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Zigmanuir [339]
3 years ago
9

How many moles of O2- ions are there in 0.450 moles of aluminum oxide, Al2O3?

Chemistry
1 answer:
Mashutka [201]3 years ago
5 0

Answer:

                     1.35 moles of O²⁻

                     21.6 grams of O²⁻

Explanation:

We know that the charge on Aluminium ion is +3 (i.e. Al³⁺) while, the charge on Oxide ion is -2 (i.e. O²⁻). Therefore, the overall neutral Al₂O₃ compound has 2 Al³⁺ ions and 3 O²⁻ ions. Since, we can say that,

                 1 mole of Al₂O3 contains  =  3 moles of O²⁻ ions

So,

                     0.450 moles of Al₂O₃ will have  =  X g of O²⁻

Solving for X,

                      X =  0.450 mol × 3 mol ÷ 1 mol

                     X =  1.35 moles of O²⁻

As the mass of an atom is mainly due to the presence of protons and neutrons hence, the addition of two electrons (-ve 2 shows two gained electron) to Oxygen will make a negligible change to the atomic masss of Oxygen because electron is said to be almost 1800 times lighter than proton. Hence, the ionic mass of O²⁻ will be 16 g/mol and the mass of given moles is calculated as,

                     Mass  =  Moles × Ionic Mass

                     Mass  =  1.35 mol × 16 g/mol

                    Mass  =  21.6 g

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Read 2 more answers
One gram-mole of methyl chloride vapor is contained in a vessel at 100°C and 10atm.(a) Use the ideal-gas equation of state to es
777dan777 [17]

Answer:

a) The volume of the system is 3.062 Liters.

b) Percentage error results from assuming ideal-gas behavior is 9.36%.

Explanation:

a)

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 10 atm

V = Volume of gas = ?

n = number of moles of gas = 1 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 100°C = 100+273K=373 K

Putting values in above equation, we get:

(10 atm)\times V=1 mol\times (0.0821L.atm/mol.K)\times 373 K

V=\frac{1 mol\times (0.0821L.atm/mol.K)\times 373 K}{10 atm}

V = 3.062 L

The volume of the system is 3.062 Liters.

b)

Volume of the container = V' = 2.8 L

System volume = V = 3.062 L

Percentage error of volume:

To calculate the percentage error, we use the equation:

\%\text{ error}=\frac{|\text{Experimental value - Accepted value}|}{\text{Accepted value}}\times 100

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Percentage error results from assuming ideal-gas behavior is 9.36%.

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3 years ago
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