Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble:
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
; Silver hydroxide at higher temperatures decomposes into black silver oxide:
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate:
. The latter precipitate is yellow.
Answer: 3390
Explanation:
Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for
.
![K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}] }](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%5Cfrac%7B%5BICl%5D%5E2%7D%7B%5BI_%7B2%7D%5D%5BCl_%7B2%7D%5D%20%20%7D)

Chose 3 you know this because you will need heat..
Helium used to be used in hot air balloons. Then, when the Hindenburg exploded, annexed file, they realized that hydrogen was not only lighter, but it also was less flammable.
Answer:
The answers are in the explanation
Explanation:
A. For the reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g); ΔH°=−41kJ.
As the reaction is exothermic ( ΔH°<0), you need to use low temperature to increase the equilibrium yield of hydrogen -LeChatelier's principle-.
We would use <em>low </em>temperature. For an <em>exothermic </em>reaction such as this, <em>decreasing </em>temperature increases the value of K and the amount of products at equilibrium.
B.
c. No. We cannot increase the equilibrium yield of hydrogen by controlling the pressure of this reaction.
It is possible to increase the equilibrium yield of reaction by controlling the amount of reactants added. As reactants and products are gases, the pressure of the reaction will not change the amount of reactants or products in the equilibrium.
I hope it helps!