Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
Answer:
467
Explanation:
ncl2 = 454.4x1/(71.0 g/mol) = 6.40 mols cl2
6.40 mols cl2 x 2molsHCL/1moleCL2 x 36.5g/1moleHCL = <u>467 g HCL</u>
Answer:
Your question is complex, because I think you wrote it wrong.
Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.
This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.
Simple junctions talks about compound organisms called ALKANS.
The double unions, in organic these compounds are called as ALQUENOS.
And as for the tertiary unions, the organic chemistry names them as ALQUINOS.
These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.
Explanation:
In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties
<h3><u>Full Question:</u></h3>
The following compound has been found effective in treating pain and inflammation (J. Med. Chem. 2007, 4222). Which sequence correctly ranks each carbonyl group in order of increasing reactivity toward nucleophilic addition?
A) 1 < 2 < 3
B) 2 < 3 < 1
C) 3 < 1 < 2
D) 1 < 3 < 2
<h3><u>Answer: </u></h3>
The rate of nucleophilic attack of carbonyl compounds is 2<3 <1.
Option B
<h3><u>Explanation. </u></h3>
Nucleophilic attack is explained as the attack of an electron rich radical to a carbonyl compound like aldehyde or a ketone. A nucleophile has a high electron density, so it searches for a electropositive atom where it can donate a portion of its electron density and become stable.
A carbonyl compound is a 
hybridized carbon atom with a double bonded oxygen atom in it. The oxygen atom pulls a huge portion of electron density from carbon being very electropositive.
In a ketone, there are two factors that make it less likely to undergo a nucleophilic attack than aldehyde. Firstly, the steric hindrance of two carbon groups being attached with the carbonyl carbon makes it harder for the nucleophile to approach. Secondly, the electron push by the carbon groups attached makes the carbonyl carbon a bit less electropositive than the aldehyde one. So aldehydes are more reactive towards a nucleophilic addition reaction.
