Предмет: Химия, автор: Lakhtinen

Помогите, тут легко. Всего лишь нужно дополнить уравнения!

Восстановите левые части уравнений химических реакций по известным правым:

1)... + ... =CH3COONa+H2O

2) ... + ... = (CH3COO)2Ca+H2

3)... + ... = (CH3COO)2Mg+H2O

4)... + ... =HCOOC2H5+H2O

И еще два маленьких задания (просьба выслать лист, где НАПИСАНО решение): номер 16 и номер 17 :З

Приложения:

Ответы

Автор ответа: Аноним

0

CH3-COOH+NaOH------>CH3-COONa+H2O;
2CH3-COOH+Ca------>(CH3-COO)2Ca+H2;
2CH3-COOH+Mg(OH)2----->(CH3-COO)Mg+2H2O;
HCOOH+C2H5OH----(t, к.H2SO4)----->HCOOC2H5+H2O;
Гомологи и изомеры выписаны попарно, последних изомеров три.

Приложения:

Автор ответа: Lakhtinen

0

MEEEEOW! СПАСИБО ОГРООООМНОЕ!!!!!!!!! :ЗЗЗ

Автор ответа: Аноним

0

Пожалуйста :3 Надеюсь, всё понятно

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Long, long ago there lived 1 rich king. His name was 2 Midas. He was 3 greedy man and liked_4_gold very much. One day he asked gods of_5_Greece to give him 6 more gold. 7_gods decided to punish him and said: "Very well; in 8 morning anything that you touch with your hand will become 9_gold." 10 Midas was very happy when he heard this. "I shall be 11 richest man in 12 world," he thought.

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