Yes just like with sugar and water temperature affects the process and warmer water causes faster dissolution.
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:
5 atoms of hydrogen will be found in the products
Explanation:
5 atoms of hydrogen will be found in the products becuase the law of conservation of mass states that mass is neither created nor destroyed in chemical reactions.
Answer:
Free energy change is -7.500248 kJ/mol
Explanation:
If we have the enthalpy change and entropy change, we can find the free energy change using the equation,
ΔG = ΔH - TΔS
Temperature in K = 25°C + 273 = 298 K
ΔH for NaI = -7.50 kJ/mol
ΔS for NaI = 74 J/K mol = 0.074 kJ/K mol
Plugin the above values in the equation, we will get,
ΔG = -7.5 kJ/mol - 0.074 kJ/K mol / 298 K
= -7.5 kJ/mol - 0.000248 kJ/mol
= -7.500248 kJ/mol