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jenyasd209 [6]
3 years ago
8

What is the answer to question 7, NH4C2H3O2

Chemistry
2 answers:
Veronika [31]3 years ago
7 0

Answer:

Ammonium acetate was not sure at first but it was Ammonium acetate sorry i was confused at first

Hope this helps :D

Sliva [168]3 years ago
4 0
The answer is Ammonium acetate
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6. Cuando se oxidan en el aire 12,120 g de vapor de Zinc se obtienen 15,084 g del óxido. ¿Cuál es la fórmula empírica del óxido?
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Answer:

 12120 g  +    O2        =    15084 g

m Zn = 12.120 Kg

m óxido = 15.084 Kg

1. calcular la masa de cinc en gramos

g = 12,120 Kg x 1000 = 12120 g de cinc

g = 15.084 Kg x 100 = 15084 g de oxígeno

2.  calcular gramos de Oxigeno

g O = 15084 g - 12120 g = 2964 g O2

3. calcular % de Zn y O

%m/m ( m soluto / m solc.) x 100

%m/m (Zn) =  ( 1210 g / 15084 g ) x 100

% m/m (Zn) = 80.35 % = 80.35 g

%m/m (O) =  ( 2964 g / 15084 g ) x 100

% m/m (Zn) = 19.65 %  = 19.65 g

4. Calcular moles de cada elemento

Zn: 80.35 g / 65.38 g/mol = 1.228 mol

O: 19.65 g / 16 g/mol = 1.228 mol

5. dividir entre el menor de los elementos

Zn: 1.228 mol / 1.228 mol = 1

O: 1.228 mol / 1.228 mol = 1

6. Fórmula empírica: ZnO

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