What is called the molar volume of gases. The volume of one mole of gas under normal conditions

Before solving problems, you should learn the formulas and rules for how to find the volume of gas. We should remember Avogadro's law. And the gas volume itself can be calculated using several formulas, choosing the appropriate one from them. When choosing the required formula, environmental conditions, in particular temperature and pressure, are of great importance.

Avogadro's law

It says that at the same pressure and the same temperature, in the same volumes of different gases, the same number of molecules will be contained. The number of gas molecules contained in one mole is Avogadro's number. It follows from this law that: 1 Kmole (kilomole) of an ideal gas, and any, at the same pressure and temperature (760 mm Hg and t = 0 * C) always occupies one volume = 22.4136 m3.

How to determine the volume of gas

• The formula V = n * Vm can most often be found in problems. Here the volume of gas in liters is V, Vm is the molar volume of gas (l / mol), which under normal conditions = 22.4 l / mol, and n is the amount of substance in moles. When there is no amount of matter under the conditions, but at the same time there is a mass of matter, then we do this: n = m / M. Here M is g / mol (molar mass of a substance), and the mass of a substance in grams is m. In the periodic table, it is written under each element, as its atomic mass. Let's add all the masses and get the desired one.
• So how to calculate the gas volume. Here's the problem: dissolve 10 g of aluminum in hydrochloric acid. Question: how much hydrogen can be released during n. at.? The reaction equation looks like this: 2Al + 6HCl (ex) = 2AlCl3 + 3H2. At the very beginning, we find aluminum (amount), which has reacted according to the formula: n (Al) = m (Al) / M (Al). We take the mass of aluminum (molar) from the periodic table M (Al) = 27g / mol. Substitute: n (Al) = 10/27 = 0.37 mol. It can be seen from the chemical equation that 3 moles of hydrogen were formed by dissolving 2 moles of aluminum. It is necessary to calculate how much hydrogen will be released from 0.4 mol of aluminum: n (H2) = 3 * 0.37 / 2 = 0.56 mol. Let's substitute the data into the formula and find the volume of this gas. V = n * Vm = 0.56 * 22.4 = 12.54 liters.

One of the basic units in the International System of Units (SI) is the unit of the amount of substance is the mole.

Moth – this is the amount of a substance that contains as many structural units of a given substance (molecules, atoms, ions, etc.) as there are carbon atoms in 0.012 kg (12 g) of the carbon isotope 12 WITH .

Considering that the value of the absolute atomic mass for carbon is m(C) = 1.99 10  26 kg, you can calculate the number of carbon atoms N A contained in 0.012 kg of carbon.

A mole of any substance contains the same number of particles of this substance (structural units). The number of structural units contained in a substance in the amount of one mole is 6.02 10 23 and called Avogadro's number (N A ).

For example, one mole of copper contains 6.02 · 10 23 copper atoms (Cu), and one mole of hydrogen (H 2) contains 6.02 · 10 23 hydrogen molecules.

Molar mass(M) is the mass of a substance taken in an amount of 1 mol.

The molar mass is designated by the letter M and has the dimension [g / mol]. In physics, the dimension [kg / kmol] is used.

In the general case, the numerical value of the molar mass of a substance numerically coincides with the value of its relative molecular (relative atomic) mass.

For example, the relative molecular weight of water is:

Мr (Н 2 О) = 2Аr (Н) + Аr (O) = 2 ∙ 1 + 16 = 18 amu

The molar mass of water has the same value, but is expressed in g / mol:

M (H 2 O) = 18 g / mol.

Thus, a mole of water containing 6.02 · 10 23 water molecules (respectively 2 · 6.02 · 10 23 hydrogen atoms and 6.02 · 10 23 oxygen atoms) has a mass of 18 grams. In water, the amount of substance is 1 mol, contains 2 mol of hydrogen atoms and one mol of oxygen atoms.

1.3.4. The relationship between the mass of a substance and its amount

Knowing the mass of a substance and its chemical formula, and hence the value of its molar mass, it is possible to determine the amount of a substance and, conversely, knowing the amount of a substance, it is possible to determine its mass. For such calculations, you should use the formulas:

where ν is the amount of substance, [mol]; m- mass of substance, [g] or [kg]; M is the molar mass of the substance, [g / mol] or [kg / kmol].

