Answer:
180.56 kilojoules of heat energy is extracted when 1.00 kg of freon-11 evaporates.
Explanation:
The molar mass of freon-11 is 137.35 g/mol
The enthalpy of vaporization for freon-11 is 24.8 kJ/mol at its normal boiling point of 24°C. Given that,
Mass of freon-11 evaporated = 1.00 kg = 1000 g
Moles of freon-11 evaporated can be calculated as
The energy removed in the form of heat when 1.00 kg of freon-11 vaporizes is:
This procedure entails diluting the 12 molar HCl. To decrease the concentration, we must create an equation to determine how much of the 12M is needed for the 3.5M solution.
12 moles HCl 3.5 moles HCl
——————— = ———————
1 Liter of Soln ‘x’ Liters of Soln
Note that the ratio of 12 moles over 1 liter corresponds to 12 molar; thus, we maintain the original concentration of the HCl. By equating it to the 3.5 over ‘x’, we are still preserving the concentration.
After computation, we determine ‘x’ to be 0.292. This value indicates that within 0.292 liters of our 12 M HCl solution, there are 3.5 moles of HCl. Yet, we are not finished.
0.292 liters of 12 M HCl can create 1 liter of 3.5 M HCl, but the inquiry demands 1.5 liters. To achieve this, multiply 0.292 liters by 1.5, resulting in 0.4375, which denotes the quantity of 12 M HCl necessary to prepare a 1500 mL 3.5 M HCl solution.
Given the mass of 
=25.6 g
The molar mass of =390.35g/mol
Conversion of mass of to moles:
Conversion of moles to moles of sulfur:
Conversion of moles of sulfur to individual sulfur atoms using Avogadro's number:
1 mol = 
Co2 is indeed the correct answer, my friend.
The production mechanism for t-butyl alcohol proceeds as follows:
The reaction and corresponding rate law for isobutene adsorption on the Amberlyst-15 catalyst can be described by:
I + S ⇄ I * S 
where 
represents the concentration of unoccupied sites
denotes the adsorption equilibrium constant
indicates the forward rate constant
are the concentrations of isobutene and the site occupied by isobutene
The equation for the adsorption of water (W) onto the Amberlyst-15 catalyst is outlined as:
W + S ⇄ W.S 
The surface reaction's rate law can be described as:
W.S + I.S ⇄ TBA. S + Sn 
The desorption rate law for TBA from the catalyst is as follows:
TBA. S ⇄ TBA + S 
The image included below illustrates the rest of the steps.
