The termination step of the free-radical chlorination of methane is the most stable one among all three steps.
The free-radical substitution reaction between chlorine and methane features three major steps:
Initiation, during which chlorine molecules undergo homolytic fission to produce chlorine free radicals. Ultraviolet radiations are typically applied to supply the energy required for breaking the chlorine-chlorine single bonds. The initiation step is thus <em>endothermic</em>.
Propagation, a process in which chlorine free radicals react with methane molecules and remove a hydrogen atom from the alkane to produce hydrogen chloride and an alkyl radical e.g., 
. The carbon-containing free radical would react with chlorine molecules to produce chloromethane and yet another chlorine free radical. This process can well repeat itself to chlorinate a significant number of methane molecules.
Termination. Free radicals combine to produce molecules. For example, two chlorine free radicals would combine to produce a chlorine molecule, whereas two alkyl free radicals would combine to produce an alkane with two-carbon atoms in its backbone.
Chemical processes that increase the stability of a substance reduces its chemical potential energy. Energy conserves, thus such processes would also release energy equal to the potential energy lost in quantity. Free radicals are unstable and- as seen in the propagation step- compete readily with neutral molecules for their electrons. The propagation step keeps the number of free radicals constant and is therefore more exothermic than the initiation step. The termination step reduces the number of free radicals, increase the stability of the system by the greatest extent, and is therefore the most exothermic step among the three.
Your answer→ <em>The lighter side absorbs less of the incident light, reflecting some of the energy. Darker materials also emit radiation more readily than light-colored materials, so they cool faster.</em>
Answer:
A inhibitor structure resembles substrate structure B inhibitor binds non covalently at site other than active site.C inhibitors bind covalently and permanently at active site.
Explanation:
A Reversible competitive inhibitors structurally resembles the substrate and competes with the substrate to bind to the active site of the target enzyme.
B Reversible noncompetitive inhibitors binds no covalently at site of the target enzyme that is different from the active site.
C irreversible inhibitors interacts very tightly to the active site of an enzyme by covalent bond which cannot be overcome.
In addition to content, elements also support attributes <span>that specify the use, the behavior, and in some cases the appearance of an element.
The other options just do not fit the blank in this question, which is why the word <em>attributes </em>is the correct one to choose here.</span>
Answer:
838 torr
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁
p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁
p₁ = p₂ × V₂/V₁ × T₁/T₂
<em>Data:
</em>
p₁ = ?; V₁ = 2.42 L; T₁ = 27.0 °C
p₂ = 754 torr; V₂ = 2.37 L; T₂ = -8.8 °C
Calculations:
(a) Convert <em>temperatures to kelvins
</em>
T₁ = (27.0 + 273.15) K = 300.15 K
T₂ = (-8.8 + 273.15) K = 264.35 K
(b) Calculate the<em> pressure
</em>
p₁ = 754 torr × (2.37 L/2.42) × (300.15/264.35)
p₁ = 754 torr × 0.979 × 1.135
p₁ = 838 torr
