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viktelen [127]
3 years ago
7

The substance gray tin is found to crystallize in a cubic lattice, with an edge length of 646.0 pm. if the density of solid gray

tin is 5.850 g/cm3, how many sn atoms are there per unit cell?
Chemistry
1 answer:
saveliy_v [14]3 years ago
8 0

First find the volume of the cubic lattice given edge length of 646 pm or 6.46x10^-8 cm.

volume = (6.46x10^-8 cm)^3

volume = 2.7x10^-22 cm^3

 

The total mass of the lattice is:

mass = (5.850 g/cm^3) * 2.7x10^-22 cm^3

mass = 1.577x10^-21 grams

 

The molar mass of tin is 118.71 g/mol and the Avogadros number is 6.022 x 10^23 atoms/mol, hence:

 

Sn atoms = [1.577x10^-21 g / (118.71 g/mol)] * 6.022 x 10^23 atoms/mol

Sn atoms = 8 atoms

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2 years ago
2L of hydrogen has an initial pressure of 750 mmHg, what is the final pressure in mmHg if the volume increases to 20 L with a co
Verdich [7]

Answer: The final pressure is 75 mm Hg.

Explanation:

According to Boyle's law, at constant temperature the pressure of a gas in inversely proportional to volume.

Since, it is given that the temperature is constant. Hence, formula used is as follows.

P_{1}V_{1} = P_{2}V_{2}

Substitute the values into above formula as follows.

P_{1}V_{1} = P_{2}V_{2}\\750 mm Hg \times 2 L = P_{2} \times 20 L\\P_{2} = \frac{750 mm Hg \times 2 L}{20 L}\\= 75 mm Hg

Thus, we can conclude that the final pressure is 75 mm Hg.

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2 years ago
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6 0
2 years ago
Read 2 more answers
What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

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