Answer:
The precipitated are option a and d.
Explanation:
2 LiI(aq) +Hg2(NO3)2(aq) → Hg2I2(s) ↓ + 2 LiNO3(aq)
Cation Hg2+ 2 in the presence of iodide, a precipitated is formed.
Zn(s) + 2AgNO3(aq) → 2 Ag(s) ↓ +Zn(NO3)2(aq)
Zinc starts to get rid, and some white particles also stick to it. Afterwards the solution becomes cloudy and a precipitate appears, which is the solid silver
Answer:
hydrogen + oxygen = water
Answer:
The answers to the question are
1. 2nd and above order order
2. 2nd order
3. 1/2 order
4. 1st order
5. 0 order
Explanation:
We have 
1. For nth order reaction half life 
∝ ![\frac{1}{[A_{0} ]^{n-1} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%20%5D%5E%7Bn-1%7D%20%7D)
Therefore for a 0 order reaction increasing concentration of the reactant there will increase
First order reaction is independent [A₀].
Second order reaction [A₀] decrease, increase.
Similarly for a third order reaction
1. 2nd order
2. 2nd order reaction
3. Order of reaction is 1/2.
4. 1st order reaction.
5. Zero order reaction.
Fe2O3 + 2Al ---> Al2O3 + 2Fe
Mole ratio Fe2O3 : Al = 1:2
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles
No. of moles of Al = 150/27 = 5.555555555 moles.
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react.
So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared.
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe
The answer is (B).
Hope this helps :).
