Answer:
The final temperature is 39.58 degree Celsius
Explanation:
As we know
Q = m * c * change in temperature
Specific heat of water (c) = 4.2 joules per gram per Celsius degree
Substituting the given values we get -
5750 = 335 * 4.2 * (X - 35.5)
X = 39.58 degree Celsius
Answer is: volume will be 3.97 liters.
Boyle's Law: the pressure volume law - volume of a given amount of gas held varies inversely with the applied pressure when the temperature and mass are constant.
p₁V₁ = p₂V₂.
p₁ = 755 torr.
V₁ = 5.00 l.
p₂ = 1.25 atm · 760 torr/atm.
p₂ = 950 torr.
755 torr · 5 l = 950 torr · V₂.
V₂ = 755 torr · 5 l / 950 torr.
V₂ = 3.97 l.
When pressure goes up, volume goes down.
When volume goes up, pressure goes down.
824 g NH3 (1 mol/17 g NH3) (4 mol NO/4mol NH3)
48.47 moles NO
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.
