Reply with anything

Very · Sep 4, 2025

Absolutely Anything

_Chilling_ · Sep 5, 2025

Very said:
Absolutely Anything

ye Correct ABSOULUTE ANYTHING
INDIRECTLY referring to NOTHINGNESS
AKA THE VOID!!!!!!!!!!!!!!!!!

AgentSpy2006 · Sep 7, 2025

I love math lol

#include<stdio.h>
int main ()
{
printf("Hey guys how are you all ")
return 0;
}
- Hey guys how are you all

xViiRuS · Sep 7, 2025

Anything

amandbren · Oct 2, 2025

AgentSpy2006 said:
I love math lol

#include<stdio.h>
int main ()
{
printf("Hey guys how are you all ")
return 0;
}
- Hey guys how are you all

Thanks for information! love you

Josuee_Gordon · Oct 3, 2025

Anything

Josuee_Gordon · Oct 3, 2025

AgentSpy2006 said:
I love math lol

#include<stdio.h>
int main ()
{
printf("Hey guys how are you all ")
return 0;
}
- Hey guys how are you all

That's Python code

_Chilling_ · Oct 7, 2025

Quick Question.. :
If 𝑦(𝑥) is the solution of the differential equation 𝑥𝑑𝑦 − (𝑦2 − 4𝑦)𝑑𝑥 = 0 for 𝑥 > 0, 𝑦(1) = 2, and the slope of the curve 𝑦 = 𝑦(𝑥) is never zero, then the value of 10 𝑦(√2 ) is __

Reply as fast as u can

the fastest gets free 'anythingness'

Meth0vas · Oct 7, 2025

it is √8²+(25-25)-(10-10)

I hope use of AI wasnt disallowed

ProDeaths__ · Oct 9, 2025

@Arrly
Kind request
Don't eat stray cats

Very · Oct 9, 2025

Anyone tried BO7?

Space242 · Oct 17, 2025

what should i do here

TsT_AnkushGaming · Oct 17, 2025

Hello.
Decided to come here because I'm bored.

_Chilling_ · 2026-01-13T19:04:42+0200

Ahh can someone tell me the answer this question ?
I am stucked guys plz help

A sealed rigid container of volume V is divided into two equal compartments by a removable, impermeable partition.

Left compartment: contains 1.00 mol of solid ammonium carbamate, NH₂COONH₄(s).
Right compartment: is evacuated.

The system is maintained at temperature TTT. Ammonium carbamate undergoes the equilibrium:

NH2COONH4(s)⇌2NH3(g)+CO2(g)\text{NH}_2\text{COONH}_4(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g)NH2COONH4(s)⇌2NH3(g)+CO2(g)
At this temperature, the equilibrium constant in terms of pressure is:

Kp=0.50K_p = 0.50Kp=0.50
Assume ideal gas behavior and that the solid phase is always present unless stated otherwise.

Part A – Initial Equilibrium

After equilibrium is established in the left compartment with the partition in place, calculate:
- (a) the partial pressures of NH₃ and CO₂
- (b) the total pressure in the left compartment

Part B – Partition Removal

The partition is now suddenly removed, allowing the gases to expand freely into the entire container of volume VVV, while temperature remains constant.

Immediately after removal (before re-equilibration):
- (a) what happens to the partial pressures of NH₃ and CO₂?
- (b) what is the reaction quotient QpQ_pQp relative to KpK_pKp?
After the system re-establishes equilibrium:
- (a) does more solid decompose, or does some gas recombine?
- (b) calculate the new equilibrium total pressure.

Part C – Thermal Perturbation

The container is now heated slowly to a higher temperature T2T_2T2, at which:

Kp(T2)=4.00K_p(T_2) = 4.00Kp(T2)=4.00

Qualitatively predict the shift in equilibrium.
Quantitatively determine the new total pressure at equilibrium, assuming excess solid remains.

Part D – Limiting Case & Conceptual Insight

Suppose instead that only 0.10 mol of ammonium carbamate was initially present.
- (a) Determine whether all the solid decomposes at temperature TTT.
- (b) If complete decomposition occurs, calculate the final total pressure and comment on whether KpK_pKp still controls the system.

Part E – Deep Olympiad Concept

Without calculation, explain why equilibria involving solids often lead to “pressure-buffering” behavior, and relate this to real industrial systems.

smokyowl · 2026-01-13T19:34:21+0200

erm .. anything?

Meth0vas · 2026-01-13T22:36:56+0200

_Chilling_ said:
NH2COONH4(s)⇌2NH3(g)+CO2(g)\text{NH}_2\text{COONH}_4(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g)NH2COONH4(s)⇌2NH3(g)+CO2(g)

I also got stuck here...
Some flex knowledge tho 1mol/L is 6.022^23 particles

Your_Death_450 · 2026-01-14T12:47:01+0200

_Chilling_ said:
Ahh can someone tell me the answer this question ?
I am stucked guys plz help

A sealed rigid container of volume V is divided into two equal compartments by a removable, impermeable partition.

Left compartment: contains 1.00 mol of solid ammonium carbamate, NH₂COONH₄(s).

Right compartment: is evacuated.

