Equation #1 - write net ionic equation for dissolution of solid NaOH in water
Equation: NaOH (s) + H2O --> Na+ (aq) + OH- (aq) + H2O (l)
Net: NaOH (s) --> Na+ (aq) + OH- (aq)
deltaH = m * c * deltaT
deltaH = 200g * 1cal/moldegcel * 0.00418kj/cal * (13degcel)
deltaH = 10.868kJ
Equation #2 - write net ionic equation for aqueous soloutions of NaOH & HCl
Equation: NaOH (aq) + HCl (aq) ---> NaCl (aq) + H2O (l)
Net: OH- (aq) + H+ (aq) --> H2O (l)
deltaH = m * c * deltaT
deltaH = 200g * 1cal/moldegcel * 0.00418kJ/cal * (10degcel)
deltaH = -8.36kJ
Equation 3: Solid NaOH and aqueous HCl
Equation: NaOH (s) + HCl (aq) ---> NaCl (aq) + H2O
Net: OH- (s) + H+ (aq) --> H2O (l)
deltaH = m * c * deltaT
is that right?
?