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Eddi Din [679]
3 years ago
7

Avogadro's number is equal to​?

Chemistry
2 answers:
Lerok [7]3 years ago
4 0

Answer:

6.273×10²³

Explanation:

hope this is useful friend

Ray Of Light [21]3 years ago
3 0

Answer:

6.023 × 10²³

Explanation:

Avogadro's number is used for calculating masses and moles and can applied to Stoichiometry. Avogadro's number is useful for atoms and molecules

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Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6
MAXImum [283]

Answer:

the energy vacancies for formation in silver is \mathbf{Q_v = 3.069*10^{-4} \ J/atom}

Explanation:

Given that:

the equilibrium  number of vacancies at 800 °C

i.e T = 800°C     is  3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

N =  \dfrac{N_A* \rho _{Ag}}{A_{Ag}}

where ;

N_A = avogadro's number = 6.023*10^{23} \ atoms/mol

\rho _{Ag} = Density of silver = 9.5 g/cm³

A_{Ag} = Atomic weight of sliver = 107.9 g/mol

N =  \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}

N = 5.30 × 10²⁸ atoms/m³

However;

The equation for equilibrium number of vacancies can be represented by the equation:

N_v = N \ e^{^{-\dfrac{Q_v}{KT}}

From above; Considering the  natural logarithm on both sides; we have:

In \ N_v =In N - \dfrac{Q_v}{KT}

Making Q_v the subject of the formula; we have:

{Q_v =  - {KT}   In( \dfrac{ \ N_v }{ N})

where;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})

\mathbf{Q_v = 2.38 \ eV/atom}

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

Q_v =  (2.38 \ * 1.602176565 * 10^{-19} ) J/atom  }

\mathbf{Q_v = 3.069*10^{-4} \ J/atom}

Thus, the energy vacancies for formation in silver is \mathbf{Q_v = 3.069*10^{-4} \ J/atom}

8 0
3 years ago
A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm
VARVARA [1.3K]

<u>Answer:</u> The nuclear equations for the given process is written below.

<u>Explanation:</u>

The chemical equation for the bombardment of neutron to U-238 isotope follows:

_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

The chemical equation for the first beta decay process of _{92}^{239}\textrm{U} follows:

_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta

The chemical equation for the second beta decay process of _{93}^{239}\textrm{Np} follows:

_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta

Hence, the nuclear equations for the given process is written above.

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<span>As for anything, 1 mole of X contains 6.022 * 10^23 of X. Multiply moles of X by Avogadro's number.</span>
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