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:
![AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)](https://tex.z-dn.net/?f=AgCl%20%28s%29%20%2B%202%20NH_3%20%28aq%29%20%2B%20H_2O%20%28l%29%5Crightarrow%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%20%2B%20HCl%20%28aq%29)
; Silver hydroxide at higher temperatures decomposes into black silver oxide: ![2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)](https://tex.z-dn.net/?f=2%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%5Crightarrow%20Ag_2O%20%28s%29%20%2B%20H_2O%20%28l%29%20%2B%204%20NH_3%20%28g%29)
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate: 
. The latter precipitate is yellow.
Answer:
Thomson to decide that atoms were not indivisible, in reality are actually composed of smaller parts? Thomson used a scanning tunneling micrograph to capture images of atoms. Thomson used a beam of negatively charged particles. Thomson used only the experimental results of other scientists
Explanation:
Answer:
T2 = 135.1°C
Explanation:
Given data:
Mass of water = 96 g
Initial temperature = 113°C
Final temperature = ?
Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)
Specific heat capacity of aluminium = 0.897 j/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
Now we will put the values in formula.
Q = m.c. ΔT
1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C
1900 j = 86.112 j/°C × T2 - 113°C
1900 j / 86.112 j/°C = T2 - 113°C
22.1°C + 113°C = T2
T2 = 135.1°C
