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Doss [256]
3 years ago
13

An equilibrium mixture of O2, SO2 and SO3 contains equal concentrations of SO2 and SO3.

Chemistry
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

[O₂(g)] = 0.0037M

Explanation:

            2SO₂(g) + O₂(g) => 2SO₃(g)

Conc:   [SO₂(g)]   [O₂(g)]     [SO₃(g)]  and  [SO₂(g)] = [SO₃(g)]

Kc =  [SO₃(g)]²/[O₂(g)][SO₂(g)]² => Kc = 1/[O₂(g)] = 270 if [SO₂(g)] = [SO₃(g)]

∴ [O₂(g)] = (1/270)M = 0.0037M  

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How many molecules are in 145.5 grams of Be(OH)2
Zanzabum

Answer:

2.04 x 10²⁴ molecules

Explanation:

Given parameters:

Mass of Be(OH)₂ = 145.5g

To calculate the number of molecules in this mass of Be(OH)₂ we follow the following steps:

>> Calculate the number of moles first using the formula below:

Number of moles = mass/molarmass

Since we have been given the mass, let us derive the molar mass of Be(OH)₂

Atomic mass of Be = 9g

O = 16g

H = 1g

Molar Mass = 9 + 2(16 + 1)

= 9 + 34

= 43g/mol

Number of moles = 145.5/43 = 3.38mol

>>> We know that a mole is the amount of substance that contains Avogadro’s number of particles. The particles can be atoms, molecules, particles etc. Therefore we use the expression below to determine the number of molecules in 3.38mol of Be(OH)₂:

Number of

molecules= number of moles x 6.02 x 10²³

Number of molecules= 3.38 x 6.02 x 10²³

= 20.37 x 10²³ molecules

= 2.04 x 10²⁴ molecules

3 0
3 years ago
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