Answer:
260 moles of Helium
Explanation:
V = 50L
T = 20°C = (20 + 273.15)K = 293.15K
P = 125 atm
R = 0.082 L.atm / mol. K
n = ?
From ideal gas equation,
PV = nRT
P = pressure of a given gas
V = volume it occupies
n = number of moles
R = ideal gas constant
T = temperature of the gas
PV = nRT
n = PV / RT
n = (125 * 50) / (0.082 * 293.15)
n = 6250 / 24.0383
n = 260.00 moles
The child inhaled 260 moles of Helium
Explanation:
D. The total mass of the reactants equals the total mass of the
products.
Solubility of barium chloride at 30 degree Celsius is 38.2g /100 g water and solubility of barium chloride at 60 degree Celsius is 46.6 g / 100 g water.
The quantity of barium chloride that is dissolved in water at 30 degree Celsius = 38.2 * [150/100] = 57.30 g.
The quantity of barium chloride that will be dissolved in water at 60 degree Celsius = 46.6 * [150/100] = 69.90 g
The difference between these quantities is the amount of barium chloride that can be dissolved by heating the barium chloride to 60 degree Celsius.
69.90 - 57.30 = 12.60 g. Therefore, 12.60 g of barium chloride can still be dissolved in the water by heating the water to 60 degree Celsius.
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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.