<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:
the answer is b mitosis in humans
To solve this, we simply equate the change in enthalpy for
the two substances since heat gained by water is equal to heat lost of aluminum.
We know that the heat capacity of aluminum is 0.089 J/g°C and that of water is
4.184 J/g°C. Therefore:
450.2 (95.2 - T) (0.089) = 60 (T – 10) (4.184)
3,814.45456 – 40.0678 T = 251.04 T – 2,510.4
291.1078 T = 6,324.85456
<span>T = 21.7°C</span>