Answer : The heat required is, 1904 calories.
Explanation :
The process involved in this problem are :
The expression used will be:
![\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3Dm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
m = mass of ice = 17 g
= specific heat of liquid water = 
= enthalpy change for fusion = 
Now put all the given values in the above expression, we get:
![\Delta H=17g\times 80.0cal/g+[17g\times 1cal/g^oC\times (32.0-0)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D17g%5Ctimes%2080.0cal%2Fg%2B%5B17g%5Ctimes%201cal%2Fg%5EoC%5Ctimes%20%2832.0-0%29%5EoC%5D)
Therefore, the heat required is, 1904 calories.
F. None of the above [Cl^(-) is oxidized]
<em>No Cl atoms are available</em> to be oxidized, only Cl^(-)ions
2Cl^(-) → Cl_2 + 2e^(-)
The substance that <em>loses electrons</em> is oxidized.
Remember <em>OIL</em> RIG (<em>O</em>xidation<em> I</em>s <em>L</em>oss of electrons) and
<em>LEO</em> the lion says GER (<em>L</em>oss of Electrons is <em>O</em>xidation).
For many solids dissolved in liquid water, the solubilityincreases with temperature. Theincrease in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.
