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Alexus [3.1K]
2 years ago
15

El acido nitrico se obtiene industrialmente por oxidacion de amoniaco (nh3), de acuerdo con la ecuacion: 4nh3 + 702 - h2o + hno2

+ 2hno3 ¿ cuanto acido nitrico del 90% de pureza se podra obtener a partir de 1.36 kg de amoniaco?
Chemistry
1 answer:
devlian [24]2 years ago
6 0

Answer:

2800 g de ácido nítrico

Explanation:

La ecuación por la oxidación de amoniaco es:

4NH₃  +  7O₂  →  4H₂O  +  2HNO₂  +  2HNO₃

Si pensamos que el oxígeno es el reactivo limitante, trabajamos con el amoniaco. Convertimos su masa a moles:

1.36 kg = 1360 g

1360 g . 1mol  /17g = 80 moles

Si 4 moles de amoniaco pueden producir 2 moles de acido nítrico

80 moles producirán, (80 . 2)/4 = 40 moles.

Convertimos los moles a gramos:

40 mol . 63g /mol = 2520 g

Si le aplicamos la pureza

2520 g . 100/90 = 2800 g

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1.
GarryVolchara [31]

Answer:

The value of the Golden Igloo is $227.4 million.

Explanation:

First, we need to find the inner and the outer volume of the half-spherical shell:

V_{i} = \frac{1}{2}*\frac{4}{3}\pi r_{i}^{3}

V_{o} = \frac{1}{2}*\frac{4}{3}\pi r_{o}^{3}

The total volume is given by:

V_{T} = V_{o} - V_{i}

Where:

V_{i}: is the inner volume

r_{i}: is the inner radius = 1.25/2 = 0.625 m

V_{o}: is the outer volume

r_{o}: is the outer radius = 1.45/2 = 0.725 m

Then, the total volume of the Igloo is:

V_{T} = \frac{2}{3}\pi r_{o}^{3} - \frac{2}{3}\pi r_{i}^{3} = \frac{2}{3}\pi [(0.725 m)^{3} - (0.625 m)^{3}] = 0.29 m^{3}

Now, by using the density we can find the mass of the Igloo:

m = 19.3 \frac{g}{cm^{3}}*0.29 m^{3}*\frac{(100 cm)^{3}}{1 m^{3}} = 5.60 \cdot 10^{6} g

Finally, the value (V) of the antiquity is:

V = \frac{\$ 1263}{oz}*5.60 \cdot 10^{6} g*\frac{1 oz}{31.1034768 g} = \$ 227.4 \cdot 10^{6}  

Therefore, the value of the Golden Igloo is $227.4 million.

I hope it helps you!  

8 0
2 years ago
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