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2 years ago

What type of crystal is halite crystal (rock salt)?

2 answers:

Alex2 years ago

8 0

Halite crystals are isometric crystals, meaning that unit cells of the crystal are cube shaped, which is why they're also called cubic crystals.

nekit [7.7K]2 years ago

6 0

Halite crystals are isometric in nature with a cube, being their unit cells.

Further explanation:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a specific manner. The unit cell is the tiniest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 major types of cubic unit cells as follows:

1. The simple cubic unit cell (sc)

2. Body-centered unit cell (bcc)

3. Face-centered unit cell(fcc)

Halite is also known as rock salt, sea salt or table salt. It is composed of sodium chloride (NaCl). NaCl is an ionic crystalline compound made up of ${\text{N}}{{\text{a}}^+}$ and ${\text{C}}{{\text{l}}^-}$ ions.

Coordination number is the number of ions surrounding an ion of opposite charge. Each ${\text{N}}{{\text{a}}^+}$ is surrounded by 6 ${\text{C}}{{\text{l}}^-}$ ions and vice-versa. So the coordination number of both ${\text{N}}{{\text{a}}^+}$ and ${\text{C}}{{\text{l}}^-}$ ions in the structure of NaCl is 6.

Halite crystals are isometric crystals. This is because they have their unit cells in the form of a cube. Due to this reason, halite crystals are also known as cubic crystals.

Learn more:

1. Determine molecules and polyatomic ions that cannot be adequately described using a single lewis structure: <u>brainly.com/question/6786947</u>

2. Determine the correct structure of ${{\text{N}}_{\text{2}}}{\text{O}}$ : <u>brainly.com/question/9422880</u>

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solid-state

Keywords: halite, rock salt, NaCl, Na+, Cl-, 6, coordination number, isometric crystals, sea salt, table salt, ions, and opposite charge.

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3 years ago

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3 years ago

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2 years ago

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What is formed when atoms of two or more elements bond in a chemical reaction? A.Mixture B.Elements C.Atoms D.Compound

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3 years ago

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

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3 years ago