is taken as 2 as value 2 is is the one in front of No2 in equilibrium equation
Answer:
doublet
Explanation:
Proton MNR is used for the determination of no. of equivalents protons in a molecule
In the molecule, single NMR signal is produced for each set of protons.
Signal splitting is called spin-spin coupling and the splitting of signals depends upon the no. of neighboring proton.
The no. of signal for a proton is equal to n+1, where n is neighboring protons.
In 1-bromo-2-methylpropane, neighboring proton for both methyl protons are one. But the chemical environment of both the methyl protons are different.
Neighboring proton for methyl protons = 1
No. of signal for methyl protons = 1+1 =2
Hence, two doublets will be generated for each set of methyl protons. protons.
Answer:
Explanation:
The ionic compound CuCl2
is a binary compound called Copper (II) Chloride.
This compound is composed of a metal cation of copper with a charge of Cu+2
and a non-metal chloride ion Cl−1. It takes two Chloride
Cl−1 to balance the Copper
Cu+2
. This makes the formula
CuCl2
The copper is the metal, it is located in the middle section of the periodic table known as transition metals. These metals may have more than one possible ionic charge. Copper can have a +1 charge
Cu+1
or a +2 charge
Cu+2
. The roman numeral in the formula name identifies the charge on the atom.
The Cu+2
ion is referred to as Cupric in a system known as Latin Naming, while the
Cu+1
ion is called Cuprous.
So this compound can be named as Copper (II) Chloride or Cupric Chloride.
Answer:30.27 g HNO3
Step by step solution:
45 g Cu x 1 mol Cu / 63.54 g Cu
= 0.1802 mol Cu
0.1802 mol Cu x 8 mol HNO3/3 mol Cu =0.4805 mol HNO3
Molar Mass HNO3
= 1.0 + 14.0 +3×16.0
= 63.0 g/mol
0.4805 moles HNO3 x 63.0 g HNO3/1 mol HNO3
= 30.27 g HNO3
Answer:
The least substituted product (anti-Markovnikov)
Explanation:
The ROOR is used in the addition reaction of HBr to an organic substance (an alkene for example).
In normal conditions (with no ROOR) the adition of the halogen will be performed in the most substituted C (following the rule of Markovnikov that says that the stability increases with the more substituted is the C).
But in presence of ROOR, the reaction takes other mechanism (free radicals), and the product in this case is the one with the Br added in the least substituted C.