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Tatiana [17]
2 years ago
10

You start with 1 L of CO2 at standard temperature and pressure in a closed container. If you raise the temperature of the gas, t

he pressure will
Increase
Remain the same
Decrease
Chemistry
1 answer:
pychu [463]2 years ago
8 0

Answer:

Increase

Explanation:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

If the initial temperature and pressure is standard,

Pressure = 1 atm

Temperature = 273.15 K

then we increase the temperature to 400.0 K, The pressure will be,

1 atm / 273.15 K = P₂/400.0K

P₂ = 1 atm × 400.0 K / 273.15 K

P₂ = 400.0 atm. K /273.15 K

P₂ = 1.46 atm

Pressure is also increase from 1 atm to 1.46 atm.

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4 0
2 years ago
A reaction has ∆H = −356 kJ and ∆S = −36 J/K. Calculate ∆G (kJ) at 25°C.
Cloud [144]

Answer: -345.2 KJ

Explanation: As we know that ,dG=dH-TdS

T=25+273=298 K

dG= -356 x1000-298(-36)= -356000+10728

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= -345.2 KJ

5 0
3 years ago
Please help i'll mark the brainlest
asambeis [7]

Answer:

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Explanation:

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6 0
3 years ago
Read 2 more answers
At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con
djyliett [7]
0.114 mol/l  
The equilibrium equation will be: 
Kc = ([Br2][Cl2])/[BrCl]^2  
The square factor for BrCl is due to the 2 coefficient on that side of the equation.  
Now solve for BrCl, substitute the known values and calculate. 
Kc = ([Br2][Cl2])/[BrCl]^2 
[BrCl]^2 * Kc = ([Br2][Cl2]) 
[BrCl]^2 = ([Br2][Cl2])/Kc 
[BrCl] = sqrt(([Br2][Cl2])/Kc)  
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142) 
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142) 
[BrCl] = sqrt(0.013021127 mol^2/l^2) 
[BrCl] = 0.114110152 mol/l  
Rounding to 3 significant figures gives 0.114 mol/l
4 0
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