The heat absorbed when 1 mol of CS₂(g) forms from its elements at these conditions is 117.4 kJ.
Given,
Number of moles CS₂ = 1 mol
Temperature = 25° = 273 +25 = 298 Kelvin
Heat absorbed = 89.7 kJ
It takes 27.7 kJ to vaporize 1 mol of the liquid
Calculate the heat absorbed:
C(s) + 2S(s) → CS₂(l) ΔH = 89.7 kJ (positive since heat is absorbed)
CS₂(l) → CS₂g) ΔH = 27.7 kJ (positive since heat is absorbed)
Prior to summing, the equations should be balanced; however, because they are balanced already, there is nothing further we need to do.
C(s) + 2S(s)---> CS₂ (g)
ΔH = 89.7 + 27.7 = 117.4 kJ
Hence, the heat absorbed is 117.4 kJ.
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Answer:
0.0685 mL
Explanation:
To find the volume of the sample, divide the mass by the density.
(1.00 g)/(14.6 g/mL) = 0.0685 mL
