Answer:
A = 2A + 3B → 5C
Explanation:
The two molecule of A and three molecules of B will react to form the five molecules of C.
2A + 3B → 5C
Other options are incorrect because,
B = A₂ + B₃ → C₅
in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.
C = 2A + 5B → 3C
in this reaction two molecules of A and five molecules of B combine to form three molecule of C.
D = A₂ + B₃ → C₃
in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.
Answer:
Single Displacement reaction
In a displacement reaction, a more reactive element replaces a less reactive element from a compound.
Change in colour takes place with no precipitate forms.
Metals react with the salt solution of another metal.
Examples:
2KI + Cl2 → 2KCl + I2
CuSO4 + Zn → ZnSO4 + Cu
Double displacement reaction
In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.
Precipitate is formed.
Salt solutions of two different metals react with each other.
Examples:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
2KBr + BaCl2 → 2KCl + BaBr2
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The answer would be 0.55 moles! good luck!
Answer:
IV
Explanation:
The complete question is shown in the image attached.
Let us call to mind the fact that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.
Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.
Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic substitution reaction by SN1 mechanism.