Question

Даны формулы веществ: SO2, Ca(OH)2,FeCl2, HNO3, ZnO, HCl, CuO, Na2SO4. Выпишите отдельно формулы оксидов, оснований кислот и солей. 2. Напишите уравнения реакций, при помощи которых можно осуществить следующие превращения: Са → СаО→ СаСl2 →СаСО3 ↓Са(ОН)2 → Са(NO)2 3. Охарактеризуйте по плану элемент №15. 4. Как изменяются (усиливаются или ослабляются) металлические (основные) свойства химических элементов в ряду: Na →Mg → Al →Si? Объясните причину. 5. Задача. Какой объем кислорода потребуется для окисления 2,7 грамм алюминия. 4Al + 3O2 = 2 Al2O3

Answer 1

2. 1) 2Ca + O2 = 2CaO

2) CaO + 2HCl = CaCl2 + H2O

3) CaCl2 + K2CO3 = 2KCl + CaCO3

4) Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O ( Ca(NO)2 нет)

3. -

4. Ослабевают, т.к. металлические свойства увеличивается справа на лево и сверху вниз

5. 4Al + 3O2 = 2Al2O3

m Al = 2,7 грамм

n al = 2,7 / 27 = 0,1 моль

n O2 = 0,1 * 3 / 4 = 0,075 моль

V O2 = 0,075 * 22,4 = 1,68л

Answer 2

Formulas of Substances:

The given formulas of substances are: - SO2: sulfur dioxide - Ca(OH)2: calcium hydroxide - FeCl2: iron(II) chloride - HNO3: nitric acid - ZnO: zinc oxide - HCl: hydrochloric acid - CuO: copper(II) oxide - Na2SO4: sodium sulfate

Formulas of Oxides, Bases, Acids, and Salts:

- Oxides: SO2, ZnO, CuO - Bases: Ca(OH)2 - Acids: HNO3, HCl - Salts: FeCl2, Na2SO4

Equations for the Given Transformations:

1. Transformation: Ca → CaO → CaCl2 → CaCO3 ↓ Ca(OH)2 → Ca(NO)2 - Equation 1: Ca → CaO (calcium reacts with oxygen to form calcium oxide) - Equation 2: CaO + 2HCl → CaCl2 + H2O (calcium oxide reacts with hydrochloric acid to form calcium chloride and water) - Equation 3: CaCl2 + Na2CO3 → CaCO3 + 2NaCl (calcium chloride reacts with sodium carbonate to form calcium carbonate and sodium chloride) - Equation 4: CaCO3 + 2HCl → CaCl2 + H2O + CO2 (calcium carbonate reacts with hydrochloric acid to form calcium chloride, water, and carbon dioxide) - Equation 5: CaCl2 + 2NaOH → Ca(OH)2 + 2NaCl (calcium chloride reacts with sodium hydroxide to form calcium hydroxide and sodium chloride) - Equation 6: Ca(OH)2 + 2HNO3 → Ca(NO)2 + 2H2O (calcium hydroxide reacts with nitric acid to form calcium nitrate and water)

2. Characterization of Element #15: Element #15 refers to phosphorus (P). Phosphorus is a nonmetallic element that belongs to Group 15 (Group VA) of the periodic table. It has an atomic number of 15 and an atomic mass of approximately 30.97 atomic mass units. Phosphorus exists in several allotropes, including white phosphorus, red phosphorus, and black phosphorus. It is highly reactive and can form various compounds, including phosphates, phosphides, and phosphoric acids. Phosphorus is an essential element for living organisms and is found in DNA, RNA, and ATP.

3. Changes in Metallic (Basic) Properties in the Sodium (Na) → Magnesium (Mg) → Aluminum (Al) → Silicon (Si) Series: In the given series, as we move from sodium (Na) to magnesium (Mg) to aluminum (Al) to silicon (Si), the metallic (basic) properties of the elements generally weaken. This is due to the following reasons: - Atomic size: As we move across the period from left to right, the atomic size decreases. This decrease in atomic size leads to a decrease in metallic character because the outermost electrons are held more tightly by the nucleus, making it more difficult for the atoms to lose electrons and exhibit metallic properties. - Electronegativity: Electronegativity increases across the period. Elements with higher electronegativity tend to attract electrons more strongly, making it harder for them to lose electrons and exhibit metallic properties. - Ionization energy: Ionization energy generally increases across the period. Elements with higher ionization energy require more energy to remove electrons, making it more difficult for them to exhibit metallic properties.

4. Calculation of Oxygen Volume Required for the Oxidation of 2.7 grams of Aluminum: The balanced equation for the reaction between aluminum and oxygen is: 4Al + 3O2 → 2Al2O3

From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

To calculate the volume of oxygen required, we need to convert the mass of aluminum to moles using its molar mass and then use the mole ratio from the balanced equation.

The molar mass of aluminum (Al) is approximately 26.98 g/mol.

Calculation: - Moles of aluminum = mass of aluminum / molar mass of aluminum - Moles of aluminum = 2.7 g / 26.98 g/mol ≈ 0.100 moles

From the balanced equation, we know that 4 moles of aluminum react with 3 moles of oxygen.

- Moles of oxygen = (moles of aluminum / 4) * 3 - Moles of oxygen = (0.100 moles / 4) * 3 ≈ 0.075 moles

Since 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP), we can calculate the volume of oxygen required.

- Volume of oxygen = moles of oxygen * 22.4 L/mol - Volume of oxygen = 0.075 moles * 22.4 L/mol ≈ 1.68 liters

Therefore, approximately 1.68 liters of oxygen would be required for the oxidation of 2.7 grams of aluminum.