Answer:
Relation between , molality and temperature is as follows.
T =
It is also known as depression between freezing point where, i is the Van't Hoff factor.
Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.
i for = 3
i for glucose = 1
i for NaCl = 2
Depression in freezing point will have a negative sign. Therefore, d
depression in freezing point for the given species is as follows.
=
=
=
Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.
Glucose < NaCl <
Explanation:
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
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1. <span>Substances in a </span>mixture<span> can be separated using different </span>methods, for example <span>distillation (separation of two liquids).
2. </span>Mass<span> is the amount of matter an object contains.
</span>Volume is t<span>he amount of space occupied by a three-dimensional object.
3. d(wallet) = 15g</span>÷5cm³ = 3g/cm³.
4. Weight of an astronaut also increases. W=m·g.
<span>iron gets hotter before copper.</span>
