Question

Задание 1. Составьте структурные формулы всех изомеров состава C4H8Br2 . Дайте им названия. Задание 2. Имеются три одинаковые колбы с бесцветными растворами. Известно, что в одной из них находится хлорид бария, в другой – соляная кислота, а третьей - карбонат натрия. Предложите способ, позволяющий определить, в какой именно колбе какой раствор находится, используя только пробирки с растворами и не применяя других реактивов. Приведите уравнения реакций. Задание 3. Содержание углерода в углеводороде 85.71%, его плотность по азоту равна 2.5. Назовите все вещества, удовлетворяющие условию задачи. Задание 4. Определите объем воздуха, необходимый для сжигания 100 м3 природного газа, содержащего 89,6 % метана, 5,6% этана и несгораемые примеси. Задание 5. Один крупный металлургический завод оштрафовали на крупную сумму денег за то, что он плохо очищал сточные воды, которые спускал в городской пруд. В сточных водах этого завода содержались ионы металлов: Fe3+, Al3+, Pb2+, Cu2+, Zn2+, Mg2+. Перед главным химиком-технологом была поставлена задачи придумать наиболее простой и безвредный способ очистки сточных вод от этих ионов. Помогите главному технологу. Предложите реактив (или реактивы), которые можно использовать для удаления данных ионов из раствора, составьте уравнения реакций. Задание 6. Напишите уравнения реакций, описывающих следующие превращения: Ca → CaO → Ca(OH)2 → CaCO3 → CaO → Ca → CaCl2 → CaCO3 →Ca(HCO3)2 → CaCO3

Answer 1

                      /BR
1)CH3-CH2-C-CH3              1,1 дибром.2-метилопропен
                     /                        1,2 дибр,2-метилопропен
                                               1,3дибр,2-метилопропан
                   BR                      

  
   br   H  H  H
    /    /   /   /
H-C-C-C-C-H          
    /    /   /   /
   br  H  H  H

2)Na2Co3+2HCl=2NACL+CO2(газ)+H2O\
    HCL+BACL-------(реакция не идет)
     NA2CO3+BACK2=2 NACL+BACO3(осадок)

ДАНО                       РЕШЕНИЕ 
3)W(C)=85,71%         1)М ве.-ва.=2,5*28=70
Д(N2)=2,5                      2)85,71          14,29
НАЙТИ                              -------    :    ---------   -7:14
М.Ф.-?                                  12               1 
                                            
                                           C=1; H=2....
                                        CH2--прост.        С5H10-конечн.

Answer 2

Task 1: Structural Formulas and Names of Isomers of C4H8Br2

The compound C4H8Br2 can have different structural isomers. Here are the possible isomers and their names:

1. 1,1-Dibromobutane: - Structural Formula: CH3CH2CH2CH2Br - IUPAC Name: 1,1-Dibromobutane

2. 1,2-Dibromobutane: - Structural Formula: CH3CHBrCH2CH3 - IUPAC Name: 1,2-Dibromobutane

3. 1,3-Dibromobutane: - Structural Formula: CH3CH2CHBrCH3 - IUPAC Name: 1,3-Dibromobutane

4. 1,4-Dibromobutane: - Structural Formula: CH3CH2CH2CH2Br - IUPAC Name: 1,4-Dibromobutane

Please note that the structural formulas provided are simplified representations of the isomers. The actual arrangement of atoms in space may differ.

Task 2: Identifying Solutions in Colored Vials

To determine which solution is present in each vial (containing barium chloride, hydrochloric acid, and sodium carbonate), you can perform a series of simple chemical reactions using additional test tubes and reagents. Here's a suggested method:

1. Take a small amount of each solution from the vials and transfer them to separate test tubes. 2. Add a few drops of silver nitrate (AgNO3) solution to each test tube. - If a white precipitate forms, it indicates the presence of chloride ions (from hydrochloric acid). - If a yellow precipitate forms, it indicates the presence of carbonate ions (from sodium carbonate). - If no precipitate forms, it indicates the presence of barium ions (from barium chloride).

