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anastassius [24]
3 years ago
11

How many titanium atoms does it contain? titanium has a density of 4.50g/cm3, a pure titanium cube has an edge length of 2.82 in

?
Chemistry
2 answers:
yKpoI14uk [10]3 years ago
5 0
2.82 inches is are equivalent to 7.16 cm.
The volume of the cube ;
 7.16 × 7.16 ×  7.16 = 367.06 cm³
The density of titanium is 4.5 g/cm³
Therefore; mass will be 4.5 × 367.06 = 1651.77 g
1 mole of titanium has 47.867 g
thus, 1651.77 g is equivalent to  1651.77/ 47.867 = 34.507 moles
According to the Avagadro's constant 1 mole = 6.022 × 10∧23 atoms 
 Therefore, we get 6.022 × 10∧23 × 34.507 = 2.078 ×10∧25 atoms
Vikentia [17]3 years ago
4 0

\boxed{2.078 \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 2.82 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( {2.82{\text{ in}}} \right)\left( {\frac{{2.54{\text{ cm}}}}{{1{\text{ in}}}}} \right)\\&= 7.16{\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 7.16 cm for edge length in equation (2).

 \begin{aligned}{\text{Volume of cube}} &= {\left( {{\text{7}}{\text{.16 cm}}} \right)^3}\\&= 367.061696{\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 367.061696{\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( {367.061696{\text{ c}}{{\text{m}}^3}} \right)\\&= 1651.77{\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 1651.77 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( {1651.77{\text{ g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{47}}{\text{.867 g}}}}} \right)\\&= 34.507{\text{ mol}}\\\end{aligned}  

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

\begin{aligned}{\text{Atoms of Ti}}&= \left({34.507{\text{ mol}}} \right)\left( {\frac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}}} \right)\\&= 2.078 \times {10^{25}}{\text{ atoms}}\\\end{aligned}  

Learn more:

  1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603
  2. Calculate the moles of ions in the solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: density, cube, mass, volume, intensive, extensive, atoms, 34.507 mol, 2.078*10^25 atoms, molar mass, Ti, edge length, 2.82 in, 2.54 cm.

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A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
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where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

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Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

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<u></u>

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