For example, to find the mass of sodium sulfate (Na 2 SO 4) in the amount of 5 mol, we find:

1) the value of the relative molecular weight of Na 2 SO 4, which is the sum of the rounded values ​​of the relative atomic masses:

Мr (Na 2 SO 4) = 2Аr (Na) + Аr (S) + 4Аr (O) = 142,

2) the numerically equal value of the molar mass of the substance:

M (Na 2 SO 4) = 142 g / mol,

3) and, finally, the mass of 5 mol of sodium sulfate:

m = ν M = 5 mol 142 g / mol = 710 g.

Answer: 710.

1.3.5. The relationship between the volume of a substance and its amount

Under normal conditions (n.o.), i.e. at pressure R equal to 101325 Pa (760 mm Hg), and a temperature T, equal to 273.15 K (0 С), one mole of different gases and vapors occupies the same volume equal to 22.4 l.

The volume occupied by 1 mole of gas or vapor at normal conditions is called molar volumegas and has a dimension of liter per mole.

V mol = 22.4 l / mol.

Knowing the amount of gaseous substance (ν ) and molar volume value (V mol) you can calculate its volume (V) under normal conditions:

V = ν V mol,

where ν is the amount of substance [mol]; V is the volume of the gaseous substance [l]; V mol = 22.4 l / mol.

And, conversely, knowing the volume ( V) of a gaseous substance under normal conditions, you can calculate its amount (ν) :

Where m is mass, M is molar mass, V is volume.

4. Avogadro's law. Installed by the Italian physicist Avogadro in 1811. The same volumes of any gases, taken at the same temperature and the same pressure, contain the same number of molecules.

Thus, we can formulate the concept of the amount of a substance: 1 mole of a substance contains a number of particles equal to 6.02 * 10 23 (called Avogadro's constant)

The consequence of this law is that 1 mole of any gas occupies under normal conditions (P 0 = 101.3 kPa and T 0 = 298K) a volume equal to 22.4 liters.

5. Boyle-Mariotte's law

At a constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is located:

6. Gay-Lussac's law

At constant pressure, the change in gas volume is directly proportional to temperature:

V / T = const.

7. The relationship between gas volume, pressure and temperature can be expressed the combined law of Boyle-Mariotte and Gay-Lussac, which is used to bring gas volumes from one condition to another:

P 0, V 0, T 0 - pressure of volume and temperature under normal conditions: P 0 = 760 mm Hg. Art. or 101.3 kPa; T 0 = 273 K (0 0 C)

8. Independent assessment of the value of molecular masses M can be done using the so-called ideal gas equations of state or the Clapeyron-Mendeleev equation :

pV = (m / M) * RT = vRT.(1.1)

where R - gas pressure in a closed system, V- the volume of the system, T - gas mass, T - absolute temperature, R - universal gas constant.

Note that the value of the constant R can be obtained by substituting the values ​​characterizing one mole of gas at normal conditions into equation (1.1):

r = (p V) / (T) = (101.325 kPa 22.4 L) / (1 mol 273K) = 8.31J / mol.K)

Examples of problem solving

Example 1. Bringing the gas volume to normal conditions.

What volume (n.u.) will take up 0.4 × 10 -3 m 3 of gas at 50 0 С and a pressure of 0.954 × 10 5 Pa?

Solution. To bring the volume of gas to normal conditions, a general formula is used that combines the laws of Boyle-Mariotte and Gay-Lussac:

pV / T = p 0 V 0 / T 0.

The gas volume (n.a.) is equal to, where T 0 = 273 K; p 0 = 1.013 × 10 5 Pa; T = 273 + 50 = 323 K;

m 3 = 0.32 × 10 -3 m 3.

At (n.o.), the gas occupies a volume equal to 0.32 × 10 -3 m 3.

Example 2. Calculation of the relative density of a gas by its molecular weight.

Calculate the density of ethane C 2 H 6 in terms of hydrogen and air.

Solution. It follows from Avogadro's law that the relative density of one gas in another is equal to the ratio of molecular masses ( M h) of these gases, i.e. D = M 1 / M 2... If M 1 C2H6 = 30, M 2 H2 = 2, the average molecular weight of air is 29, then the relative density of ethane with respect to hydrogen is D H2 = 30/2 =15.

