The complete reaction along with intermediates is given below, with each step highlighted in different color.
Step 1: In this step an acidic proton at alpha position is abstracted from lactone moiety and corresponding enolate is formed, which is resonance stabilized. Both structures are shown. In this case LDA (<span>Lithium diisopropylamide) acts as a base.
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Step 2: The enolate formed attacks on Methyl Iodide, as Iodide being greater in size is a good leaving group and alpha methyl moiety is generated.
Step 3: This step is acid catalyzed Bromination. Bromine is added at alpha position.
Step 4: This is elimination reaction and is according to <span>Hofmann's Rule. Here less substituted alkene is generated.</span>
According to
Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".
r₁ / r₂ =

---- (1)
r₁ = Rate of effusion of He
r₂ = Rate of Effusion of O₃
M₁ = Molecular Mass of He = 4 g/mol
M₂ = Molecular Mass of O₃ = 48 g/mol
Putting values in eq. 1,
r₁ / r₂ =

r₁ / r₂ =

r₁ / r₂ =
3.46
Result: Therefore, Helium will effuse
3.46 times more faster than Ozone.
To answer this question, you need to know the aluminum acetate weight and formula. Molecular weight of aluminum acetate is 204.11g/mol, so the concentration should be: (25.20/204.11 )/(0.185l) = 0.1234 mol/0.185l= 0.6673 M<span>
</span>The aluminum acetate Al(C2H3O2)3 which mean for one aluminum acetate there will be 3 acetate ion. The concentration of acetate should be: 0.6673 M= 2M
Answer:
°C = -33.39 °C
K = 239.76 K
Explanation:
- °C = ( °F -32 ) / 1.8
- K = 273.15 + °C
⇒ °C = ( -28.1 - 32 ) / 1.8
⇒ °C = - 33.39 °C
⇒ K = 273,15 + (-33.39)
⇒ K = 239.76 K
Answer:
Ag⁺ (aq) + Br⁻ (aq) ⇒ AgBr (s)
Explanation:
The reaction between silver nitrate and lithium bromide is as follows:
AgNO₃ (aq) + LiBr (aq)⇒ AgBr (s) + LiNO₃ (aq)
The aqueous compounds will dissociate into their ions, so the reaction may be rewritten as:
Ag⁺ (aq) + NO₃⁻ (aq) + Li⁺ (aq) + Br⁻ (aq) ⇒ AgBr (s) + Li⁺ (aq) + NO₃⁻ (aq)
In order to form the net ionic equation, we remove from the reaction any ions that did not undergo a chemical change (ions that are exactly the same on the reactants side and on the products side). These ions are referred to as spectator ions. The spectator ions are NO₃⁻, Li⁺, and Br⁻. The net ionic equation is then:
Ag⁺ (aq) + Br⁻ (aq) ⇒ AgBr (s)