C. liquid to solid i hope this helps
No more solute will dissolve at that temperature, the temperature would have to be increased in order for more solute to dissolve.
Over the ocean, the temperature rises much slower, because the water evaporates causing the hot molecules to go into the atmosphere, and the overall temperature of the water doesn't increase much, this causes the area without the water to be much hotter.
Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ
Explanation:
Using this formular, q = [mCpΔT] and = [nΔHfusion]
The energy that is needed in the different physical changes is thus:
The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:
q = [mCpΔT]
q = 52.0 x 2.09 x 10
q = 1.09 kJ
While from 0°C to 100°C is calculated as:
q = [mCpΔT]
q = 52.0 x 4.18 x 100
q = 21.74 kJ
And for fusion at 0°C is called Heat of fusion and would be given as:
q = n ΔHfusion
q = 52.0 / 18.02 x 6.02
q = 17.38 kJ
And that required for vaporization at 100°C is called Heat of vaporization and it's given as:
q = n ΔHvaporization
q = 52.0 / 18.02 x 40.7
q = 117.45 kJ
Add up all the energy gives 157.8 kJ
The density is 3.144 g / cm^3.
<u>Explanation</u>:
If effective number of atom in NaCl type structure, z = 4
a = 705.2 pm ⇒ In centimeter = 705.2 
10^-10
Na = 6.023 10^23
density = (molecular weight) (z) / (Na) (a^3)
where molecular weight of KI is 166 g,
Z represents the atomic number
density = (molecular weight) (z) / (Na) (a^3)
= (166 4) / (6.023 10^23) (705.2 10^-10)
density = 3.144 g / cm^3.
