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Natasha_Volkova [10]
3 years ago
15

A água oxigenada é uma solução que contém a substância peróxido de hidrogênio.Ela é usada como bactericida.Quando aplicamos a ág

ua oxigenada em um ferimento,podemos observar a intensa formação de bolhas.Isso ocorre porque temos uma enzima chamada catalase. Ela não participa da reação química,mas leva a rápida reação de decomposição do peróxido de hidrogênio em água e em gás oxigênio.Escreva a equação química que representa a transformação descrita no texto
Chemistry
1 answer:
Svetllana [295]3 years ago
8 0

Answer:

Ver explicacion

Explanation:

Como se mencionó en la pregunta, un catalizador aumenta la velocidad de una reacción pero él mismo no participa en la reacción.

La descomposición del peróxido de hidrógeno se muestra mediante la ecuación;

2H2O2 (l) ------> 2H2O (l) + O2 (g)

La enzima involucrada en el proceso es la catalasa.

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Can someone explain me why the molar mass become 90.12 g/mol and when I calculated it's 90 g/mol?​
e-lub [12.9K]

Answer:

It could be because of rounding error somewhere in your calculations. Another thing to note is 90 +/- 4 g/mol.

Your calculations don't have to be spot on.

8 0
3 years ago
The product of 2 × 104 cm and 4 × 10–12 cm, expressed in scientific notation is ____.
Hatshy [7]

Answer:

b,,

Explanation:

7 0
3 years ago
The final volume of the solution is 284 mL. What is the concentration of CuSO4 in the final solution, in mol/L?
DochEvi [55]

Answer:

0.0252mol/L

Explanation:

The following data were obtained obtained from the question:

Volume of solution = 284mL = 284/1000 = 0.284L

Mole of CuSO4 = 7.157 × 10^-3 mol

Molarity =?

Molarity = mole/Volume

Molarity = 7.157x10^-3 /0.284

Molarity = 0.0252mol/L

The concentration of the solution is 0.0252mol/L

7 0
3 years ago
Which substance is a homogeneous mixture?
AlladinOne [14]
An example of the substnce would be steel
8 0
3 years ago
The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

7 0
3 years ago
Read 2 more answers
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