The system is maintained at temperature TTT. Ammonium carbamate undergoes the equilibrium:

NH2COONH4(s)⇌2NH3(g)+CO2(g)\text{NH}_2\text{COONH}_4(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g)NH2COONH4(s)⇌2NH3(g)+CO2(g)
At this temperature, the equilibrium constant in terms of pressure is:

Kp=0.50K_p = 0.50Kp=0.50
Assume ideal gas behavior and that the solid phase is always present unless stated otherwise.

Part A – Initial Equilibrium

After equilibrium is established in the left compartment with the partition in place, calculate:

(a) the partial pressures of NH₃ and CO₂

(b) the total pressure in the left compartment

Part B – Partition Removal
The partition is now suddenly removed, allowing the gases to expand freely into the entire container of volume VVV, while temperature remains constant.

Immediately after removal (before re-equilibration):

(a) what happens to the partial pressures of NH₃ and CO₂?

(b) what is the reaction quotient QpQ_pQp relative to KpK_pKp?

After the system re-establishes equilibrium:

(a) does more solid decompose, or does some gas recombine?

(b) calculate the new equilibrium total pressure.

_Chilling_ said:
Part C – Thermal Perturbation
The container is now heated slowly to a higher temperature T2T_2T2, at which:

Kp(T2)=4.00K_p(T_2) = 4.00Kp(T2)=4.00

Qualitatively predict the shift in equilibrium.

Quantitatively determine the new total pressure at equilibrium, assuming excess solid remains.

Part D – Limiting Case & Conceptual Insight

Suppose instead that only 0.10 molof ammonium carbamate was initially present.

(a) Determine whether all the solid decomposes at temperature TTT.

(b) If complete decomposition occurs, calculate the final total pressure and comment on whether KpK_pKp still controls the system.

Part E – Deep Olympiad Concept

Without calculation, explain why equilibria involving solids often lead to “pressure-buffering” behavior, and relate this to real industrial systems.

Part a
Kp=Partial pressure of (Nh3)²×Partial pressure of(CO2)
P(Nh3)=mole fraction(Nh3)×P(total)=2/3×P(total)
P(CO2)=1/3×P(Total)
Kp=4/27P(total)

Part B
Since temperature is constant it's isothermal process meaning PV=Constant
Since V increased to twice P total is reduced and partial pressure of Nh3 and CO2 also reduced and since temperature is same Kp won't change (equilibrium constant depends only on temperature) so to increase the pressure back to what it was before. Therefore Qp<Kp as reaction has to go forward more of the solid will decompose.

Part C
Since temp is increased and Kp(t2)>Kp(t1) reaction will go forward leading to more decomposition
Kp(t2)=4/27P(Total) to find new equilibrium pressure

Part D a) ig more information should be there as after finding partial pressure in part A to find no. of moles We can use PV=nRT and there is no V and T given so I can't find there could be way do it but I am not able to find it

b) if complete decomposition occurs then Kp won't control the total pressure as it already reached the maximum possible and there is no more solid to decompose

Part E
I never heard these terms
Disclaimer: I am just learning these things I might get this wrong xd
I am too lazy to do calculation put the values yourself xd

_Chilling_ · 2026-01-14T13:27:50+0200

Your_Death_450 said:
Part a
Kp=Partial pressure of (Nh3)²×Partial pressure of(CO2)
P(Nh3)=mole fraction(Nh3)×P(total)=2/3×P(total)
P(CO2)=1/3×P(Total)
Kp=4/27P(total)

Part B
Since temperature is constant it's isothermal process meaning PV=Constant
Since V increased to twice P total is reduced and partial pressure of Nh3 and CO2 also reduced and since temperature is same Kp won't change (equilibrium constant depends only on temperature) so to increase the pressure back to what it was before. Therefore Qp<Kp as reaction has to go forward more of the solid will decompose.

Part C
Since temp is increased and Kp(t2)>Kp(t1) reaction will go forward leading to more decomposition
Kp(t2)=4/27P(Total) to find new equilibrium pressure

Part D a) ig more information should be there as after finding partial pressure in part A to find no. of moles We can use PV=nRT and there is no V and T given so I can't find there could be way do it but I am not able to find it

b) if complete decomposition occurs then Kp won't control the total pressure as it already reached the maximum possible and there is no more solid to decompose

Part E
I never heard these terms
Disclaimer: I am just learning these things I might get this wrong xd
I am too lazy to do calculation put the values yourself xd

damn thnx for the information ... Don't let this spirit die !

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Part A – Initial Equilibrium​

Part B – Partition Removal​

Part C – Thermal Perturbation​

Part D – Limiting Case & Conceptual Insight​

Part E – Deep Olympiad Concept​

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Part D – Limiting Case & Conceptual Insight​

Part E – Deep Olympiad Concept​

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Part A – Initial Equilibrium

Part B – Partition Removal

Part C – Thermal Perturbation

Part D – Limiting Case & Conceptual Insight

Part E – Deep Olympiad Concept

Part A – Initial Equilibrium

Part B – Partition Removal

Part D – Limiting Case & Conceptual Insight

Part E – Deep Olympiad Concept