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mezya [45]
3 years ago
13

Help

Chemistry
1 answer:
uranmaximum [27]3 years ago
7 0

I believe the answer is A

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Define natural and artificial hazards?<br><br><br>wrong and silly answer will be reported​
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If there are 20.0g of KOH available, how many grams Ca(OH)2 will form?
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Brianliest for whoever gets this right with step by step method
Sedaia [141]
The Mr is the mass numbers of each element added up so…. Fe = 56, O=16, H=1 … now add these up with the number of each element -> there’s 1 Fe, and 3 Os and 3 Hs as they are in brackets with a 3 outside-> (56+16+16+16+1+1+1=107) … your answer is 107
8 0
3 years ago
Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/ato
nexus9112 [7]

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by N_v.

It is depends on

  • total no. of atomic number(N)
  • energy required for vacancy
  • Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
  • temperature (T).

N_v=Ne^{-\frac{Q_v}{kT} }

To find  equilibrium number of of vacancies we have find N.

N=\frac{N_A\ \rho}{A_{cu}}

Here ρ= 8.45 g/cm³  =8.45 ×10⁶m³

N_A= Avogadro Number = 6.023×10²³

A_{Cu}= 63.5 g/mole

N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}

   =8.01\times 10^{28 g/mole

Here Q_v=0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})

   =2.34×10²⁴ vacancies/ mole

3 0
3 years ago
Given the following thermodynamic data, calculate the lattice energy of LiCl:
tiny-mole [99]

Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

3 0
3 years ago
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