Question

А1.Электронная формула внешнего энергетического уровня иона S2-: А) 3s23p2 Б) 3s23p4 В) 3s23p6 Г) 4s24p6 А2. В ряду химических элементов F- Cl – S – P неметаллические свойства: А) Уменьшаются В) Не изменяются Б) Увеличиваются Г) Изменяются периодически А3.Азот не может проявлять в соединениях степень окисления: А) +5 Б) +4 В) -3 Г) -4 А4. Восстановительные свойства простых веществ от кислорода к теллуру: А) усиливаются Б) ослабевают В) изменяются периодически Г) не изменяются А5. Взаимодействию оксида серы (VI) с кислородом соответствует схема превращения: А) S+4 → S +6 Б) S+4→ S 0 В) S0→ S +4 Г) S0 → S -2 А6.Неметаллические свойства элементов усиливаются в ряду: А) Cl – S – Se - As В) O – S – Se - Br Б) F – Cl – Br - I Г) As – P – S - O А7. Атомную кристаллическую решетку имеет: А) Na2O; Б) SiO2; В) CaF2; Г) H2O. А8. Массовая доля кислорода в гидроксиде кальция равна: А) 34% Б) 43% В) 27% Г) 46% Часть В В1. Оксид серы (IV) взаимодействует с: 1) Сероводородом 4) Азотом 2) Концентрированной серной кислотой 5) Азотной кислотой 3) Раствором карбоната натрия 6) Оксидом углерода (IV) В2. Установите соответствие между схемой реакции и формулой недостающего вещества: Схема реакции Формула вещества А) Сr + …→ Cr(NO3)2 + NO2 + H2O 1) H2SО4 (разб) Б) Cr + …→ Cr2(SO4)3 + SO2 + H2O 2) H2SО4 (конц) В) CrO + … → CrSO4 + H2O 3) Сr(NO3)2 Г) CrO + HNO3(конц) → … + NO2 + H2O 4) Сr(NO3)3 5) HNO3(разб) 6) HNO3(конц) Часть С С1. Напишите уравнения реакций, с помощью которых можно осуществить следующие превращения: Ca3(PO4)2 → P→ H3PO4 →K3PO4 С2. К 200 г 15 %-го раствора сульфата железа (III) добавили избыток гидроксида лития. Вычислите массу образовавшегося осадка.

Answer 1

А
1)в, 2)б,3)г,4)а,5)-,6)г,7)б,8)б
В
1)4
2)
А-6
Б-2
В-1
Г-3
С
1)2Ca3(PO4)2 + 6SiO2 + 10C = 6CaSiO3 + P4 +10CO
2) Fe2(SO4)3+ 6LiOH=2Fe(OH)3+ 3Li2SO4
если 15 г в 100г,то 30 г в 200г раствора.
х=30·214(М= 2Fe(OH)3)= 27,91304
230(М=Fe2(SO4)3)
Вроде все

Answer 2

Question 1: The electronic formula of the outer energy level of the S2- ion is: A) 3s^2 3p^2 B) 3s^2 3p^4 C) 3s^2 3p^6 D) 4s^2 4p^6

The correct answer is C) 3s^2 3p^6. The S2- ion has gained two electrons, resulting in a complete 3s and 3p subshell with a total of 8 electrons.

Question 2: In the series of chemical elements F, Cl, S, and P, the non-metallic properties: A) Decrease B) Remain unchanged C) Increase D) Periodically change

The correct answer is A) Decrease. As we move from left to right across the periodic table, the non-metallic properties tend to decrease. Fluorine (F) is the most non-metallic element, while phosphorus (P) is less non-metallic compared to chlorine (Cl) and sulfur (S).

Question 3: Nitrogen cannot exhibit the oxidation state of: A) +5 B) +4 C) -3 D) -4

The correct answer is B) +4. Nitrogen can exhibit oxidation states of -3, -2, -1, 0, +1, +2, +3, +4, and +5. However, it cannot exhibit an oxidation state of +4.

Question 4: The reducing properties of simple substances from oxygen to tellurium: A) Increase B) Decrease C) Periodically change D) Remain unchanged

The correct answer is C) Periodically change. As we move from left to right across the periodic table, the reducing properties of simple substances tend to change periodically. Oxygen is a strong oxidizing agent, while tellurium has weaker reducing properties compared to oxygen.

Question 5: The interaction of sulfur oxide (VI) with oxygen corresponds to the transformation scheme: A) S^4+ → S^6+ B) S^4+ → S^0 C) S^0 → S^4+ D) S^0 → S^-2

The correct answer is B) S^4+ → S^0. The interaction of sulfur oxide (VI) with oxygen results in the reduction of sulfur from the +6 oxidation state to the 0 oxidation state.

