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Pepsi [2]
2 years ago
15

Copper crystallizes in a face-centered cubic unit cell. the density of copper is 8.94 g/cm3. calculate the length of the edge of

the unit cell in pm.
Chemistry
1 answer:
Ilia_Sergeevich [38]2 years ago
4 0
<span>361.4 pm is the length of the edge of the unit cell. First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight. Atomic weight copper = 63.546 Now divide the mass by the atomic weight, getting 8.94 g / 63.546 g/mol = 0.140685488 mol And multiply by Avogadro's number to get the number of atoms: 0.140685488 * 6.022140857x10^23 = 8.472278233x10^22 Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume. 8.472278233x10^22 / 4 = 2.118069558x10^22 Now calculate the volume each unit cell occupies. 1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3 Let's get the cube root to get the length of an edge. (4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm Now let's convert from cm to pm. 3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
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A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution wa
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Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

<em>Change in heat:</em>

q = m*S*ΔT

<em>Where q is heat in J,</em>

<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

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3830.87J/0.1048 moles KNO3 =

36554J/mol =

<h3>36.55kJ/mol</h3>

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