Pls hlp me with these questions: brainliest, ratting,

Ответ:

billydeck30

16.11.2020

$\mathbf {density \ d =4.4845 \ g/cm^3}$

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

$\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}$

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

$\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}$

$\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}$

$\mathbf {density \ d =4.4845 \ g/cm^3}$

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