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:
Please, see attached two figures:
- The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.
- The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.
Explanation:
The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>
From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.
Assuming density 1.0 g/mol for water, 10 mL of water is:
Thus, the solutibily is:

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Combustion Have a great day!