Question 6: The non-metallic properties of elements are enhanced in the series: A) Cl, S, Se, As B) O, S, Se, Br C) F, Cl, Br, I D) As, P, S, O

The correct answer is A) Cl, S, Se, As. As we move down a group in the periodic table, the non-metallic properties tend to increase. Chlorine (Cl) is more non-metallic compared to sulfur (S), selenium (Se), and arsenic (As).

Question 7: The atomic crystal lattice is present in: A) Na2O B) SiO2 C) CaF2 D) H2O

The correct answer is C) CaF2. CaF2 has an atomic crystal lattice structure.

Question 8: The mass fraction of oxygen in calcium hydroxide is: A) 34% B) 43% C) 27% D) 46%

The correct answer is A) 34%. The mass fraction of oxygen in calcium hydroxide (Ca(OH)2) is 34%.

Part B

B1: Sulfur oxide (IV) interacts with: 1) Hydrogen sulfide 2) Concentrated sulfuric acid 3) Sodium carbonate solution 4) Carbon dioxide 5) Nitrogen 6) Nitric acid

The correct answer is 4) Carbon dioxide. Sulfur oxide (IV) can react with carbon dioxide.

B2: Match the reaction scheme with the formula of the missing substance: Reaction Scheme | Substance Formula ----------------|----------------- A) Cr^3+... → Cr(NO3)2^+...NO2^+...H2O | 1) H2SO4 (diluted) B) Cr^3+... → Cr2(SO4)3^+...SO2^+...H2O | 2) H2SO4 (concentrated) C) CrO3^+... → CrSO4^+...H2O | 3) Cr(NO3)2 D) CrO3^+...HNO3 (concentrated) → ...NO2^+...H2O | 4) Cr(NO3)3 E) HNO3 (diluted) → ...NO2^+...H2O | 5) HNO3 (decomposed) F) HNO3 (concentrated) → ...NO2^+...H2O | 6) HNO3 (concentrated)

The correct matches are: A) Cr^3+... → Cr(NO3)2^+...NO2^+...H2O | 3) Cr(NO3)2 B) Cr^3+... → Cr2(SO4)3^+...SO2^+...H2O | 2) H2SO4 (concentrated) C) CrO3^+... → CrSO4^+...H2O | 4) Cr(NO3)3 D) CrO3^+...HNO3 (concentrated) → ...NO2^+...H2O | 1) H2SO4 (diluted) E) HNO3 (diluted) → ...NO2^+...H2O | 6) HNO3 (concentrated) F) HNO3 (concentrated) → ...NO2^+...H2O | 5) HNO3 (decomposed)

Part C

C1: Write the equations of reactions that can be used to carry out the following transformations: Ca3(PO4)2 → P → H3PO4 → K3PO4

The equations are: 1) Ca3(PO4)2 + 6HCl → 2H3PO4 + 3CaCl2 2) 2H3PO4 + 6KOH → K3PO4 + 6H2O

C2: Calculate the mass of the precipitate formed when 200 g of a 15% solution of iron(III) sulfate is added to an excess of lithium hydroxide.

To calculate the mass of the precipitate, we need to determine the limiting reactant in the reaction between iron(III) sulfate and lithium hydroxide. Since there is an excess of lithium hydroxide, iron(III) sulfate is the limiting reactant.

The molar mass of iron(III) sulfate (Fe2(SO4)3) is 399.88 g/mol. From the chemical equation, we know that 1 mole of Fe2(SO4)3 reacts with 6 moles of LiOH to form 2 moles of Fe(OH)3.

First, calculate the number of moles of Fe2(SO4)3: moles of Fe2(SO4)3 = mass / molar mass moles of Fe2(SO4)3 = 200 g / 399.88 g/mol ≈ 0.500 mol

Since 1 mole of Fe2(SO4)3 forms 2 moles of Fe(OH)3, the number of moles of Fe(OH)3 formed is twice the number of moles of Fe2(SO4)3: moles of Fe(OH)3 = 2 * moles of Fe2(SO4)3 moles of Fe(OH)3 = 2 * 0.500 mol = 1.000 mol

The molar mass of Fe(OH)3 is 106.87 g/mol. Calculate the mass of Fe(OH)3 formed: mass of Fe(OH)3 = moles of Fe(OH)3 * molar mass mass of Fe(OH)3 = 1.000 mol * 106.87 g/mol ≈ 106.87 g

Therefore, the mass of the precipitate formed is approximately 106.87 g.