Answer: The enthalpy change is 34.3 kJ
Explanation:
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of water = 72.0 g
= specific heat of ice = 
= specific heat of liquid water = 
n = number of moles of water = 
= enthalpy change for fusion = 6010 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B72.0g%5Ctimes%202.09J%2Fg%5E0C%5Ctimes%20%280-%28-18%29%5E0C%5D%2B4.00mole%5Ctimes%206010J%2Fmole%2B%5B72.0g%5Ctimes%204.184J%2Fg%5E%29C%5Ctimes%20%2825-0%29%5E0C%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change is 34.3 kJ
Answer:
588.2 mL
Explanation:
- FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)
First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):
- moles = molarity * volume
- 187 mL * 0.692 M = 129.404 mmol Fe⁺²
Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:
- 129.404 mmol Fe⁺² *
= 258.808 mmol KOH
Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:
- volume = moles / molarity
- 258.808 mmol KOH / 0.440 M = 588.2 mL
N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol
n = 2.47 / 197
n = 0.01253 moles.
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