23. The integral heats of solution at 18°C for the various solid modifications of calcium chloride in the indicated quantities of water are given by CaCl2(s) + 400 H;0(1) = CaCl2(400 H;0) СаCla - 2 H:0(s) + 398 H:0(1) — СаCli:(400 H:0) СаCli 4 H,0(8) + 396 H,0(1) — СаCl:(400 H,0) CaCl2 • 6 H,0(s) + 394 H;0(1) = CaCla(400 H;O) AH, = -17,990 cal AH: = -10,030 cal AH3 = -1830 cal ΔΗ %3D +4560 cal From these data find the heats of the following hydration reactions: (a) CaCl2(s) + 2 H¿0(1) = CaCl2 · 2 H,0(s) (b) СаCl; - 2 H0(8) + 2 H:0(1) (c) CaCl, · 4 H20(s) + 2 H¿0(1) = CaCl, · 6 H,0(s) (d) CaCl (s) + 6 H;0(1) CaClą · 4 H20(s) CaCl2 · 6 H,0(s)
23. The integral heats of solution at 18°C for the various solid modifications of calcium chloride in the indicated quantities of water are given by CaCl2(s) + 400 H;0(1) = CaCl2(400 H;0) СаCla - 2 H:0(s) + 398 H:0(1) — СаCli:(400 H:0) СаCli 4 H,0(8) + 396 H,0(1) — СаCl:(400 H,0) CaCl2 • 6 H,0(s) + 394 H;0(1) = CaCla(400 H;O) AH, = -17,990 cal AH: = -10,030 cal AH3 = -1830 cal ΔΗ %3D +4560 cal From these data find the heats of the following hydration reactions: (a) CaCl2(s) + 2 H¿0(1) = CaCl2 · 2 H,0(s) (b) СаCl; - 2 H0(8) + 2 H:0(1) (c) CaCl, · 4 H20(s) + 2 H¿0(1) = CaCl, · 6 H,0(s) (d) CaCl (s) + 6 H;0(1) CaClą · 4 H20(s) CaCl2 · 6 H,0(s)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:23. The integral heats of solution at 18°C for the various solid modifications
of calcium chloride in the indicated quantities of water are given by
CaCl2(s) + 400 H;0(1) = CaCl2(400 H;0)
СаCla - 2 H:0(s) + 398 H:0(1) — СаCli:(400 H:0)
СаCli 4 H,0(8) + 396 H,0(1) — СаCl:(400 H,0)
CaCl2 • 6 H,0(s) + 394 H;0(1) = CaCla(400 H;O)
AH, = -17,990 cal
AH: = -10,030 cal
AH3 = -1830 cal
ΔΗ
%3D
+4560 cal
From these data find the heats of the following hydration reactions:
(a) CaCl2(s) + 2 H¿0(1) = CaCl2 · 2 H,0(s)
(b) СаCl; - 2 H0(8) + 2 H:0(1)
(c) CaCl, · 4 H20(s) + 2 H¿0(1) = CaCl, · 6 H,0(s)
(d) CaCl (s) + 6 H;0(1)
CaClą · 4 H20(s)
CaCl2 · 6 H,0(s)
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