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GarryVolchara [31]
3 years ago
9

Which of the following statements concerning a metal crystallized in a face-centered cubic cell is/are CORRECT? 1. One metal ato

m is located on each face of the unit cell, where it is shared equally between four unit cells. 2. One metal atom is located at the center of the unit cell. 3. A metal atom is located at each of the eight lattice points, where it is shared equally between eight unit cells.
Chemistry
1 answer:
AleksAgata [21]3 years ago
5 0

Answer:

Only 3 is correct.

Explanation:

The crystal of a metal or an ionic compound is called a cell, and there are 7 types of unit cells: cubic, tetragonal, orthorhombic, monoclinic, hexagonal, rhombohedral, and triclinic.

In a face-centered cubic cell (FCC) all angles are 90º and all lengths are equal. Each cubic cell has 8 atoms in each corner of the cube, and that atom is shared with 8 neighboring cells. So for a metal crystal, the atom is located at each of the eight lattice points, where it is shared equally between eight unit cells.

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____________ rays show areas where large quantities of energy are released.
sladkih [1.3K]

Answer:

Gamma

Explanation:

3 0
2 years ago
How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a
Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of water = 72.0  g

c_{s} = specific heat of ice = 2.09J/g^0C

c_{l} = specific heat of liquid water = 4.184J/g^0C

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles

\Delta H_{fusion} = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]\Delta H=34279.8J=34.3kJ        (1 KJ = 1000 J)

Therefore, the enthalpy change is 34.3 kJ

5 0
3 years ago
Which half reaction is associated with the anode?
Marta_Voda [28]

Answer:

vgf

Explanation:

3 0
3 years ago
Calculate the number of milliliters of 0.440 M KOH required to precipitate all of the Fe2+ ions in 187 mL of 0.692 M FeSO4 solut
EleoNora [17]

Answer:

588.2 mL

Explanation:

  • FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)

First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):

  • moles = molarity * volume
  • 187 mL * 0.692 M = 129.404 mmol Fe⁺²

Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:

  • 129.404 mmol Fe⁺² * \frac{2mmolKOH}{1mmolFeSO_4} = 258.808 mmol KOH

Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:

  • volume = moles / molarity
  • 258.808 mmol KOH / 0.440 M = 588.2 mL
6 0
3 years ago
Gold has a molar (atomic) mass of 197 g/mol. consider a 2.47 g sample of pure gold vapor. (a) calculate the number of moles of g
adell [148]
N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol

n = 2.47 / 197 
n = 0.01253 moles.
I'm sure you wanted to ask more than this. Just put some comments in. I can do the same.
3 0
3 years ago
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