Hydrogen cyanide is prepared commercially by the reaction of methane, ch4(g), ammonia, nh3(g), and oxygen, o2(g), at high temper
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Answer:
The enthlapy of solution is -55.23 kJ/mol.
Explanation:
Mass of water = m
Density of water = 1 g/mL
Volume of water = 50.0 mL
m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g
Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C
Heat capacity of water,c =4.186 J/g°C
Heat gained by the water when an unknown compound is dissolved be Q
Q= mcΔT
heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.
Q'=-Q
Q'= -983.71 J =-0.98371 kJ
Moles of unknown compound =
The enthlapy of solution :
The enthlapy of solution is -55.23 kJ/mol.
Explanation:
The given data is as follows.
m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)
Initial temp. of block 1, = (100 + 273) K = 373 K
Initial temp. of block 2, = (0 + 273) K = 273 K
So, heat released by block 1 = heat gained by block 2
Convert temperature into kelvin as (50 + 273) K = 323 K.
Also, we know that the relation between enthalpy and temperature change is as follows.
=
= 1243550 J
or, = 1243.5 kJ
Now, calculate entropy change for block 1 as follows.
=
=
= -554.12 J/K
Now, entropy change for block 2 is as follows.
=
=
= 647.49 J/K
Hence, total entropy will be sum of entropy change of both the blocks.
= -554.12 J/K + 647.49 J/K
= 93.37 J/K
Thus, we can conclude that for the given reaction is 1243.5 kJ and
is 93.37 J/K.