Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK
</span>
Now, energy required for melting of ICE = <span> 334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC = 4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span> 2257X 5.25 = 11849.25 J ......(3)
</span><span>
Thus, total heat energy required for entire process = (1) + (2) + (3)
= 1753.5 + 2195.18 + 11849.25
= </span><span>15797.93 J
</span><span> = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
Answer:
all the elements in the same period have the same valence electrons.
Answer: 292.54g of Ag
Explanation:
Cu + 2AgNO3 →Cu(NO3)2 + 2 Ag
mass conc. Of Ag = n x molar Mass
Mass conc. Of Ag = 2 x 108 = 216g
From the equation,
63.5g of Cu produced 216g of Ag
Therefore, 86g of Cu will produce Xg of Ag. i.e
Xg of Ag = (86 x 216)/63.5 = 292.54g
Determining on the temperature, ice could melt, water could freeze or evaporate. Just an example.
Answer:
solubility of X in water at 17.0 
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = 
= 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
