Question

Визначте масові частки елементів у натрій оксиді, нітроген(ІІІ) оксиді, купрум(І) оксиді, сульфатній кислоті.

Answer 1

Ответ:

Объяснение:

Для визначення масових часток елементів у сполуках, спершу розглянемо склад кожної з них:

Натрій оксид (Na₂O):

Натрій (Na) має атомну масу приблизно 22,99 г/моль.

Оксиген (O) має атомну масу приблизно 16,00 г/моль.

Масова частка натрію (Na):

(2 * 22,99 г/моль) / [(2 * 22,99 г/моль) + 16,00 г/моль] ≈ 0,74 (або 74%).

Масова частка оксигену (O):

(16,00 г/моль) / [(2 * 22,99 г/моль) + 16,00 г/моль] ≈ 0,26 (або 26%).

Нітроген(ІІІ) оксид (NO):

Нітроген (N) має атомну масу приблизно 14,01 г/моль.

Оксиген (O) має атомну масу приблизно 16,00 г/моль.

Масова частка нітрогену (N):

(14,01 г/моль) / [(14,01 г/моль) + 16,00 г/моль] ≈ 0,47 (або 47%).

Масова частка оксигену (O):

(16,00 г/моль) / [(14,01 г/моль) + 16,00 г/моль] ≈ 0,53 (або 53%).

Купрум(І) оксид (Cu₂O):

Мідь (Cu) має атомну масу приблизно 63,55 г/моль.

Оксиген (O) має атомну масу приблизно 16,00 г/моль.

Масова частка міді (Cu):

(2 * 63,55 г/моль) / [(2 * 63,55 г/моль) + 16,00 г/моль] ≈ 88%.

Масова частка оксигену (O):

(16,00 г/моль) / [(2 * 63,55 г/моль) + 16,00 г/моль] ≈ 12%.

Сульфатна кислота (H₂SO₄):

Водень (H) має атомну масу приблизно 1,01 г/моль.

Сірка (S) має атомну масу приблизно 32,07 г/моль.

Оксиген (O) має атомну масу приблизно 16,00 г/моль.

Масова частка водню (H):

(2 * 1,01 г/моль) / [(2 * 1,01 г/моль) + 32,07 г/моль + (4 * 16,00 г/моль)] ≈ 2,02% (або 2%).

Масова частка сірки (S):

(32,07 г/моль) / [(2 * 1,01 г/моль) + 32,07 г/моль + (4 * 16,00 г/моль)] ≈ 32,07% (або 32%).

Масова частка оксигену (O):

[(4 * 16,00 г/моль)] / [(2 * 1,01 г/моль) + 32,07 г/моль + (4 * 16,00 г/моль)] ≈ 65,91% (або 65,91%).

Отже, ось масові частки елементів у розглянутих сполуках:

Натрій оксид (Na₂O): Na - 74%, O - 26%

Нітроген(ІІІ) оксид (NO): N - 47%, O - 53%

Купрум(І) оксид (Cu₂O): Cu - 88%, O - 12%

Сульфатна кислота (H₂SO₄): H - 2,02%, S - 32,07%, O - 65,91%

Answer 2

Mass Fractions of Elements in Sodium Oxide (Na2O):

The mass fractions of elements in sodium oxide (Na2O) can be determined by calculating the ratio of the mass of each element to the total mass of the compound.

Sodium oxide (Na2O) consists of two sodium atoms (Na) and one oxygen atom (O). The molar mass of sodium is 22.99 g/mol, and the molar mass of oxygen is 16.00 g/mol.

To calculate the mass fractions, we need to determine the molar mass of sodium oxide (Na2O). The molar mass of Na2O is calculated as follows:

Molar mass of Na2O = (2 * molar mass of Na) + (1 * molar mass of O) = (2 * 22.99 g/mol) + (1 * 16.00 g/mol) = 45.98 g/mol + 16.00 g/mol = 61.98 g/mol

Now, we can calculate the mass fractions of sodium (Na) and oxygen (O) in sodium oxide (Na2O):

Mass fraction of sodium (Na) = (2 * molar mass of Na) / molar mass of Na2O = (2 * 22.99 g/mol) / 61.98 g/mol = 45.98 g/mol / 61.98 g/mol = 0.741

Mass fraction of oxygen (O) = (1 * molar mass of O) / molar mass of Na2O = (1 * 16.00 g/mol) / 61.98 g/mol = 16.00 g/mol / 61.98 g/mol = 0.259

Therefore, the mass fraction of sodium (Na) in sodium oxide (Na2O) is approximately 0.741, and the mass fraction of oxygen (O) is approximately 0.259. [[1]]

Mass Fractions of Elements in Nitrogen (III) Oxide (N2O3):

The mass fractions of elements in nitrogen (III) oxide (N2O3) can be determined in a similar manner.

