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Fiesta28 [93]
3 years ago
14

The Al2O3 crystal structure (corundum) consists of an HCP arrangement of O2- ions; the Al3 ions occupy octahedral positions. Wha

t fraction of the available octahedral positions are filled with Al3 ions
Chemistry
1 answer:
ddd [48]3 years ago
8 0

Answer:

2/3

Explanation:

Crystals structures can also be seen when two elements combines together and the perfect example is Al₂O₃ which is given in the question above. Just like it is given in the question above, the kind of arrangement in the crystal structure for Al₂O₃ is called HCP which stands for Hexagonally Closed Pack.

The aluminum ions which is in form of Al³⁺ occupies the two-third[2/3] positions while the position that the oxygen ion occupies is one[1].

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Answer:

Step-by-step explanation: We are asked to find the distance covered by a man in 15 minutes at a speed of 16 km/hr. 15 minutes = 15/60 hour. Therefore, the person can run 4 km in 15 minutes.

Explanation:

4 0
3 years ago
A sample of gold has a mass of 67.2 g and a volume of 3.5 cm3. the density of gold is
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67.2g/3.5cm3= 19.2g/cm3
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Help please!!  Why do astronauts' muscles weaken in space?
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B. They work against a strong gravitational force

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4vir4ik [10]

Answer: 25.8 g of Cl_2 will be produced from the decomposition of 73.4 g of AuCl_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles

The balanced chemical reaction is:

2AuCl_3\rightarrow 2Au+3Cl_2  

According to stoichiometry :

2 moles of AuCl_3 produce =  3 moles of Cl_2

Thus 0.242 moles of  will produce= \frac{3}{2}\times 0.242=0.363mol of Cl_2

Mass of Cl_2= moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g

Thus 25.8 g of Cl_2 will be produced from the decomposition of 73.4 g of AuCl_3

5 0
2 years ago
If 3.31 moles of argon gas occupies a volume of 100 L what volume does 13.15 moles of argon occupy under the same temperature an
kumpel [21]

Answer:

397 L

Explanation:

Recall the ideal gas law:

\displaystyle PV = nRT

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

\displaystyle \frac{P}{RT} = \frac{n}{V}

The left-hand side is simply some constant. Hence, we can write that:

\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}

Substitute in known values:

\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})}  = \frac{(13.15\text{ mol })}{V_2}

Solving for <em>V</em>₂ yields:

\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.

(Assuming 100 L has three significant figures.)

3 0
2 years ago
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