Relative density of ethane in air: D air= 30/29 = 1.03, i.e. ethane is 15 times heavier than hydrogen and 1.03 times heavier than air.

Example 3. Determination of the average molecular weight of a gas mixture by relative density.

Calculate the average molecular weight of a gas mixture of 80% methane and 20% oxygen (by volume) using the relative hydrogen densities of these gases.

Solution. Calculations are often made according to the mixing rule, which is that the ratio of the volumes of gases in a two-component gas mixture is inversely proportional to the differences between the density of the mixture and the densities of the gases that make up this mixture. Let us denote the relative density of the gas mixture with respect to hydrogen through D H2. it will be greater than the density of methane, but less than the density of oxygen:

80D H2 - 640 = 320 - 20 D H2; D H2 = 9.6.

The hydrogen density of this gas mixture is 9.6. average molecular weight of a gas mixture M H2 = 2 D H2 = 9.6 × 2 = 19.2.

Example 4. Calculation of the molar mass of a gas.

The mass of 0.327 × 10 -3 m 3 of gas at 13 0 C and a pressure of 1.040 × 10 5 Pa is 0.828 × 10 -3 kg. Calculate the molar mass of the gas.

Solution. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:

where m- gas mass; M- molar mass of gas; R- molar (universal) gas constant, the value of which is determined by the accepted units of measurement.

If the pressure is measured in Pa, and the volume in m 3, then R= 8.3144 × 10 3 J / (kmol × K).

Where m is mass, M is molar mass, V is volume.

4. Avogadro's law. Installed by the Italian physicist Avogadro in 1811. The same volumes of any gases, taken at the same temperature and the same pressure, contain the same number of molecules.

Thus, we can formulate the concept of the amount of a substance: 1 mole of a substance contains a number of particles equal to 6.02 * 10 23 (called Avogadro's constant)

The consequence of this law is that 1 mole of any gas occupies under normal conditions (P 0 = 101.3 kPa and T 0 = 298K) a volume equal to 22.4 liters.

5. Boyle-Mariotte's law

At a constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is located:

6. Gay-Lussac's law

At constant pressure, the change in gas volume is directly proportional to temperature:

V / T = const.

7. The relationship between gas volume, pressure and temperature can be expressed the combined law of Boyle-Mariotte and Gay-Lussac, which is used to bring gas volumes from one condition to another:

P 0, V 0, T 0 - pressure of volume and temperature under normal conditions: P 0 = 760 mm Hg. Art. or 101.3 kPa; T 0 = 273 K (0 0 C)

8. Independent assessment of the value of molecular masses M can be done using the so-called ideal gas equations of state or the Clapeyron-Mendeleev equation :

pV = (m / M) * RT = vRT.(1.1)

where R - gas pressure in a closed system, V- the volume of the system, T - gas mass, T - absolute temperature, R - universal gas constant.

Note that the value of the constant R can be obtained by substituting the values ​​characterizing one mole of gas at normal conditions into equation (1.1):

r = (p V) / (T) = (101.325 kPa 22.4 L) / (1 mol 273K) = 8.31J / mol.K)

Examples of problem solving

Example 1. Bringing the gas volume to normal conditions.

What volume (n.u.) will take up 0.4 × 10 -3 m 3 of gas at 50 0 С and a pressure of 0.954 × 10 5 Pa?

Solution. To bring the volume of gas to normal conditions, a general formula is used that combines the laws of Boyle-Mariotte and Gay-Lussac:

pV / T = p 0 V 0 / T 0.

The gas volume (n.a.) is equal to, where T 0 = 273 K; p 0 = 1.013 × 10 5 Pa; T = 273 + 50 = 323 K;

M 3 = 0.32 × 10 -3 m 3.

At (n.o.), the gas occupies a volume equal to 0.32 × 10 -3 m 3.

Example 2. Calculation of the relative density of a gas by its molecular weight.

Calculate the density of ethane C 2 H 6 in terms of hydrogen and air.

