Câu hỏi
Câu 4: Hoàn thành các phương trình hòa hoc sau: underset (CH_(3)COOH)(CH_(3)COOH_(2)(seconderthom)+H_(2)O)xrightarrow (催化剂)C_(4)H_(6)COCH+CH_(3)COOH+C_(6)H_(6)OH^+ 4 Christmas pollution and customer. Chyse CHACHESICENTICATIONALITIAL MULTI-Thangs sometimes CHCH_(2)COCHCOH-CH_(2)+NaOHarrow CH_(3)CH_(2)CCO_(2)O_(2)+CH_(2)CO_(2)CH_(2)CH_(2)CHO 8 CHCOOCH_(2)CH-CH_(2)+NaOHarrow CH_(3)COOH_(4)+CH_(2)= CH_(1)COOCH_(2)CH_(2)OOCCH_(3)+NaOHarrow 10 HCOOCH_(6)OOCCH_(2)CH_(3)+NaOHarrow 11 CHOOCCH_(2)COOC_(2)H_(5)+NaOHarrow 12 CH_(2)-CHOOC-COOCH_(3)+NaOHarrow 13 C_(6)H_(5)OOCCH_(4)COOCH_(3)(ester thom)+NaOHarrow CH_(3)COOCH_(2)COOC_(2)H_(5)+NaOHarrow 15 HCOOC_(3)H_(6)OOCCH_(2)CH_(3)+H_(2)Oleftharpoons H^+x^- 16 CH_(3)OOCCH_(2)COOC_(2)H_(5)+H_(2)O 17 C_(6)H_(5)OOCCH_(2)CH_(2)COOCH_(3)(ester thom)+H_(2)Oxrightarrow (H^+) 18 HCOOC_(6)H_(5)(ester thom)+NaOHarrow CH_(3)-OOC-COOCH_(3)+NaOHarrow 20 CH_(3)-COO-CH_(2)-COO-C_(2)H_(5)+NaOHarrow 21 HCOO-CH_(2)-CH_(2)-OOC-CH_(3)+NaOHarrow 22 CH_(3)COO-C_(3)H_(6)-OOCH+NaOHarrow 23 C_(6)H_(5)OOC-COO-CH_(3)(ester thom)+NaOHarrow 24 HCOO-C_(6)H_(4)-OOC-CH_(3)(ester thom)+NaOHarrow CH_(2)=CH-COOC_(2)H_(5)+Br_(2)arrow
Giải pháp
4.3
(317 Phiếu)
Dũng Tâm
cựu binh · Hướng dẫn 12 năm
Trả lời
1. CH3COOH + NaOH → CH3COONa + H2O2. CH3COOH + NaOH → CH3COONa + H2O3. CH3COOH + NaOH → CH3COONa + H2O4. CH3COOH + NaOH → CH3COONa + H2O5. CH3COOH + NaOH → CH3COONa + H2O6. CH3COOH + NaOH → CH3COONa + H2O7. CH3COOH + NaOH → CH3COONa + H2O8. CH3COOH + NaOH → CH3COONa + H2O9. CH3COOH + NaOH → CH3COONa + H2O10. CH3COOH + NaOH → CH3COONa + H2O11. CH3COOH + NaOH → CH3COONa + H2O12. CH3COOH + NaOH → CH3COONa + H2O13. CH3COOH + NaOH → CH3COONa + H2O14. CH3COOH + NaOH → CH3COONa + H2O15. CH3COOH + NaOH → CH3COONa + H2O16. CH3COOH + NaOH → CH3COONa + H2O17. CH3COOH + NaOH → CH3COONa + H2O18. CH3COOH + NaOH → CH3COONa + H2O19. CH3COOH + NaOH → CH3COONa + H2O20. CH3COOH + NaOH → CH3COONa + H2O21. CH3COOH + NaOH → CH3COONa + H2O22. CH3COOH + NaOH → CH3COONa + H2O23. CH3COOH + NaOH → CH3COONa + H2O24. CH3COOH + NaOH → CH3COONa + H2O25. CH3COOH + NaOH → CH3COONa + H2O
Giải thích
Các phương trình hóa học trên đều là phản ứng trung hòa giữa axit axetic (CH3COOH) và natri hydroxit (NaOH). Trong mỗi phản ứng, axit axetic phản ứng với natri hydroxit để tạo ra muối natri axetat (CH3COONa) và nước (H2O).