Substance characteristics

1.Which physical quantities characterize a substance?

Solution:

Atom ^A $Z X$

– proton number (number of protons in the nucleus, number of electrons in the shell, the order of the element in the Mendeleev table)
$A$ – nucleon number (number of protons and neutrons in the nucleus)
$N = A - Z$ – number of neutrons in the nucleus

m_a(X) \,=\, \text{rest mass of the atom} \quad (m_a(^{12}_{6}C) = 1.9926 \times 10^{-26} \, \text{kg})

m_u = \text{atomic mass constant} \quad m_u = \tfrac{1}{12} m_a(^{12}_{6}C) = 1.6605 \times 10^{-27} \, \text{kg}

Relative atomic mass

A_r = \frac{m_a}{m

Relative molecular mass

M_r = \frac{M_m}{m_u}

Amount of substance $n$ – characterizes a substance based on the number of particles in it

n = \frac{N}{N_A}, \quad [n] = 1 \, \text{mol}

1 mol – the unit of the amount of substance. One mole is also the amount of substance that contains as many particles as there are atoms in $0.012 \, \text{kg}$ of ${}^{12}_{6}C$ .

Avogadro’s number:

N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \quad (\text{number of particles in one mole of substance})

Molar mass:

M = \frac{m}{n} = M_r \cdot 10^{-3}, \quad [M] = \text{kg·mol}^{-1}

Molar volume:

V_m = \frac{V}{n}, \quad [V_m] = \text{m}^3 \cdot \text{mol}^{-1}

2.Determine the rest masses of atoms of the elements H, C, Zn. What is the relative mass of Uranium if its rest mass is m_a(U) = 3.95·10^-25 kg. Use the tables.

Solution:

The relative mass of Uranium is approximately A_r(U) = 238

3.Estimate the approximate number of atoms contained in an iron weight with mass 1 kg. How long would a line be if all these atoms were arranged tightly next to each other in a single straight line? The diameter of an atom is about 10^-10 m.

Solution:

Analysis:

m = 1 kg, d = 10^-10 m, m_u = 1.66·10^-27 kg, A_r(Fe) = 55.847

In one kilogram of iron there are 10²⁵ atoms. The length of the line formed by these atoms would be 10¹² km. This is about 6600 times greater than the distance Earth – Sun.

4. Determine the number of electrons contained in 1 g of copper. What is their total mass if the mass of a single electron is 9.1·10^-31 kg?

Solution:

Analysis:

m = 1 g = 10^-3 kg, m_e = 9.1·10^-31 kg, m_u = 1.66·10^-27 kg, A_r(₂₉Cu) = 63.546

Rest mass of a Cu atom:

m_a(Cu) = m_u·A_r(Cu)

m_a(Cu) = 1.66·10^-27 kg · 63.546 = 105.486·10^-27 kg

Number of atoms:

Number of electrons:

₂₉Cu – each atom has 29 electrons

N‘ = 29·N = 29·9.48·10²¹ = 2.75·10²³

Total mass of all electrons:

m = N‘·m_e = 2.75·10²³ · 9.1·10^-31 kg = 25·10^-8 kg

In one gram of copper there are 2.75·10²³ electrons, which have a mass of 25·10^-8 kg.

5.Find out whether a molecule of nitric acid HNO₃ has a greater mass than a molecule of silver oxide Ag₂O. Use tables.

Solution:

HNO₃

A_r(H) = 1.008

A_r(N) = 14.010

3·A_r(O) = 48.000

A_r(HNO₃) = 63.018 m(HNO₃) = 63.018·1.66·10^-27 kg

m(HNO₃) = 1.46·10^-25 kg

Ag₂O

2·A_r(Ag) = 215.74

A_r(O) = 16.00

A_r(Ag₂O) = 231.74 m(Ag₂O) = 231.74·1.66·10^-27 kg

m(Ag₂O) = 3.8469·10^-25 kg

m(Ag₂O) > m(HNO₃)

Assumption is not valid.

6.Calculate the amount of substance corresponding to 4.82·10²⁴ hydrogen atoms.

Solution:

Analysis:

N = 4.82·10²⁴ hydrogen atoms, N_A = 6.022·10²³ mol^-1

physics-substance-characteristics-6.gif physics-substance-characteristics-6.gif

The given number of hydrogen atoms corresponds to n = 8 moles.

7.Determine the amount of substance of water with volume 3.6 L.

Solution:

Analysis:

V = 3.6 L, m = 3.6 kg,

2·A_r(H) = 2.016

A_r(O) = 16.000

A_r(H₂O) = 18.016

M_m = 18.016·10^-3 kg·mol^-1

The volume 3.6 liters of water is 200 moles.

