Chemistry, 03.07.2020 23:01 harmonytaylor13

Above 882oC, zirconium has a BCC crystal structure with a = 0.332 nm. Below this temperature, zirconium has an HCP structure with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC zirconium transforms to HCP zirconium. Is that contraction or expansion?

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Answer from: demetriascott20

$\mathbf{\Delta V = -0.63 \%}$

Contraction

Explanation:

From the given information;

Above 88° C

zirconium has a BCC crystal structure with a = 0.332 nm

Below this temperature

zirconium has an HCP structure with a = 0.2978 nm and c = 0.4735 nm

the volume of BCC can now be:

$V_{BCC} = (a)^3$

$V_{BCC} = (0.332 \ nm)^3$

$V_{BCC} =0.03660 \ nm^3$

the volume of HCP can now be:

$V_{HCP} = (a)^2 (c) cos \ 30^0$

$V_{HCP} = (0.2978)^2 (0.4735) \ cos 30$

$V_{HCP} =0.03637 \ nm^3$

Ths; the volume percent change when BCC zirconium transforms to HCP zirconium can be calculated as:

$\Delta V = \dfrac{V_{HCP}-V_{BCC}}{V_{BCC}} * 100 \%$

$\Delta V = \dfrac{0.03637 \ nm^3-0.03660\ nm^3}{0.03660\ nm^3}} * 100 \%$

$\mathbf{\Delta V = -0.63 \%}$

Hence; it is contraction due to what the negative sign portray, The negative sign signifies that there is contraction during cooling

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

Its a hope this

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Above 882oC, zirconium has a BCC crystal structure with a = 0.332 nm. Below this temperature, zircon...