Answer:
c
Explanation:
A B and D are all fosil fuses c is not
There are one antibonding molecular orbitals present in molecular orbital model of c.
The cyclobutadiene has a pi system comprised of four individual atomic p - orbital and thus should have a four pi molecular orbitals. The compound is the prototypical antiaromatic hydrocarbon with 4
- electrons . Its rectangular structure is the result of jahn teller reaction which disorder the molecule and lowers its symmetry , converting the triplet to a singlet ground state. It is a small annulene . The delocalisation energy of the
electrons of the cyclobutene is predicted to be zero .
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Answer:
Explanation:
The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized
The correct answer is A, Water is not used up during this process. This is because when cellular respiration occurs oxygen and glucose combine. When this takes place water is left behind when carbon is separated from glucose. Because water is being left behind it is not being used up in this process.
Answer:
Rate constant of the reaction is
.
Explanation:
A + B + C → D + E
Let the balanced reaction be ;
aA + bB + cC → dD + eE
Expression of rate law of the reaction will be written as:
![R=k[A]^a[B]^b[C]^c](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5Ea%5BB%5D%5Eb%5BC%5D%5Ec)
Rate(R) of the reaction in trail 1 ,when :
![[A]=0.30 M,[B]=0.30 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.30%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.30%20M)

...[1]
Rate(R) of the reaction in trail 2 ,when :
![[A]=0.30 M,[B]=0.30 M,[C]=0.90 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.30%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.90%20M)

...[2]
Rate(R) of the reaction in trail 3 ,when :
![[A]=0.60 M,[B]=0.30 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.60%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.30%20M)

...[3]
Rate(R) of the reaction in trail 4 ,when :
![[A]=0.60 M,[B]=0.60 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.60%20M%2C%5BB%5D%3D0.60%20M%2C%5BC%5D%3D0.30%20M)

...[4]
By [1] ÷ [2], we get value of c ;
c = 1
By [3] ÷ [4], we get value of b ;
b = 0
By [2] ÷ [3], we get value of a ;
a = 2
Rate law of reaction is :
![R=k[A]^2[B]^0[C]^1](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E2%5BB%5D%5E0%5BC%5D%5E1)
Rate constant of the reaction = k
![9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1](https://tex.z-dn.net/?f=9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%3Dk%5B0.30%20M%5D%5E2%5B0.30%20M%5D%5E0%5B0.30%20M%5D%5E1)
![k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%7D%7B%5B0.30%20M%5D%5E2%5B0.30%20M%5D%5E0%5B0.30%20M%5D%5E1%7D)
