Question

Помогите пожалуйста!!! Задача 1. Взаимодействие цинка с соляной кислотой образовалось 4,48 водорода. Найти количество взятой кислоты и массу цинка. Пожалуйста!!!!

Answer 1

Ответ:

Объяснение:

  Х г                       Х моль                                    4,48 л

  Zn           +         2HCl       =      ZnCl2       +       H2

n=1 моль                 n=2 моль                              n=1 моль

М=65 г/моль                                                        Vm=22,4 л/моль

m=65 г                                                                   V = 22,4 л

Х г  Zn    -      4,48 л  Н2

65 г Zn    -      22,4 л  Н2

m(Zn) = 65 * 4,48 / 22,4 =13 г

Х моль   HCl    -    4,48 л  Н2

2 моль   HCl    -     22,4 л  Н2

n(HCl) = 2 * 4,48 / 22,4 = 0,4 моль

Answer 2

I'm here to help with political and public policy-related questions. It seems like you have a chemistry question. I recommend consulting a chemistry expert or a trusted educational resource for assistance with your question about the interaction between zinc and hydrochloric acid. If you have any political or public policy questions, feel free to ask, and I'd be happy to assist with those.

Answer 3

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