Explanation:
The given data is as follows.
= 1.5 atm,
= 20 L,
= (28 + 273) K = 301 K
= 5 atm,
= ?,
= (50 + 273) K = 323 K
Formula to calculate the volume will be as follows.
= 
Putting the given values into the above formula as follows.
=
=
= 0.64 L
Thus, we can conclude that the change in volume of the balloon will be 0.64 L.
The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.
Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
<span>Molality(m) or molal concentration is a measure
of concentration and it refers to amount of substance in a specified amount of
mass of the solvent. Used unit for molality is mol/kg which is also
sometimes denoted as 1 molal. It is equal to the moles of solute (the substance
being dissolved) divided by the kilograms of solvent (the substance used to
dissolve).</span>
Molarity(M) or molar concentration is also a
measure of concentration and represents the amount of substance per unit volume
of solution(number of moles per litre of solution. Used unit for molarity is
mol/L or M. A solution with a concentration of 1 mol/L is equivalent to 1 molar
(1 M).
Molality is preferred when
the temperature of the solution varies, because it does not depend on
temperature, (neither number of moles of solute nor mass of solvent will be affected
by changes of temperature), while molarity changes as temperature changes(volume
of solution changes as temperature changes).
Absorbance is related to the concentration of a substance using the Beer-Lambert's Law. According to this law, absorbance is linearly related to concentration. However, this is only true up to a certain concentration depending on the substance. For this case, we assume that the said law is applicable.
A = kC
Using the first conditions, ewe solve for k.
0.26 = k (0.10)
k = 2.6
A = kC
A = 2.6 (0.20) = 0.52
Therefore, the absorbance at a concentration of 0.20 M and wavelength of 500nm is 0.52.
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.