Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.
Answer:
The pressure will increase due ot expnasion of gasses in a closed sealed tube tube .
Explanation:
If volume remains the same while the mass of a substance increases, the density of the substance will increase.
So if the volume remains the same while the mass of a substance decreases, the density of the substance will decrease, too.
Answer:
The possible structures are ketone and aldehyde.
Explanation:
Number of double bonds of the given compound is calculated using the below formula.
=Number of double bonds
= Number of carbon atoms
= Number of hydrogen atoms
= Number of nitrogen atoms
The number of double bonds in the given formula - 
The number of double bonds in the compound is one.
Therefore, probable structures is as follows.
(In attachment)
The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.
alkene compounds I and II shows signal less than 140 ppm.
Hence, the probable structures III and IV are given as follows.
The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.
Hence, the molecular formula of the compound having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.
The rate constant, k, for the decomposition reaction : k = 0.0124 / days
<h3>Further explanation</h3>
Given
The half-life of 56 days
Required
The rate constant, k
Solution
For first-order, rate law : ln[A]=−kt+ln[A]o
The half-life : the time required to reduce to half of its initial value.
The half life :
t1/2 = (ln 2) / k
k = (ln 2) / t1/2
k = 0.693 / 56 days
k = 0.0124 / days
