Further Studies On Phenolic Hexamethylenetetramine Compounds
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88 To represent the above results we can write the equation (CH 2 ) C N 4 + 18. 55(0) = 6C0 2 = 5. 89H 2 O + 1. 89N 2 + 0. 22 HNO 3 + 1036. 9 Cal.
We know that 6(C) + 12(0) = 6C0 2 + 6 X 96. 98 Cal. (1) (Land, and Born. , 4th Ed. , p. 855) 11. 78(H) + 5. 89(0) = 5. 89H 2 O + 68. 357 X 5. 89 Cal. (2) (Land, and Born. , 4th Ed. , p. 850) 0. 22(H) -f- 0. 22(N) -f 0. 66(O) = 0. 22HNO 3 + 41. 60 Cal. (3) (Land, and Born. , 4th Ed. , p. 854) Substituting these three equations in the found equa- tion ...above, we have: 6(C) + 12(H) +4(N) = (CH 2 )eN 4 43. 18 Cal. Therefore the heat of formation of hexamethylene- tetramine = 43. 18 Cal.
(e) Heat of formation of phenol Berthelot 1 gives the heat of combustion of phenol at constant pressure and 18 C. As 736. 00 Cal. Per mole.
Required: 6(C) + 6(H) + (O) = C 6 H 6 O + x Cal. Found: C 6 H 6 O + 14(O) = 60 O 2 + 3H 2 O + 736. 00 Cal. We know that: 6(C) + 12(0) = 60 2 + 96. 98 Cal. * (1) 3H 2 + 30 = 3H 2 O + 68. 36 X 3 Cal. (2) Substituting Equations 1 and 2 in the found equa- tion we obtain 6(C) + 6(H) + O = C 6 H 6 O + 50.
What to read after Further Studies On Phenolic Hexamethylenetetramine Compounds?
You can find similar books in the "Read Also" column, or choose other free books by Mortimer Thomas Harvey to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
88 To represent the above results we can write the equation (CH 2 ) C N 4 + 18. 55(0) = 6C0 2 = 5. 89H 2 O + 1. 89N 2 + 0. 22 HNO 3 + 1036. 9 Cal.
We know that 6(C) + 12(0) = 6C0 2 + 6 X 96. 98 Cal. (1) (Land, and Born. , 4th Ed. , p. 855) 11. 78(H) + 5. 89(0) = 5. 89H 2 O + 68. 357 X 5. 89 Cal. (2) (Land, and Born. , 4th Ed. , p. 850) 0. 22(H) -f- 0. 22(N) -f 0. 66(O) = 0. 22HNO 3 + 41. 60 Cal. (3) (Land, and Born. , 4th Ed. , p. 854) Substituting these three equations in the found equa- tion ...above, we have: 6(C) + 12(H) +4(N) = (CH 2 )eN 4 43. 18 Cal. Therefore the heat of formation of hexamethylene- tetramine = 43. 18 Cal.
(e) Heat of formation of phenol Berthelot 1 gives the heat of combustion of phenol at constant pressure and 18 C. As 736. 00 Cal. Per mole.
Required: 6(C) + 6(H) + (O) = C 6 H 6 O + x Cal. Found: C 6 H 6 O + 14(O) = 60 O 2 + 3H 2 O + 736. 00 Cal. We know that: 6(C) + 12(0) = 60 2 + 96. 98 Cal. * (1) 3H 2 + 30 = 3H 2 O + 68. 36 X 3 Cal. (2) Substituting Equations 1 and 2 in the found equa- tion we obtain 6(C) + 6(H) + O = C 6 H 6 O + 50.
What to read after Further Studies On Phenolic Hexamethylenetetramine Compounds?
You can find similar books in the "Read Also" column, or choose other free books by Mortimer Thomas Harvey to read onlineMoreLess
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