Solution. It follows from Avogadro's law that the relative density of one gas in another is equal to the ratio of molecular masses ( M h) of these gases, i.e. D = M 1 / M 2... If M 1 C2H6 = 30, M 2 H2 = 2, the average molecular weight of air is 29, then the relative density of ethane with respect to hydrogen is D H2 = 30/2 =15.

Relative density of ethane in air: D air= 30/29 = 1.03, i.e. ethane is 15 times heavier than hydrogen and 1.03 times heavier than air.

Example 3. Determination of the average molecular weight of a gas mixture by relative density.

Calculate the average molecular weight of a gas mixture of 80% methane and 20% oxygen (by volume) using the relative hydrogen densities of these gases.

Solution. Calculations are often made according to the mixing rule, which is that the ratio of the volumes of gases in a two-component gas mixture is inversely proportional to the differences between the density of the mixture and the densities of the gases that make up this mixture. Let us denote the relative density of the gas mixture with respect to hydrogen through D H2. it will be greater than the density of methane, but less than the density of oxygen:

80D H2 - 640 = 320 - 20 D H2; D H2 = 9.6.

The hydrogen density of this gas mixture is 9.6. average molecular weight of a gas mixture M H2 = 2 D H2 = 9.6 × 2 = 19.2.

Example 4. Calculation of the molar mass of a gas.

The mass of 0.327 × 10 -3 m 3 of gas at 13 0 C and a pressure of 1.040 × 10 5 Pa is 0.828 × 10 -3 kg. Calculate the molar mass of the gas.

Solution. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:

where m- gas mass; M- molar mass of gas; R- molar (universal) gas constant, the value of which is determined by the accepted units of measurement.

If the pressure is measured in Pa, and the volume in m 3, then R= 8.3144 × 10 3 J / (kmol × K).

3.1. When performing measurements of atmospheric air, air in the working area, as well as industrial emissions and hydrocarbons in gas lines, there is a problem of bringing the volumes of measured air to normal (standard) conditions. Often, in practice, when measuring air quality, the conversion of measured concentrations to normal conditions is not used, as a result of which unreliable results are obtained.

Here is an excerpt from the Standard:

“The measurements are brought to standard conditions using the following formula:

C 0 = C 1 * P 0 T 1 / P 1 T 0

where: С 0 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume of air, mol / cu. m, at standard temperature and pressure;

С 1 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume

air, mol / cu. m, at temperature T 1, K, and pressure P 1, kPa. "

The formula for reduction to normal conditions in a simplified form has the form (2)

C 1 = C 0 * f, where f = P 1 T 0 / P 0 T 1

standard conversion factor for normalization. The parameters of air and impurities are measured at different values ​​of temperature, pressure and humidity. The results lead to standard conditions for comparing measured air quality parameters in different locations and different climates.

3.2 Industry normal conditions

Normal conditions are the standard physical conditions to which the properties of substances (Standard temperature and pressure, STP) are usually related. Reference conditions are defined by IUPAC (International Union of Practical and Applied Chemistry) as follows: Atmospheric pressure 101325 Pa = 760 mm Hg. Air temperature 273.15 K = 0 ° C.

Standard conditions (Standard Ambient Temperature and Pressure, SATP) are the normal ambient temperature and pressure: pressure 1 Bar = 10 5 Pa = 750.06 mm T. st .; temperature 298.15 K = 25 ° C.

Other areas.

Air quality measurements.

The results of measuring the concentrations of harmful substances in the air of the working area lead to the following conditions: temperature 293 K (20 ° C) and pressure 101.3 kPa (760 mm Hg).

Aerodynamic parameters of pollutant emissions must be measured in accordance with applicable government standards. The volumes of waste gases obtained from the results of instrumental measurements should be brought to normal conditions (n.o.): 0 ° С, 101.3 kPa ..

Aviation.

The International Civil Aviation Organization (ICAO) defines an International Standard Atmosphere (ISA) at sea level with a temperature of 15 ° C, an atmospheric pressure of 101,325 Pa, and a relative humidity of 0%. These parameters are used when calculating the movement of aircraft.

Gas facilities.

The gas industry of the Russian Federation, when making settlements with consumers, uses atmospheric conditions in accordance with GOST 2939-63: temperature 20 ° C (293.15K); pressure 760 mm Hg. Art. (101325 N / m²); humidity is 0. Thus, the mass of a cubic meter of gas according to GOST 2939-63 is slightly less than under "chemical" normal conditions.

