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Citrus2011 [14]
3 years ago
8

A reaction in the laboratory yields 5.98 g KAI(SO2)2, How many moles of potassium aluminum

Chemistry
1 answer:
Ivahew [28]3 years ago
6 0

Answer:

0.0308 mol

Explanation:

In order to convert from grams of any given substance to moles, we need to use its molar mass:

  • Molar mass of KAI(SO₂)₂ = MM of K + MM of Al + (MM of S + 2*MM of O)*2
  • Molar mass of KAI(SO₂)₂ = 194 g/mol

Now we <u>calculate the number of moles of KAI(SO₂)₂ contained in 5.98 g</u>:

  • 5.98 g ÷ 194 g/mol = 0.0308 mol
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Since the given equation is not balanced properly.

Since oxygen and hydrogen atoms are not balanced.

There should be 6 H2O (g) molecules and 14 mol H2 (g)

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Explain why a can implodes
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boom

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Se ha añadido un evaporador para una alimentación de 11500 kg/dia de zumo de pomelo de forma que evapore 3000 kg/dia de agua por
mamaluj [8]

Answer:

37 %.

Explanation:

¡Hola!

En este caso, para el problema descrito, conocemos la corriente de entrada y la de salida del agua, por lo que podemos obtener el flujo de la corriente que contiene el zumo a la salida una vez el agua fue evaporada:

F_{sol}=11500kg/dia-3000kg/dia=8500 kg/dia

Luego, por medio de un balance de zumo de limón en el evaporador en el cual la cantidad que entra es igual a la que sale con sus respectivas concentraciones:

x_z^{entra}*11500kg/dia=x_z^{sale}*8500kg/dia

Como la concentración del zumo a la salida es del 50 % (0.50), la de entrada es:

x_z^{entra}=\frac{x_z^{sale}*8500kg/dia}{11500kg/dia} =\frac{0.50*8500kg/dia}{11500 kg/dia}\\ \\x_z^{entra}=0.37

Que es igual al 37%.

¡Saludos!

6 0
4 years ago
Which two energy sources were used the most in
Naily [24]

Answer:

The 1st and 2nd ones on the top

Explanation:

Hope this helps:)

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3 years ago
A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
a_sh-v [17]

Answer:

m_{H_2O}=39.0g

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g

Best regards.

8 0
3 years ago
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