<h3>1</h3>
Species shown in bold are precipitates.
- Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
- Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃
- Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
- Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
- Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃
- Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
- Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃
- Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
- Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃
<h3>2</h3>
A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.
Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.
<h3>3</h3>
Compare the first and last row:
Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.
As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.
<h3>4</h3>
Compare the second and third row:
Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.
Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,
The expression for rate of reaction for the reactant :
![\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DNH_3%3D-%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DO_2%3D-%5Cfrac%7B1%7D%7B7%7D%5Ctimes%20%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The expression for rate of reaction for the product :
![\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DNO_2%3D%2B%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DH_2O%3D%2B%5Cfrac%7B1%7D%7B6%7D%5Ctimes%20%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
From this we conclude that, all the options are correct.
Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
