Answer:
- Part a) 0.0104 moles copper(II) nitrate.
i) 0.0418 mole Cu
ii) 0.0209 mol Ag NO₃
Explanation:
<u>1) Balanced chemical reaction (single replacement):</u>
In a single replacement reaction a more acitve metal (Cu) replaces a less active metal (Ag)
- Cu + 2 Ag NO₃ → Cu (NO₃)₂ + 2 Ag
<u>2) Mole ratio: </u>
- 1 mole Cu : 2 mole Ag NO₃ : 2 mole Ag
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<u>3) Moles of Ag</u>
- n = mass in grams / atomic mass
- atomic mass of Ag: 107.868 g/mol
- n = 2.25 g / 107.868 g/mol = 0.0209 mol Ag
<u>4) Moles of copper(II) nitrate:</u>
- Set the proportion using the mole ratio:
- 2 mole Ag / 1 mole Cu (NO₃)₂ = 0.0209 mole Ag / x
- Solve: x = 0.0209 / 2 mole Cu (NO₃)₂ = 0.0104 moles Cu(NO₃)₂
That is the answer of part a: 0.0104 moles copper(II) nitrate.
<u>5) Moles of each reactant</u>
i) Cu:
- Set a proportion using the theoretical mole ratio
1 mole Cu / 2 mole Ag = x / 0.0209 mol Ag
- Solve for x: x = 0.0209 / 2 mole Cu = 0.0418 mole Cu
ii) Ag NO₃
- Set a proportion using the teoretical mole ratio
2 mole Ag NO₃ / 2 mole Ag = x / 0.0209 mole Ag
- Solve for x: x = 0.0209 mol Ag NO₃
Answer:
The answer is F
Explanation: Only because Henrieta wasn't the sister that got kicked out
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:
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Explanation:
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