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2 years ago

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In a coffee-cup calorimeter, 1.60 g of NH4NO3 was mixed with 75.0 g of water at an initial temperature of 25.008C. After dissolu

tion of the salt, the final temperature of the calorimeter contents was 23.348C.
Required:
a. Assuming the solution has a heat capacity of 4.18 J 8C21 g21, and assuming no heat loss to the calorimeter, calculate the enthalpy of solution (DHsoln) for the dissolution of NH4NO3 in units of kJ/mol.
b. If the enthalpy of hydration for NH4NO3 is 2630. kJ/mol, calculate the lattice energy of NH4NO3.

1 answer:

jekas [21]2 years ago

7 0

Answer:

Explanation:

mass of the solution m = 1.6 + 75 = 76.6 g

fall in temperature = 25 - 23.34 = 1.66°C

heat absorbed = mass x specific heat x fall in temperature

= 76.6 x 1.66 x 4.18

= 531.5 J .

= .5315 kJ .

mol weight of ammonium nitrate = 80 g

heat absorbed by 1.6 g = .5315 kJ

heat absorbed by 80 g or one mole = 26.575 kJ

enthalpy change ΔH = +26.575 kJ

b )

enthalpy of hydration = 2630 kJ / mol

lattice energy = enthalpy of hydration + enthalpy change

= 2630 + 26.575

= 2656.575 kJ .

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1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

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