Testing

To test machines, instruments and other technical products, the following are taken for normal values ​​of climatic factors during product testing (normal climatic test conditions):

Temperature - plus 25 ° ± 10 ° С; Relative humidity - 45-80%

Atmospheric pressure 84-106 kPa (630-800 mm Hg)

Verification of measuring instruments

The nominal values ​​of the most common normal influencing quantities are chosen as follows: Temperature - 293 K (20 ° C), atmospheric pressure - 101.3 kPa (760 mm Hg).

Rationing

In the guidelines for the establishment of air quality standards, it is indicated that MPCs in the ambient air are set under normal indoor conditions, i.e. 20 C and 760 mm. rt. Art.

Acid names are formed from the Russian name for the central acid atom with the addition of suffixes and endings. If the oxidation state of the central acid atom corresponds to the group number of the Periodic Table, then the name is formed using the simplest adjective from the name of the element: H 2 SO 4 - sulfuric acid, HMnO 4 - manganic acid. If acid-forming elements have two oxidation states, then the intermediate oxidation state is indicated by the suffix -ist-: H 2 SO 3 - sulfurous acid, HNO 2 - nitrous acid. Various suffixes are used for the names of halogen acids having many oxidation states: typical examples - HClO 4 - chlorine n th acid, HClO 3 - chlorine novat th acid, HClO 2 - chlorine ist acid, HClO - chlorine novatist th acid (anoxic acid HCl is called hydrochloric acid - usually hydrochloric acid). Acids can differ in the number of water molecules that hydrate the oxide. The acids containing the largest number of hydrogen atoms are called ortho acids: H 4 SiO 4 - orthosilicic acid, H 3 PO 4 - orthophosphoric acid. Acids containing 1 or 2 hydrogen atoms are called metacids: H 2 SiO 3 - metasilicic acid, HPO 3 - metaphosphoric acid. Acids containing two central atoms are called di acids: H 2 S 2 O 7 - disulfuric acid, H 4 P 2 O 7 - diphosphoric acid.

Complex names are formed in the same way as salt names, but the complex cation or anion is given a systematic name, that is, it is read from right to left: K 3 - potassium hexafluoroferrate (III), SO 4 - tetraammine copper (II) sulfate.

Oxides names are formed using the word "oxide" and the genitive case of the Russian name for the central atom of the oxide, indicating, if necessary, the oxidation state of the element: Al 2 O 3 - aluminum oxide, Fe 2 O 3 - iron (III) oxide.

Base names are formed using the word "hydroxide" and the genitive case of the Russian name for the central hydroxide atom indicating, if necessary, the oxidation state of the element: Al (OH) 3 - aluminum hydroxide, Fe (OH) 3 - iron (III) hydroxide.

Hydrogen compound names are formed depending on the acid-base properties of these compounds. For gaseous acid-forming compounds with hydrogen, the following names are used: H 2 S– sulfane (hydrogen sulfide), H 2 Se– selane (hydrogen selenide), HI– hydrogen iodide; their solutions in water are called, respectively, hydrosulfuric, hydroselenic and hydroiodic acids. For some compounds with hydrogen, special names are used: NH 3 - ammonia, N 2 H 4 - hydrazine, PH 3 - phosphine. Compounds with hydrogen having an oxidation state of –1 are called hydrides: NaH - sodium hydride, CaH 2 - calcium hydride.

Salt names are formed from the Latin name of the central atom of the acid residue with the addition of prefixes and suffixes. The names of binary (two-element) salts are formed using the suffix - id: NaCl - sodium chloride, Na 2 S - sodium sulfide. If the central atom of the oxygen-containing acid residue has two positive oxidation states, then the highest oxidation state is indicated by the suffix - at: Na 2 SO 4 - sulf at sodium, KNO 3 - nitr at potassium, and the lowest oxidation state - with the suffix - it: Na 2 SO 3 - sulf it sodium, KNO 2 - nitr it potassium. Prefixes and suffixes are used to name oxygenated halogen salts: KClO 4 - lane chlorine at potassium, Mg (ClO 3) 2 - chlorine at magnesium, KClO 2 - chlorine it potassium, KClO - hypo chlorine it potassium.