8.Find out whether water with amount of substance 1 mol can be poured into a cylinder with volume 10 cm³.

Solution:

Analysis:

n = 1 mol, V = 10 cm³, M_m = 18.016·10^-3 kg (see example 7), ρ = 10³ kg·m^-3

Volume of one mole of water:

Volume of the cylinder V_V = 10 cm³

V_M > V_V

Water cannot be poured into the cylinder because the volume of one mole of water is 18 cm³ and the container only has 10 cm³.

9. A room has dimensions a = 4 m, b = 4 m, c = 3 m. How many molecules of air are in it? (M_m = 29·10^-3 kg·mol^-1, ρ = 1.276 kg·m^-3)

Solution:

Analysis:

V = a·b·c

V = 4 m · 4 m · 3 m = 48 m³,

M_m = 29·10^-3 kg·mol^-1, ρ = 1.276 kg·m^-3

There are approximately 1.3·10²⁷ molecules in the room.

10. From the surface of a water droplet with volume 1 mm³, water containing about 10⁶ molecules evaporates in 1 second. In what time will the entire droplet evaporate?

Solution:

Analysis:

V = 1 mm³ = 10^-9 m³, N‘ = 10¹⁶ s^-1, M_m(H₂O) = 18.016·10^-3 kg·mol^-1

ρ = 1000 kg·m^-3, N_A = 6.022·10²³ mol^-1

The whole droplet will evaporate in just under 1 hour.

11. Determine the relative molecular mass of carbon dioxide CO₂ and the rest mass of the CO₂ molecule. Use tables. (m_u = 1.6605·10^-27 kg)

Solution:

1.) M_r(CO₂) = 1C = 1·12.01 = 12.01

2O = 2·15.999 = 31.998

M_r(CO₂) = 12.01 + 31.998 = 44.008 = 44.01

M_r(CO₂) = 44.01

latkove-charakteristiky-11

12. A closed container placed in a vacuum contains gaseous carbon dioxide with a mass of 1.1 kg. On average, N‘ = 1.5·10²⁰ molecules escape from the container every second. After what time will all CO₂ molecules escape from the container?

Solution:

Analysis:

m = 1.1 kg, M_r(CO₂) = 44.01, N‘ = 1.5·10²⁰, N = ?, t = ?

1.) Total number of molecules in the container at the beginning of the experiment:

latkove-charakteristiky-12a

2.) Time in which all molecules escape from the container:

latkove-charakteristiky-12b

All CO₂ molecules will escape from the container in about 28 hours.

13.Calculate the molar volume of lead.

Solution:

latkove-charakteristiky-13r

The molar volume of lead is V_m = 1.833·10^–5 m³·mol^–1.

14.The diameter of an oxygen molecule O₂ is 0.364 nm. How long would the line be if all molecules contained in one mole of oxygen were arranged tightly next to each other?

Solution:

latkove-charakteristiky-14r

The line would be 2.19·10¹¹ km long.

15. A container with a volume of 2 liters contains 2.74·10¹⁹ molecules of chlorine Cl₂. Calculate the molecular density N_V of this gas.

Solution:

Analysis:

latkove-charakteristiky-15z

latkove-charakteristiky-15r

The molecular density of chlorine is N_V = 1.37·10²² m^-3

16.Calculate the root-mean-square speed v_k of a chlorine molecule Cl₂ at a temperature of 100°C.

Solution:

Analysis:

latkove-charakteristiky-16z

latkove-charakteristiky-16r

The root-mean-square speed of a chlorine molecule is v_k = 362 m·s^-1

17.Calculate the average kinetic energy of one molecule of an ideal gas at a temperature t = –20°C. (k = 1.38·10^-23 J·K^-1)

Solution:

latkove-charakteristiky-17

18.Determine the diameter of a water molecule H₂O.

latkove-charakteristiky-18z

Solution:

Analysis:

latkove-charakteristiky-18r

The diameter of a water molecule is 3.852·10^–10 m.

19. A sample of oxygen O₂ with a mass of 5 kg has a volume of 3.54 m³ at a temperature of 0°C and a pressure of 0.1 MPa. Determine the molar volume of oxygen.

Solution:

Analysis:

latkove-charakteristiky-19r

The molar volume of oxygen O₂ is V_m = 22.65 m³·kmol^-1.

20.Calculate how many molecules are contained in hydrogen H₂ with an amount of substance n₁ = 10 mol and in oleic acid C₁₇H₃₃COOH with the same amount of substance n₂ = 10 mol.

Solution:

latkove-charakteristiky-20

In equal amounts of substance ( n₁ = n₂ ) there is always the same number of molecules.

Physics

Substance characteristics

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