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BartSMP [9]
3 years ago
11

De acuerdo con la Ley de la Conservación de la Materia postulada por Lavoisier, ¿cuál ecuación química está balanceada? A C2H5OH

(l) + O2(g) → CO2 + H2O B C2H5OH(l) + O2(g) → 2CO2 + H2O C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O D C2H5OH(l) + O2(g) → 2CO2 + 3H2O E 2C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
6 0

Answer:

C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O

Explanation:

La ley de conservación de la materia implica que en una reacción química los átomos en los reactivos son iguales a la cantidad de átomos en los productos:

A C2H5OH(l) + O2(g) → CO2 + H2O

En los reactivos hay dos átomos de carbono pero en los productos solo 1. De esta manera, no cumple la ley de conservación.

B C2H5OH(l) + O2(g) → 2CO2 + H2O

En los reactivos hay 6 átomos de hidrógeno pero en los productos solo 2. No cumple la ley de conservación.

C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O

En los reactivos y productos hay: 2 átomos de carbono, 6 átomos de hidrógeno y 7 átomos de oxígeno. Cumple la ley de conservación.

D C2H5OH(l) + O2(g) → 2CO2 + 3H2O

En los reactivos hay 3 átomos de oxígeno y en los productos 7. No cumple la ley de conservación.

E 2C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O

En los reactivos hay 4 átomos de carbono y en los productos solo dos. No cumple la ley de conservación.

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A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),
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Answer:        

[KOH] : 1.47 M

[KOH] : 1.22 m

[KOH]: 6.42 % mass percent.      

Explanation:

First of all we must determine the volume of solution. We have to work with the density

Density = mass / volume

1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

155g / 56.1 g/ mol = 2.76 moles

Now we can calculate molarity

2.76 mol / 1.87 L = 1.47 M

To calculate molality we have to find out the mass of solvent

mass solute + mass solvent = mass solution

155 g + mass solvent = 2412.3 g

2412.3g - 155g = 2257.3g

We have to convert the 2257.3 g to kg

2257.3 g = 2.25 kg

molality = 2.76 moles / 2.25 kg = 1.22 m

To find out the % mass percentation, we have to calculate the mass of solute in 100 g of solution.

In 2412.3 g of solution we have 155 g of KOH

In 100 g of solution, we would have (100 . 155) / 2412.3 = 6.42 %mass percent.

7 0
3 years ago
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