Chemistry, 25.09.2019 03:00 kadeWayne

Titanium has an hcp unit cell for which the ratio of the lattice parameters cais 1.58. if the radius of the be atom is 0.1445 nm, (a) determine the unit cell volume, and (b) calculate the theoretical density of ti, given that its atomic weight is 47.87 g/mol

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Answer from: stunnakid9094

Answer :

(a) The volume of unit cell is, $9.91\times 10^{-23}cm^3/\text{ unit cell}$

(b) The theoretical density of Ti is, 4.81g/cm^3

Explanation :

(a) First we have to calculate the volume of the unit cell.

Formula used :

$V=6r^2c\sqrt {3}$

where,

V = volume of unit cell = ?

r = atomic radius = $0.1445nm=1.445\times 10^{-8}cm$

conversion used : $(1nm=10^{-7}cm)$

Ratio of lattice parameter = c : a = 1.58 : 1

So, c = 1.58 a

And, a = 2r

c = 1.58 × 2r

Now put all the given values in this formula, we get:

$V=6\times r^2\times (1.58\times 2r)\sqrt {3}$

$V=6\times r^3\times (1.58\times 2)\sqrt {3}$

$V=6\times (1.445\times 10^{-8}cm)^3\times (1.58\times 2)\sqrt {3}$

$V=9.91\times 10^{-23}cm^3/\text{ unit cell}$

The volume of unit cell is, $9.91\times 10^{-23}cm^3/\text{ unit cell}$

(b) Now we have to calculate the density of Ti.

Formula used for density :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

$\rho=\frac{Z\times M}{N_{A}\times V}$ ..........(1)

where,

$\rho$ = density of Ti = ?

Z = number of atom in unit cell = 6 atoms/unit cell (for HCP)

M = atomic mass = 47.87 g/mol

$(N_{A})$ = Avogadro's number = $6.022\times 10^{23}atoms/mole$

a^3=V = volume of unit cell = $9.91\times 10^{-23}cm^3/\text{ unit cell}$

Now put all the values in above formula (1), we get:

$\rho=\frac{6\times 47.87}{(6.022\times 10^{23})\times (9.91\times 10^{-23})}$

$\rho=4.81g/cm^3$

The theoretical density of Ti is, 4.81g/cm^3

Answer from: Quest

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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