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NeX [460]
3 years ago
11

Describe in detail what you know about the enthalpy, entropy, and free energy changes when a sample of gas condenses to a liquid

. How does temperature affect these changes?.
Chemistry
2 answers:
ycow [4]3 years ago
8 0
The keyword here is gas condenses to a liquid, which mean we're talking about condensation process

The enthalpy energy in condensation process is negative because it releases energy
The entropy in general will also decreases .

Temperature affect this change because it will create free energy if added with this result

hope this helps
Vera_Pavlovna [14]3 years ago
6 0

When a gas condenses to liquid the process is exothermic with the release energy and the enthalpy change is negative. As liquid has more ordered structure than gas, the entropy, which is the measure of disorder decreases or ∆ S has a negative value.  

Free energy is given by the Gibb's free energy equation,  

∆G = ∆H –T∆S

For a process to be spontaneous ∆ G must be negative.

∆H is negative and ∆S is negative, so the magnitude of |T∆S| should be less than the magnitude of |∆H| in that way ∆G is always negative.

|T∆S| should be less than the magnitude of |∆H|, so to favor that T should be low. The process is spontaneous at a lower temperature than at a higher temperature.  

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A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume
lbvjy [14]
We have that every gas satisfies the fundamental gas equation, PV=nRT where P is the Pressure, V is the volume of the gas, n are the moles of the gas, R is a universal constant and T is the Temperature in Kelvin. We have that PV/T=nR and during our process, the moles of the gas do not change (no argon enters or escapes our sample). See attached.

6 0
3 years ago
A compound ‘X’ is used for drinking, has pH =7.Its acidified solution undergoes decomposition in presence of electricity to prod
umka21 [38]

Answer:

X = Water (H2O) ; Y = Hydrogen ; Z = Oxygen

Explanation:

2(H2O) -------> 2H2 + O2

6 0
3 years ago
Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.
Alborosie

<u>Given:</u>

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

<u>To determine:</u>

The Ka of HNO3

<u>Explanation:</u>

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

            HNO3   ↔    H+      +      NO3-

I            7.50               0                 0

C          -2.48          +2.48              +2.48

E            5.02            2.48              2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

6 0
3 years ago
When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for
balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-

Whereas the half reactions are:

Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O

Next, we exchange the transferred electrons:

1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow  4N^{4+}O^{2-}_2+4H_2O

Afterwards, we add them to obtain:

Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O

By adding and subtracting common terms we obtain:

Sn^0+4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O

Finally, by removing the oxidation states we have:

Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0
3 years ago
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adoni [48]

Answer: assume pathogens are present and treat the samples accordingly

Explanation:

When investigators are unable to conclusively ascertain the source of a biological sample found at a crime scene, the correct thing to do is to treat it as if pathogens are present in it and handle it according to set rules on how to handle pathogens.

This is done to ensure that if a pathogen is indeed present, it would not cause a health emergency by infecting those who come in contact with the samples at the scene.

3 0
3 years ago
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