They undergo nuclear fission.
The
correct answer is A. In the combined gas law, if the volume is decreased and
the pressure is constant, then the temperature decreases.
<span>P1V1/
T1 = P2V2 / T2</span>
<span>Assume
the volume decrease by half; V2 = V1/2</span>
<span>P1V1/
T1 = P2V1 /2 T2</span>
<span>Cancelling
terms,</span>
<span>1/T1
= 1/2 T2</span>
T2
= T1/2
<span>Thus,
the temperature decreased.</span>
Answer:
The initial volume in mL is 5959.2 mL
Explanation:
As the number of moles of a gas increases, the volume also increases. Hence, number of moles and volumes are directly proportional i.e
n ∝ V
Where n is the number of moles and V is the volume
Then, n = cV
c is the proportionality constant
∴n/V = c
Hence n₁/V₁ = n₂/V₂
Where n₁ is the initial number of moles
V₁ is the initial volume
n₂ is the final number of moles
and V₂ is the final volume.
From the question,
n₁ = 0.693 moles
V₁ = ?
n₂ = 0.928 moles
V₂ = 7.98 L
Putting the values into the equation
n₁/V₁ = n₂/V₂
0.693 / V₁ = 0.928 / 7.98
Cross multiply
∴ 0.928V₁ = 0.693 × 7.98
0.928V₁ = 5.53014
V₁ = 5.53014/0.928
V₁ = 5.9592 L
To convert to mL, multiply by 1000
∴ V₁ = 5.9592 × 1000 mL
V₁ = 5959.2 mL
Hence, the initial volume in mL is 5959.2 mL
Answer:
The change in entropy of the surrounding is -146.11 J/K.
Explanation:
Enthalpy of formation of iodine gas = 
Enthalpy of formation of chlorine gas = 
Enthalpy of formation of ICl gas = 
The equation used to calculate enthalpy change is of a reaction is:
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28ICl%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28I_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Cl_2%29%7D%29%5D)
![=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol](https://tex.z-dn.net/?f=%3D%5B2%5Ctimes%2017.78%20kJ%2Fmol%5D-%5B1%5Ctimes%200%20kJ%2Fmol%2B1%5Ctimes%2062.436%20kJ%2Fmol%5D%3D-26.878%20kJ%2Fmol)
Enthaply change when 1.62 moles of iodine gas recast:

Entropy of the surrounding = 

1 kJ = 1000 J
The change in entropy of the surrounding is -146.11 J/K.