C. ΔH for Reaction of HCl(aq) and
NaOH(s):
Volume of HCl: 0.055 L
Volume of water: 0.045 L
Volume total: 0.1 L
Molarity of HCl: 2.08 M
Mass of NaOH: 4 g
ΔT for reaction C: 23°C
number of moles of NaOH : 0.100 moles
molarity of the resulting sodium chloride solution : 1.00 M
ΔT : 23.0 °C
Calculate the value (calories) for the heat of reaction.
Calculate the ΔH for this reaction.
total volume of solution = 0.1L = 100ml
mass of solution = volume * density
= 100*1 = 100g
ΔT : 23.0 °C
q = mcT
= 100*1*23
= 2300cal
HCl(aq) + NaOH(aq) -------------->NaCl(aq) + H2O(l)
MOLARITY of NaCl = 0.1 moles
total volume of solution = 0.1L
no of moles of NaCl = molarity * volume in L
= 0.1*0.1 = 0.01moles
q = 2300Cal
= 2300*4.184J
= 9623.2J
H = q = 9623.2J
= 9623.2/0.01
= 962320J/mole
= 962.32KJ/mole >>>>answer
C. ΔH for Reaction of HCl(aq) and NaOH(s): Volume of HCl: 0.055 L Volume of water:...
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