Nitrogen (III) oxide (N2O3) consists of two nitrogen atoms (N) and three oxygen atoms (O). The molar mass of nitrogen is 14.01 g/mol, and the molar mass of oxygen is 16.00 g/mol.

To calculate the mass fractions, we need to determine the molar mass of nitrogen (III) oxide (N2O3). The molar mass of N2O3 is calculated as follows:

Molar mass of N2O3 = (2 * molar mass of N) + (3 * molar mass of O) = (2 * 14.01 g/mol) + (3 * 16.00 g/mol) = 28.02 g/mol + 48.00 g/mol = 76.02 g/mol

Now, we can calculate the mass fractions of nitrogen (N) and oxygen (O) in nitrogen (III) oxide (N2O3):

Mass fraction of nitrogen (N) = (2 * molar mass of N) / molar mass of N2O3 = (2 * 14.01 g/mol) / 76.02 g/mol = 28.02 g/mol / 76.02 g/mol = 0.368

Mass fraction of oxygen (O) = (3 * molar mass of O) / molar mass of N2O3 = (3 * 16.00 g/mol) / 76.02 g/mol = 48.00 g/mol / 76.02 g/mol = 0.632

Therefore, the mass fraction of nitrogen (N) in nitrogen (III) oxide (N2O3) is approximately 0.368, and the mass fraction of oxygen (O) is approximately 0.632. [[2]]

Mass Fractions of Elements in Copper (I) Oxide (Cu2O):

The mass fractions of elements in copper (I) oxide (Cu2O) can be determined in a similar manner.

Copper (I) oxide (Cu2O) consists of two copper atoms (Cu) and one oxygen atom (O). The molar mass of copper is 63.55 g/mol, and the molar mass of oxygen is 16.00 g/mol.

To calculate the mass fractions, we need to determine the molar mass of copper (I) oxide (Cu2O). The molar mass of Cu2O is calculated as follows:

Molar mass of Cu2O = (2 * molar mass of Cu) + (1 * molar mass of O) = (2 * 63.55 g/mol) + (1 * 16.00 g/mol) = 127.10 g/mol + 16.00 g/mol = 143.10 g/mol

Now, we can calculate the mass fractions of copper (Cu) and oxygen (O) in copper (I) oxide (Cu2O):

Mass fraction of copper (Cu) = (2 * molar mass of Cu) / molar mass of Cu2O = (2 * 63.55 g/mol) / 143.10 g/mol = 127.10 g/mol / 143.10 g/mol = 0.888

Mass fraction of oxygen (O) = (1 * molar mass of O) / molar mass of Cu2O = (1 * 16.00 g/mol) / 143.10 g/mol = 16.00 g/mol / 143.10 g/mol = 0.112

Therefore, the mass fraction of copper (Cu) in copper (I) oxide (Cu2O) is approximately 0.888, and the mass fraction of oxygen (O) is approximately 0.112. [[3]]

Mass Fractions of Elements in Sulfuric Acid (H2SO4):

The mass fractions of elements in sulfuric acid (H2SO4) can be determined in a similar manner.

Sulfuric acid (H2SO4) consists of two hydrogen atoms (H), one sulfur atom (S), and four oxygen atoms (O). The molar mass of hydrogen is 1.01 g/mol, the molar mass of sulfur is 32.07 g/mol, and the molar mass of oxygen is 16.00 g/mol.

To calculate the mass fractions, we need to determine the molar mass of sulfuric acid (H2SO4). The molar mass of H2SO4 is calculated as follows:

Molar mass of H2SO4 = (2 * molar mass of H) + (1 * molar mass of S) + (4 * molar mass of O) = (2 * 1.01 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol) = 2.02 g/mol + 32.07 g/mol + 64.00 g/mol = 98.09 g/mol

Now, we can calculate the mass fractions of hydrogen (H), sulfur (S), and oxygen (O) in sulfuric acid (H2SO4):

Mass fraction of hydrogen (H) = (2 * molar mass of H) / molar mass of H2SO4 = (2 * 1.01 g/mol) / 98.09 g/mol = 2.02 g/mol / 98.09 g/mol = 0.021

Mass fraction of sulfur (S) = (1 * molar mass of S) / molar mass of H2SO4 = (1 * 32.07 g/mol) / 98.09 g/mol = 32.07 g/mol / 98.09 g/mol = 0.327

Mass fraction of oxygen (O) = (4 * molar mass of O) / molar mass of H2SO4 = (4 * 16.00 g/mol) / 98.09 g/mol = 64.00 g/mol / 98.09 g/mol = 0.653

Therefore, the mass fraction of hydrogen (H) in sulfuric acid (H2SO4) is approximately 0.021, the mass fraction of sulfur (S) is approximately 0.327, and the mass fraction of oxygen (O) is approximately 0.653. [[4]]