KE=6,250Joules
You plug in your given numbers and remember to square 5 because that’s the speed
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.
Hey there! :D
This is a true statement. Gamma radiation comes from electromagnetic energy from radioactive decay. This decay has the shortest electromagnetic wave lengths and therefore has the highest photon energy. It is extremely dangerous. Radiation in general is something to be cautious of!
I hope this helps!
~kaikers
The reaction is:
<span>4Li(s) + O2 (g) = 2Li+ + 2O-2(s).
The oxidizing agent is the one that is being reduced which is oxygen where the charge changed from neutral to -2 while the reducing agent is the on being oxidized which is lithium where the charge change from neutral to +1.</span>