Answer:
I think these are it
Explanation:
The amount of current is the same at every point in a series circuit.
All of the parts of a series circuit—power source, wires, and devices—are connected along the same pathway
If one device in a series circuit burns out or is disconnected, the entire circuit is broken
The electrons have multiple pathways to travel.
Each time there is damage (break) in any one of the resistors the entire circuit will not function.
Answer:
2.63*10^23
Explanation:
1 mol rhodium = 102.91
44.8g/1 mol * 1 mol/ 102.91mol * 6.022*10^23/1 mol =
2.63*10^23
?? Is that the whole question?
The balanced equation is 2
AlI
3
(
a
q
)
+
3
Cl
2
(
g
)
→
2
AlCl
3
(
a
q
)
+
3
I
2
(
g
)
.
<u>Explanation:</u>
- Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- .
Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.
- Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.
Balancing the equation, we get:
2AlI
3( aq ) + 3Cl2
( g ) → 2AlCl3
( aq )
+ 3
I
2 ( g )
-
Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.
-
Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.
-
We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
