Chemistry, 16.09.2019 09:20 blackboy21

Apure titanium cube has an edge length of 2.84 in how many titanium atoms does it contain? titanium has a density of 4.50g/cm3.

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Answer from: jazzy4211

There are $2.12 \times 10^{25}$ titantium atoms in the cube.

Further Explanation

This problem involves unit conversion and mole conversion. The steps are:

Convert the length of the cube from inches to centimeter since the density is given in g/cm³.Determine the mass of the cube using the given density.Convert the mass of the cube to number of atoms using Avogadro's number.

STEP 1. Converting inches to cm

$length \ in \ cm \ = 2.84 \ in \times (\frac{2.54 \ cm}{1 \ in})\\\boxed {length \ in \ cm \ = 7.2136 \ cm}$

STEP 2: Determining mass of cube from the density

$density = \frac{mass}{volume}\\mass = density \times \ volume\\\\mass \ = 4.50 \frac{g}{cm^3} \times (7.2136 \ cm)^3\\\boxed {mass \ = 1689.15 \ g}$

STEP 3: Converting mass of titanium cube to number of atoms

$no. \ of \ Ti \ atoms = 1689.15 \ g \ Ti \times (\frac{1 \ mol \ Ti}{47.867 \ g})(\frac{6.022 \times 10^{23} \ Ti \ atoms}{1 \ mol \ Ti}) \\\\\boxed {no. \ of \ Ti \ atoms = 2.125 \times 10^{25} \ atoms}$

Since the given values have 3 significant figures, the final answer must be:

$\boxed {\boxed {no. \ of \ atoms \ = 2.12 \times 10^{25} \ Ti \ atoms}}$

Learn MoreMole Conversions Stoichiometry Dimensional Analysis

Keywords: Avogadro's number

Answer from: musicismylove2340

$\boxed{2.125 \times {{10}^{25}}{\text{ atoms}}}$ of titanium are present if it has a density of $4.50{\text{ g/c}}{{\text{m}}^3}$ and edge length of 2.84 in.

Further Explanation:

Density is an intensive property. It is the ratio of mass of substance to the volume of substance. Both these quantities are extensive in nature so their ratio comes out to be an intensive quantity. Density depends only on the nature of the substance, not on the quantity of the substance. The expression to calculate the density of a cube is,

${\text{Density of cube}} = \dfrac{{{\text{Mass of cube}}}}{{{\text{Volume of cube}}}}$ …… (1)

The edge of cube is to be converted into cm. The conversion factor for this is,

$1{\text{ in}} = {\text{2}}{\text{.54 cm}}$

Therefore the edge length of cube is calculated as follows:

$\begin{aligned}{\text{Edge length of cube}} &= \left( {2.84{\text{ in}}} \right)\left( {\frac{{2.54{\text{ cm}}}}{{1{\text{ in}}}}} \right)\\&= 7.2136{\text{ cm}}\\\end{aligned}$

The formula to calculate the volume of cube is as follows:

${\text{Volume of cube}} = {\left( {{\text{Edge length}}} \right)^3}$ …… (2)

Substitute 7.2136 cm for edge length in equation (2).

$\begin{aligned}{\text{Volume of cube}}&={\left( {{\text{7}}{\text{.2136 cm}}} \right)^3}\\&= 375.36707{\text{ c}}{{\text{m}}^3}\\\end{aligned}$

Rearrange equation (1) to calculate the mass of cube.

${\text{Mass of cube}} = \left( {{\text{Density of cube}}} \right)\left( {{\text{Volume of cube}}} \right)$ …… (3)

Substitute $4.50{\text{ g/c}}{{\text{m}}^3}$ for density of cube and $375.36707{\text{ c}}{{\text{m}}^3}$ for volume of cube in equation (3).

$\begin{aligned}{\text{Mass of cube}} &= \left( {\frac{{{\text{4}}{\text{.50 g}}}}{{1{\text{ c}}{{\text{m}}^3}}}} \right)\left( {375.36707{\text{ c}}{{\text{m}}^3}} \right)\\&= 1689.1519{\text{ g}}\\\end{aligned}$

The formula to calculate the moles of Ti is as follows:

${\text{Moles of Ti}} = \dfrac{{{\text{Given mass of Ti}}}}{{{\text{Molar mass of Ti}}}}$ ...... (4)

Substitute 1689.1519 g for given mass of Ti and 47.867 g/mol for molar mass of Ti in equation (4).

$\begin{aligned}{\text{Moles of Ti}} &= \left( {1689.1519{\text{ g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{47}}{\text{.867 g}}}}} \right)\\&= 35.288{\text{ mol}}\\\end{aligned}$

The number of atoms of Ti can be calculated as follows:

$\begin{aligned}{\text{Atoms of Ti}} &= \left( {35.288{\text{ mol}}} \right)\left( {\frac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}}} \right)\\&= 2.125 \times {10^{25}}{\text{ atoms}}\\\end{aligned}$