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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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Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)→C
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Answer:

m=8.79kg

Explanation:

First of all we need to calculate the heat that the water in the cooler is able to release:

Q=\rho * V*Cp*\Delta T

Where:

  • Cp is the mass heat capacity of water
  • V is the volume
  • \rho is the density

Q=1 g/cm^3 *15000 cm^3*4.184 \frac{J}{g*^{\circ}C}*(10-90)^{\circ}C

Q=-5020800 J=-5020.8 kJ

To calculate the mass of CO2 that sublimes:

-Q=\Delta H_{sub}*m

Knowing that the enthalpy of sublimation for the CO2 is: \Delta H_{sub}=571 kJ/kg

5020.8 kJ=571 kJ/kg*m

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6 0
2 years ago
I need help with this assignment
yuradex [85]

The answers for the following sums is given below.

1.Pd_{2}H_{2}

2.C_{2} H_{6}

3.C_{2} H_{2} O_{2} Cl_{2}

4.C_{3}Cl_{3} N_{3}

5.Tl_{2 } C_{4} H_{4}O_{6}

6.C_{8}H_{8}

7. N_{2}O_{5}

8.P_{4}O_{6}

9.C_{4}H_{8}  O_{2}

Explanation:

1.Given:

        Molar mass=216.8g

Molecular formula=PdH_{2}

       we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of PdH_{2}:

Pd=106.4×1=106.4u

H=1×2=2

molar mass of PdH_{2}=106.4+2=108.4

n=\frac{216.8}{106.4}

<u><em>n=2</em></u>

Molecular formula=2(PdH_{2})

Molecular formula=Pd_{2}H_{2}

Therefore the molecular formula of the compound is Pd_{2}H_{2}

2. Given:

        Molar mass=30.0g

Molecular formula=CH_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of CH_{3}:

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of CH_{3}=12.01 + 3 =15.01u

n=\frac{30.0}{15.01}

<u><em>n=2</em></u>

Molecular formula=2(CH_{3})

Molecular formula=C_{2} H_{6}

Therefore the molecular formula of the compound is C_{2} H_{6}

3. Given:

        Molar mass=129g

Molecular formula=CHOCl

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n=\frac{129}{64.5}

<u><em>n=2</em></u>

Molecular formula=2(CHOCl)

Molecular formula=C_{2} H_{2} O_{2} Cl_{2}

Therefore the molecular formula of the compound is C_{2} H_{2} O_{2} Cl_{2}

5. Given:

        Molar mass=577g

Molecular formula=TlC_{2} H_{2}O_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of TlC_{2} H_{2}O_{3} :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of TlC_{2} H_{2}O_{3}=204.3+24.02+1+48.00=278.32u

n=\frac{577}{278.32}

<u><em>n=2</em></u>

Molecular formula=2 (TlC_{2} H_{2}O_{3})

Molecular formula=Tl_{2 } C_{4} H_{4}O_{6}

Therefore the molecular formula of the compound is Tl_{2 } C_{4} H_{4}O_{6}

4. Molar mass=184.5g

Molecular formula=CClN

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n=\frac{184.5}{61.5}

<u><em>n=3</em></u>

Molecular formula=3 (CClN)

Molecular formula=C_{3}Cl_{3} N_{3}

Therefore the molecular formula of the compound is C_{3}Cl_{3} N_{3}

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula=C_{8}H_{8}

Molecular formula =C_{8}H_{8}

Molar mass of C_{8}H_{8}:

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of C_{8}H_{8}=96.08+8=104.08u

n=104.08÷78.0

<em><u>n=1</u></em>

Molecular formula = n(Empirical formula)

Molecular formula = 1(C_{8}H_{8})

Molecular formula =C_{8}H_{8}

Therefore the molecular formula of a compound is C_{8}H_{8}

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen=\frac{4.04}{14.01} = 0.289 moles

moles of oxygen=\frac{11.46}{15.999} =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula  of N_{2} O{5}.

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is N_{2}O_{5}.

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula=P_{2} O_{3}

molecular formula= 2(Empirical formula)

Molecular formula =2(P_{2} O_{3})

Molecular formula =P_{4}O_{6}

Therefore the molecular formula of the compound is P_{4}O_{6}

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula =C_{2} H_{4} O

Molecular formula = 2(Empirical formula)

Molecular formula =2(C_{2} H_{4} O)

Molecular formula =C_{4}H_{8}  O_{2}

Therefore the molecular formula of the compound is C_{4}H_{8}  O_{2}

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