<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
If there is NO wind resistance, just plain math, this is the answer.
To get the answer you would divide 800 by 650.
800/650 = 1.230769231 (exactly)
That would translate to exactly <span>1 hour 13 minutes and 50 seconds </span>
B. Fluorine (F) is the right answer
The molarity and normality of 5.7 g of Ca(OH)2 in 450ml 0f solution is calculated as follows
molarity = moles/volume in liters
moles =mass/molar mass
= 5.7g/74g/mol = 0.077moles
molarity = 0.077/450 x1000= 0.17M
Normality = equivalent point x molarity
equivalent point of Ca(OH)2 is 2 since it has two Hydrogen atom
normality is therefore = 0.17 x2 = 0.34 N
