The graph above shows the heating curve of water. Use this graph to answer the f

Question

The graph above shows the heating curve of water. Use this graph to answer the following question.

How much heat is required to change 17 g of ice at -8.00oC (Celsius) to 17 g of steam at 109.0oC?

For your answer, enter the value of q.

spec. heat of H2O = 4.184 J/goC
spec. heat of ice = 2.087 J/goC
spec. heat of steam = 1.996 J/goC
Heat of Fusion = 6.02 kJ/mol
Heat of Vaporization = 40.7 kJ/mol
1 mol of water = 18.0g

Heati ng Curve of Water EvaporaonSteam 100 Condensation 50- Water Melting AYIce Freezing Joules The graph above shows the heating curve of water. Use this graph to answer the following question. How much heat is required to change 17 g of ice at -8.00°C (Celsius) to 17 g of steam at 109.0°C? For your answer, enter the value of q. spec. heat of H2O = 4.184 J/g°C spec, heat of ice = 2.087 J/g°C spec. heat of steam = 1.996 J/g°C Heat of Fusion = 6.02 kJ/mol Heat of Vaporization = 40.7 kJ/mol 1 mol of water = 18.0g q = 794. KJ

Explanation / Answer

K (s) + 1/2 Cl2 (g) KCl (s)            H =-436 kJ

K (s) K (g)              H1 = 90.1 kJ

K (g) K+ (g) + e-          H2 = 419 kJ

1/2 Cl2 (g) Cl (g)           H3 = 122 kJ

Cl (g) + e- Cl- (g)                          H4 =-349 kJ

K+ (g) + Cl- (g) ---------> KCl(s)         H5 =                lattice energy

According to Hess law

H    = H1 + H2 + H3+ H4 + H5

-436 = 90.1+419 +122-349 + H5

H5   = -718.1 Kj

Lattice energy of KCl is -718.1KJ

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