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3 years ago

Barium has a body-centered cubic structure. If the atomic radius of barium is 222 pm, calculate the density of solid barium?

Chemistry

1 answer:

Flauer [41]3 years ago

5 0

Answer: density =3.377g/cm³

Explanation:

Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)

D= (M ×n) /(V×A)

M= 137g/mol

n= 2 (For BCC)

V=a³ , where a= 4r/√3

a= (4×222)/√3

a=512.69pm

a= 512.69×10^-10cm

V= ( 512.69×10^-10)^3

V= 1.3476×10^-22cm³

D= (137×2)/(1.3476×10^-22 × 6.02^23)

D= 3.377g/cm³

Therefore the density of barium is 3.377g/cm³

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Answer:

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Explanation:

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3 years ago

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Korvikt [17]

Answer:

$55.2kgNa_{3}AlF_{6}$

Explanation:

1. First balance the equation for the synthesis of cryolite:

$Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}$

2. Find the limiting reagent between the $Al_{2}O_{3},NaOH$ and $HF$

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}$

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- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

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The $Al_{2}O_{3}$ is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}$

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4 years ago

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