Answer:
Density of unit cell ( rhodium) = 12.279 g/cm³
Explanation:
Given that:
The radius (r) of a rhodium atom = 135 pm
The atomic mass of rhodium = 102.90 amu
For a face-centered cubic unit cell,
![r = \dfrac{a}{2\sqrt{2}}](https://tex.z-dn.net/?f=r%20%3D%20%5Cdfrac%7Ba%7D%7B2%5Csqrt%7B2%7D%7D)
where;
a = edge length.
Making "a" the subject of the formula:
![a = 2 \sqrt{2} \times r](https://tex.z-dn.net/?f=a%20%3D%202%20%5Csqrt%7B2%7D%20%5Ctimes%20r)
![a = 2 \times 1.414 \times 135 \ pm](https://tex.z-dn.net/?f=a%20%3D%202%20%5Ctimes%201.414%20%5Ctimes%20135%20%5C%20pm)
a = 381.8 pm
to cm, we get:
a = 381.8 × 10⁻¹⁰ cm
However, recall that:
where;
mass of unit cell = mass of atom × numbers of atoms per unit cell
Also;
![mass\ of\ atom =\dfrac{ atomic \ mass}{Avogadro \ number}](https://tex.z-dn.net/?f=mass%5C%20%20of%5C%20atom%20%3D%5Cdfrac%7B%20atomic%20%5C%20mass%7D%7BAvogadro%20%20%5C%20%20number%7D)
![mass\ of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}](https://tex.z-dn.net/?f=mass%5C%20%20of%5C%20atom%20%3D%5Cdfrac%7B%20102.9%7D%7B6.023%20%5Ctimes%2010%5E%7B23%7D%7D)
Recall also that number of atoms in a unit cell for a face-centered cubic = 4
So;
![mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4](https://tex.z-dn.net/?f=mass%20%5C%20of%20%5C%20unit%20%5C%20cell%3D%20%5Cdfrac%7B102.90%7D%7B6.023%20%5Ctimes%2010%5E%7B23%7D%7D%5Ctimes%204)
mass of unit cell = 6.83380375 × 10⁻²² g
![Density \ of \ unit \ cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}](https://tex.z-dn.net/?f=Density%20%20%5C%20of%20%20%5C%20unit%20%5C%20%20cell%20%3D%20%5Cdfrac%7B6.83380375%20%5Ctimes%2010%5E%7B-22%7D%7D%7B%28381.8%5Ctimes%2010%5E%7B-10%7D%29%5E3%7D)
Density of unit cell ( rhodium) = 12.279 g/cm³
Answer:
0.99 kg O₂
1.9 kg SO₂
Explanation:
Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.
S + O₂ → SO₂
The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:
1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂
The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:
1 kg S × (64.07 kg SO₂/32.07 kg S) = 1.99 kg SO₂
<span>false - sodium is not a member of the transition elements, however </span><span>copper is a </span><span>member of the transition elements.</span>