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kirza4 [7]
3 years ago
7

In a solution, the substance that changes phase is the _____, while the substance that does not change phase is the____.

Chemistry
2 answers:
iren [92.7K]3 years ago
6 0
Phase change is the solute
the one that doesn't phase change is the solvent.
A
Nezavi [6.7K]3 years ago
5 0

Answer:   solute, solvent

Explanation:  Solute is the substance which is present in smaller quantity and solvent is the substance which is present in larger quantity.

Thus when solid solute is mixed with the liquid solvent, after the dissolution of the former, it will change its state from solid to liquid  but the phase of the solvent is kept the same.

Thus , In a solution, the substance that changes phase is the <u>solute</u>, while the substance that does not change phase is the <u>solvent</u>.

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Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota
Ghella [55]

Answer:

(a) ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = -2.881 J/K; total change in entropy = 0

(b)ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = 0 ; total change in entropy = 2.881 J/K

(c) ΔS_{sys}  = 0 ; ΔS_{sur}  = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = V_{1}

Final volume = V_{2} = 2V_{1}

(a) Change in entropy of the system ΔS_{sys} = nRIn\frac{V_{2} }{V_{1} }

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

ΔS_{sys} = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding ΔS_{sur} = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, ΔS_{sys}  = 2.881 J/K

Since surrounding does not change in this process ΔS_{sur} = 0.

total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

ΔS_{sys}  = 0

Since heat energy is not transferred from the system to the surrounding

ΔS_{sur}  = 0

total change in entropy = ΔS_{sys}+ΔS_{sur} = 0

6 0
2 years ago
A substance that loses electron and itself is oxidized is called a?​
Verdich [7]

Answer:

oxidation-reduction or redox reaction

Explanation:

n

8 0
3 years ago
Read 2 more answers
Need help with the 1st one
lutik1710 [3]

If I’m right this should be the answer

7 0
2 years ago
An object has a mass of 55 grams and a density of 3.23g/ml ? what is the volume
lara [203]
Volume= mass divided by density
V= m/d
55/3.23
= 17.03
8 0
2 years ago
Which member of each pair is more soluble in diethyl ether? Why?<br> (c) MgBr₂(s) or CH₃CH₂MgBr(s)
White raven [17]

CH3CH2MgBr is more soluble in diethyl ether .

We know that polar solvent dissolve in polar solvent very perfectly . as diethyl ether is a polar solvent so it have dipole -dipole interaction .

Hence the compound with similar interaction can dissolve in diethyl  ether .

Here , MgBr2  is an ionic compound . there is ion-ion interactions occurs which is not similar to dipole -dipole interaction in diethyl ether .hence the solubility of MgBr2 in diethyl ether is less .

but in case of CH3CH2MgBr there are both polar and nonpolar end .CH3CH2 is the nonpolar end and MgBr is the polar end .

thus with the nonpolar end solute interact using depression forces and with polar end solute interact using dipole-dipole interaction . so CH3CH2MgBr is more soluble .

Learn more about polar solvent here :

brainly.com/question/3184550

#SPJ4

4 0
1 year ago
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