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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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The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p
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Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen               40300 g

Carbon                11160 g

Hydrogen            6200 g

Nitrogen              1860 g

Calcium                992 g

Phosphorus         744 g

Other elements   744 g

It is possible to check it adding all the components, the total is 62000 g.  

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

                            Percentage ( %)

Oxygen               65.0

Carbon               18.0

Hydrogen            10.0

Nitrogen              3.0

Calcium               1.6

Phosphorus         1.2

Other elements   1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

                           Percentage ( %)      Decimal number

Oxygen               65.0                         65/100 = 0.65

Carbon               18.0                           18/100 = 0.18

Hydrogen            10.0                         10/100 = 0.10

Nitrogen              3.0                           3/100 = 0.03

Calcium               1.6                           1.6/100 = 0.016

Phosphorus         1.2                           1.2/100 = 0.012

Other elements   1.2                           1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

                           Percentage ( %)      Decimal number      Mass in grams (g)        

Oxygen               65.0                         65/100 = 0.65         0.65*62000=40300

Carbon               18.0                           18/100 = 0.18         0.18*62000=  11160

Hydrogen            10.0                         10/100 = 0.10         0.10*62000=   6200

Nitrogen              3.0                           3/100 = 0.03          0.03*62000= 1860

Calcium               1.6                           1.6/100 = 0.016        0.016*62000= 992

Phosphorus         1.2                           1.2/100 = 0.012        0.012*62000=744

Other elements   1.2                           1.2/100 = 0.012        0.012*62000=744  

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3 years ago
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