<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:
5.355 g
Explanation:
first you have 30.6 g from ammonium nitrate ( NH4NO3 )
molecular weight for NH4NO3 is 80 g/mole
and molecular weight for nitrogen gas N2 is 14 g/ mole
make this
NH4NO3 --------------> N2
80 g/mol --------------> 14 g/mol
30.6 g ---------------> x
So X = 14 x 30.6 ÷ 80 = 5.355 g of N2
Good Luck
Answer:
Option D
Explanation:
analyzing cells is something I'm doing in science class right now!
Hope this helped, have a nice day! :)
