<u>Given:</u>
Mass of Ag = 1.67 g
Mass of Cl = 2.21 g
Heat evolved = 1.96 kJ
<u>To determine:</u>
The enthalpy of formation of AgCl(s)
<u>Explanation:</u>
The reaction is:
2Ag(s) + Cl2(g) → 2AgCl(s)
Calculate the moles of Ag and Cl from the given masses
Atomic mass of Ag = 108 g/mol
# moles of Ag = 1.67/108 = 0.0155 moles
Atomic mass of Cl = 35 g/mol
# moles of Cl = 2.21/35 = 0.0631 moles
Since moles of Ag << moles of Cl, silver is the limiting reagent.
Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles
Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol
Ans: Formation enthalpy = 126.5 kJ/mol
Answer chocies .......................
When the substance is creating gases. Sometimes when it’s bubbling up
Answer:
false
Explanation:
Elements cannot be broken down into a simpler substance.
Molecular mass of Carbon in compound = 29.784% * 241.94 g/mol
= 72.05 g/mol
Number of moles = 72.05 / 12 = 6 [ Molar mass of Carbon = 12 ]
Now, Molecular mass of Hydrogen in compound = 4.166% * 241.94 g/mol
= 10.07 g/mol
Number of moles = 10.07 / 1 = 10 [ Molar mass of Hydrogen = 1 ]
Now, Molecular mass of Bromine in compound = 66.05% * 241.94 g/mol
= 159.80
Number of moles = 159.80 / 79.90 = 2
Empirical Formula = C₆H₁₀Br₂
Hope this helps!