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allochka39001 [22]
4 years ago
8

What is the equilibrium expression for the reatcion bellow 2SO3(g)<=>O2(g)+2SO2(g)

Chemistry
1 answer:
Luda [366]4 years ago
8 0

Answer:

[O2(g)][SO2(g)]^2/[SO3(g)]^2

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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

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

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1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

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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

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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

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