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Vlad [161]
3 years ago
12

What is a business cycle?

Engineering
1 answer:
Anestetic [448]3 years ago
5 0

Answer:

a cycle or series of cycles of economic expansion and contraction.

Explanation:

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The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are
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Answer:

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum  = 9.93*10^{-8}

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the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = 6.022*10^{23

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

5.854*10^{15} = 9.93*10^{-8}  × Total number of sites(N)

Total number of sites (N) = \dfrac{5.854*10^{15}}{9.93*10^{-8}}

Total number of sites (N) = 5.895*10^{22}

From the expression of the total number of sites; we can determine the density of the electrum;

N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}

where ;

N_A = Avogadro's Number

\rho_{electrum} = density of the electrum

A_{electrum} = Atomic mass

5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}

5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}

8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}

\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}

\mathbf{  \rho _{electrum}=14.30 \ g/cm^3}

Thus; the density of the electrum is 14.30 g/cm³

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3÷3=1

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