Question

Exercise 2.58. Part A: Determine the number of Kr atoms in a 5.55 −mg sample of krypton.

Part B:Determine the molar mass, M, of an element if the mass of a 2.80×1022−atom sample of the element is 2.09 g.

Part C: Determine the identity of an element if the mass of a 2.80×1022−atom sample of the element is 2.09 g. (Answer choices: titanium. potassium, scandium, calcium.)

Exercise 2.56. Without doing detailed calculations, indicate which of the following quantities contains the greatest number of atoms: 6.022×1023Ni atoms, 25.0 g nitrogen, 52.0 g Cr, 10.0cm3 Fe (d=7.86g/cm3). (answer choices: Ni, Nitrogen, Cr, Fe)

Excercise 2.62. A particular lead–cadmium alloy is 8.0% cadmium by mass. What mass of this alloy, in grams, must you weigh out to obtain a sample containing 6.30×10²³ Cdatoms? Express your answer using two significant figures.

Exercise 3.6 Determine the mass, in grams, of,

Part A 7.32 mol N2O4

Part B 3.22×10²⁴ O2 molecules

Part C 18.8 mol CuSO4⋅5H2O

Part D 4.14×10²⁴ molecules of C2H4(OH)2

Excercise 3.11

Part A moles of N2O4 in a 145 −g sample

Part B N atoms in 43.5 g of Mg(NO3)2

Part C N atoms in a sample of C7H5(NO2)3 that has the same number of O atoms as 12.4 g C6H12O6

thank you in advance ^^

Answer 1

Exercise 2.58. Part A: Determine the number of Kr atoms in a 5.55 −mg sample of krypton.

Solution:

The atomic weight of Kr = 83.798g/mole

Moles in 5.55mg = Mass / Atomic weight = 0.06623 moles

Number of atoms = Moles X Avagadro’s number = 0.06623 X 6.023 X 10^23

                            = 3.98 X 10^22 atoms

Part B:Determine the molar mass, M, of an element if the mass of a 2.80×1022−atom sample of the element is 2.09 g.

Solution:

Mass of 2.80×10²² atoms = 2.09grams

Mass of 6.023 X 10^23 atoms = 44.96 g

Molecular weight = 44.96g / mole

Part C: Determine the identity of an element if the mass of a 2.80×1022−atom sample of the element is 2.09 g. (Answer choices: titanium. potassium, scandium, calcium.)

Solution: It should be scandium

Exercise 2.56. Without doing detailed calculations, indicate which of the following quantities contains the greatest number of atoms: 6.022×10²³Ni atoms, 25.0 g nitrogen, 52.0 g Cr, 10.0cm3 Fe (d=7.86g/cm3). (answer choices: Ni, Nitrogen, Cr, Fe)

Solution:

6.022×10²³Ni atoms

Number of atoms in 25g Nitrogen (molecular weight of N2= 28g / mole)

= 25/28 X 6.022×10^{23 =} 5.377 X 10²³

The moles in 52g Cr = 1 moles

So number of atoms = 6.022×10²³

Mass of 10cm3 Fe = volume X density = 10cm3 X 7.86 = 78.6 grams

Atomic weight of Fe= 56g / mole

So number of atoms = 78.6 / 56 X 6.023 X 10²³ = 8.45 X 10²³

So highest number of atoms in Fe

Excercise 2.62. A particular lead–cadmium alloy is 8.0% cadmium by mass. What mass of this alloy, in grams, must you weigh out to obtain a sample containing 6.30×10²³ Cdatoms? Express your answer using two significant figures.

Solution:

The mass of cadmium in an alloy is 8%

We wish to have 6.3 X 10^23 atoms (almost one mole)

Atomic weight of Cd = 112 g / mole

Mass required = 112 grams

112 = 8% of x grams of alloy

x = mass of alloy = 1400 grams

Exercise 3.6 Determine the mass, in grams, of,

Part A 7.32 mol N2O4

Mass = Moles X molecular weight = 7.32 X 92 = 673.44 grams

Part B 3.22×10²⁴ O2 molecules

Mass of O2 = 32 X 3.22×10²⁴ / 6.023 X 10²³ = 171.1 grams

Part C 18.8 mol CuSO4⋅5H2O

Molecular weight of CuSO4.5H2O = 249.72 g / mole

Mass of 18.8moles = 249.72 X 18.8 g = 4694.736 grams

Part D 4.14×10²⁴ molecules of C2H4(OH)2

Mass of 6.023 X 10²³ = 62g

Mass of 4.14×10²⁴ molecules = 62 X 4.14×10²⁴ / 6.023 X 10²³ = 426.17 grams

Excercise 3.11

Part A moles of N2O4 in a 145 −g sample

Solution:

Molecular weight of N2O4 = 92g / mole

Moles in 145g = mass / Molecular weight = 145 / 92= 1.58 moles

Part B N atoms in 43.5 g of Mg(NO3)2

Solution:

Molecular weight of Mg(NO3)2 = 148.3 g / mole

Moles in 43.5g of Mg(NO3)2 = 43.5 / 148.3 = 0.293 moles

Moles of Nitrogen in 0.293 moles = 2 X 0.293 = 0.586 moles

Number of atoms = 0.586 X 6.023 X 10²³ = 3.53 X 10²³