Number 16.the correct answer is B l encrgy incrcases. Based on the figure, which
ID: 701553 • Letter: N
Question
Number 16.the correct answer is B
l encrgy incrcases. Based on the figure, which of the following is NOT true? A 12 , At-140°C, this substance can be liquefied by compression. This substance exists in three phases in equilibrium at-180 C and 0.1 atm 10-2 The normal melting point of this substance is approximately -180 °C. This substance melts only above 10 atm Pentanol and hexanol are mixed. Do you expect the enthalpy of solution to be? 13. 100 100 Tenuperature( CHMOH, MM 88 pentan-1-o GHpHNM 102 hexan-1-a need more information B) a large positive number a large negative numberD)close to zero 14. Which of the following is true about solvation? A) When organic liquids are used as solvents, no energy input is necessary to "separate" them because they are considered open based on their size. B) When a solid solute mixes with a liquid solvent, no "separating" is required in order for mixing to occur C) When two liquids mix, only the solute must experience some "separating" in order for mixing to occur When a gas solute mixes with water, insignificant energy input is necessary to "separate" them because they are considered "open" based on their size. (4) 15. How many atoms are contained in a face centered cubic unit cell? A) 14 B) 8 C2D4 aply change to convert 100 g of water vapor at 150 °C to ice at -30 °C under a constant pressure of I atm? for water of ice, liquid, and steam are 2.03, 4.18, and 1.84 JgK, respectively. The enthalpy of fusion for water is 6.01 kJ/mol while that for vaporization is 40.67 kimo A) .145k/ B)-316k/ (C))-104k/T)-6200 kJ l . 150 115.94 100Explanation / Answer
Ans 16
Heat released from 150 C to 100 c
Q1 = m x Cp x (T2-T1)
= 100 g x 1.84 J/gK x (-150+100) K
= - 9200 J
Phase change at 100c from steam to liquid
Q2 = 100 g x (-40.67 kJ/mol) x 1mol/18g
= - 225.944 x 1000 = - 225944.44 J
Heat released from 100 C to 0 c
Q3 = m x Cp x (T2-T1)
= 100 g x 4.18 J/gK x (-100) K
= - 41800 J
Phase change at 100c from liquid to ice
Q4 = 100 g x(- 6.01 kJ/mol) x 1mol/18g
= - 33.3888 x 1000 = - 33388.88 J
Heat released from 0 c to - 30c
Q5 = m x Cp x (T2-T1)
= 100 g x 2.03 J/gK x (-30) K
= - 6090 J
Total enthalpy change
= - (9200 + 225944.44 + 41800 + 33388.88 + 6090)
= - 316423.32 J x 1kJ/1000J
= - 316.423 kJ
Option B is the correct answer
Negative sign represents the heat released from the process
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