Saturation is covalentoopsconnectionher- it is manifested in the fact that there are no unpaired electrons in the compounds of s- and p-elements, that is, all unpaired electrons of atoms form bonding electron pairs (the exceptions are NO, NO 2, ClO 2 and ClO 3).

Lonely electron pairs (LEP) are electrons that occupy atomic orbitals in pairs. The presence of LEP determines the ability of anions or molecules to form donor-acceptor bonds as donors of electron pairs.

Unpaired electrons are the electrons of an atom contained one at a time in the orbital. For s- and p-elements, the number of unpaired electrons determines how many bonding electron pairs a given atom can form with other atoms by the exchange mechanism. In the method of valence bonds, it is assumed that the number of unpaired electrons can be increased due to lone electron pairs if there are vacant orbitals within the valence electronic level. In most compounds of the s- and p-elements, there are no unpaired electrons, since all unpaired electrons of the atoms form bonds. However, molecules with unpaired electrons exist, for example NO, NO 2, they are highly reactive and tend to form dimers of the N 2 O 4 type at the expense of unpaired electrons.

Normal concentration - this is the number of moles equivalents in 1 liter of solution.

Normal conditions - temperature 273K (0 o C), pressure 101.3 kPa (1 atm).

Exchange and donor-acceptor mechanisms of chemical bond formation... The formation of covalent bonds between atoms can occur in two ways. If the formation of a bonding electron pair occurs due to the unpaired electrons of both bound atoms, then this method of forming a bonding electron pair is called the exchange mechanism - atoms exchange electrons, and the bonding electrons belong to both bound atoms. If the bonding electron pair is formed due to the lone electron pair of one atom and the vacant orbital of another atom, then this formation of the bonding electron pair is a donor-acceptor mechanism (see. method of valence bonds).

Reversible ionic reactions - these are reactions in which products are formed that can form initial substances (if we bear in mind the written equation, then about reversible reactions we can say that they can proceed in both directions with the formation of weak electrolytes or poorly soluble compounds). Reversible ionic reactions are often characterized by incomplete conversion; since during a reversible ionic reaction, molecules or ions are formed, which cause a shift towards the initial products of the reaction, that is, as it were, "inhibit" the reaction. Reversible ionic reactions are described using the ⇄ sign, and irreversible ones - by the → sign. An example of a reversible ionic reaction is the reaction H 2 S + Fe 2+ ⇄ FeS + 2H +, and an example of an irreversible reaction is S 2- + Fe 2+ → FeS.

Oxidants – substances in which oxidation states of some elements decrease during redox reactions.

Redox duality - the ability of substances to act in redox reactions as an oxidizing or reducing agent, depending on the partner (for example, H 2 O 2, NaNO 2).

Redox reactions(OVR) - these are chemical reactions during which the oxidation states of the elements of the reacting substances change.

Redox potential - a value characterizing the redox capacity (strength) of both the oxidizing agent and the reducing agent making up the corresponding half-reaction. Thus, the redox potential of the Cl 2 / Cl - pair, equal to 1.36 V, characterizes molecular chlorine as an oxidizing agent and chloride ion as a reducing agent.

Oxides - compounds of elements with oxygen, in which oxygen has an oxidation state equal to –2.

Orientation interactions- intermolecular interactions of polar molecules.

Osmosis - the phenomenon of transfer of solvent molecules on a semi-permeable (permeable only for the solvent) membrane towards a lower concentration of the solvent.

Osmotic pressure - physicochemical property of solutions due to the ability of membranes to pass only solvent molecules. Osmotic pressure from the side of a less concentrated solution equalizes the rates of penetration of solvent molecules into both sides of the membrane. The osmotic pressure of the solution is equal to the pressure of the gas, in which the concentration of molecules is the same as the concentration of particles in the solution.

Arrhenius bases - substances that in the process of electrolytic dissociation split off hydroxide ions.

Bronsted bases - compounds (molecules or ions of the type S 2-, HS -) that can add hydrogen ions.

Foundations according to Lewis (Lewis bases) – compounds (molecules or ions) with lone electron pairs capable of forming donor-acceptor bonds. The most common Lewis base is water molecules, which have strong donor properties.