Solution

when carbonium is formed by isopropyl group it is more stable than that formed by n propyl group thus there reactivity will be more.

SOLUTION

2O₃ → 3O₂
2mole O₃ ≡ 3mole O₂=3×4eq.O₂
≡12 eq.O2
∴1 moleO3=6eq.O₂

Solution

A catalyst is a substance which when added to a reaction, finds an alternative route (hence, altering the reaction mechanism) with lower activation energy as shown in the diagram. Remember that with a catalyst, the average kinetic energy of the molecules remains the same but the required energy decreases.

Add equations
C+ O₂ → CO₂ , ΔH_f^o=- 393.0 kJ/mol and
2( S+ O₂ → SO₂ , ΔH_f^o=- 287.0 kJ/mol)
Resultant is
C+ 2S + 3O₂ → CO₂ + 2SO₂ ( -393 - 2×287 = -987 kJ )
By subtracting C + 2S → CS₂ , ΔH_f^o = + 117.0 kJ/mol
from above equation we get
CS₂ + 3O₂ → CO₂ + 2SO₂ ( -987 - 117=-1104 kJ /mol)

Solution

Superphosphate lime is produced by treating rock phosphate with sulfuric acid. It is the principal carrier of phosphate, the form of phosphorous usable by plants.

SOLUTION

Three chiral carbon ∴ 2³ = 8 isomers

Solution

Let $$1$$ $$g$$ of both $$A$$ and $$B$$ are taken initially. $${ W }_{ A }$$ and $${ W }_{ B }$$ are the amounts left after $$20$$ days.
$${ \lambda  }_{ A }=\dfrac { 2.303 }{ 20 } \log { \dfrac { 1 }{ { W }_{ A } }  } $$
$${ \lambda  }_{ B }=\dfrac { 2.303 }{ 20 } \log { \dfrac { 1 }{ { W }_{ B } }  } $$
$${ \lambda  }_{ A }-{ \lambda  }_{ B }=\dfrac { 2.303 }{ 20 } \log { \dfrac { { W }_{ B } }{ { W }_{ A } }  } $$
               $$=\dfrac { 2.303 }{ 20 } \log { 4 } =0.0693$$
$${ \lambda  }_{ B }={ \lambda  }_{ A }-0.0693=\dfrac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } -0.0693$$
               $$=\dfrac { 0.693 }{ 1 } -0.0693=0.6237$$
$${ t }_{ { 1 }/{ 2 } }B=\dfrac { 0.693 }{ 0.6237 } =1.11$$ day

SOLUTION

CrO₂Cl₂

Cr+2(O)+2(Cl)=0

Cr+2(−2)+2(1)=0

Cr=+6