반응식
[OH-].[Na+:2].[C:3]([OH:14])(=[O:13])[C:4]1[CH:12]=[CH:11][C:7]([C:8]([OH:10])=[O:9])=[CH:6][CH:5]=1>O>[C:3]([O-:14])(=[O:13])[C:4]1[CH:12]=[CH:11][C:7]([C:8]([O-:10])=[O:9])=[CH:6][CH:5]=1.[Na+:2].[Na+:2]참가 물질
실험 절차
A solution was prepared containing 383 grams of 50% sodium hydroxide in seven liters of water. To this solution was added 400 grams of terephthalic acid thereby producing a solution of sodium terephthalate. A solution of stannic chloride was prepared by adding 342 grams of stannic chloride to 0.4 liter water. The stannic chloride solution was then added to the sodium terephthalate solution at about 50° C. with continuous agitation, causing the development of a white precipitate. The precipitate was filtered, washed with water and dried. The stannic terephthalate produced by the above method was analyzed at 26.8% Sn; 42.5% C; 1.65% H. This analysis is within the experimental error of stannic terephthalate, i.e., Sn(C8H4O4)2 formula which theoretically provides 26.7% Sn, 42.8% C and 1.8% H. Thermogravimetric analysis for the stannic terephthalate indicated decomposition at 360° C. to a residue of SnO2. The infrared spectrum of this salt, obtained in a KBr showed strong carbon-oxygen stretching absorption at 1685 and 1290/cm, typical of metal carboxylates having the M+(C=O)O--structure.
후처리
- customA solution was prepared
출처 및 근거
- ORD 기록
- ord-14390cd0f4e54a918c1028a755c7cbfa
- 출처 데이터셋
- uspto-grants-1990_10
- 특허 번호
- US04963127
연결된 노트
불러오는 중…