Based on the observations, you can determine the contents of each vial: - The vial with the white precipitate contains hydrochloric acid. - The vial with the yellow precipitate contains sodium carbonate. - The vial with no precipitate contains barium chloride.

The reactions involved are as follows: 1. AgNO3 + HCl → AgCl (white precipitate) + HNO3 2. AgNO3 + Na2CO3 → Ag2CO3 (yellow precipitate) + 2NaNO3 3. AgNO3 + BaCl2 → AgCl (white precipitate) + Ba(NO3)2

Task 3: Substances with 85.71% Carbon and Density of 2.5 (relative to nitrogen)

To find substances that satisfy the given conditions (85.71% carbon content and a density of 2.5 relative to nitrogen), we can consider hydrocarbons with different molecular formulas. Here are a few examples:

1. Octane (C8H18): - Carbon content: (8/10) * 100% = 80% - Density relative to nitrogen: Approximately 2.5

2. Decane (C10H22): - Carbon content: (10/12) * 100% = 83.33% - Density relative to nitrogen: Approximately 2.5

3. Dodecane (C12H26): - Carbon content: (12/14) * 100% = 85.71% - Density relative to nitrogen: Approximately 2.5

Please note that these are just a few examples, and there may be other substances that satisfy the given conditions.

Task 4: Volume of Air Required to Burn 100 m3 of Natural Gas

To determine the volume of air required to burn 100 m3 of natural gas, we need to consider the stoichiometry of the combustion reaction. Natural gas primarily consists of methane (CH4) and smaller amounts of ethane (C2H6) and other impurities.

Assuming complete combustion, the balanced equation for the combustion of methane is:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that one mole of methane requires two moles of oxygen to completely burn. The molar volume of any gas at standard temperature and pressure (STP) is approximately 22.4 L.

Given that natural gas contains 89.6% methane and 5.6% ethane, we can calculate the moles of methane and ethane in 100 m3 of natural gas:

- Moles of methane: 89.6% of 100 m3 = 89.6 m3 - Moles of ethane: 5.6% of 100 m3 = 5.6 m3

Now, we can calculate the moles of oxygen required to burn the methane and ethane:

- Moles of oxygen for methane: 2 * 89.6 m3 = 179.2 m3 - Moles of oxygen for ethane: 2 * 5.6 m3 = 11.2 m3

To convert the moles of oxygen to volume at STP, we use the molar volume:

- Volume of oxygen for methane: 179.2 m3 * 22.4 L/mole = 4014.08 L - Volume of oxygen for ethane: 11.2 m3 * 22.4 L/mole = 250.88 L

Therefore, the total volume of air (oxygen + nitrogen) required to burn 100 m3 of natural gas is approximately 4264 L (4014.08 L + 250.88 L).

Task 5: Removing Metal Ions from Wastewater

To remove metal ions (Fe3+, Al3+, Pb2+, Cu2+, Zn2+, Mg2+) from wastewater, you can use a precipitation method. One common reagent that can be used is sodium hydroxide (NaOH). Here's how you can proceed:

1. Take a sample of the wastewater and transfer it to a beaker. 2. Slowly add a solution of sodium hydroxide (NaOH) to the wastewater while stirring. - If a precipitate forms, it indicates the presence of metal ions. - The color and appearance of the precipitate can help identify the specific metal ions present. 3. Allow the mixture to settle, and then carefully decant the supernatant liquid. 4. Collect the precipitate by filtration using filter paper and a funnel. 5. Wash the precipitate with distilled water to remove any impurities. 6. Dry the precipitate and analyze it further if necessary.

The reactions involved in the precipitation of metal ions with sodium hydroxide are as follows:

1. Fe3+ + 3OH- → Fe(OH)3 (reddish-brown precipitate) 2. Al3+ + 3OH- → Al(OH)3 (white gelatinous precipitate) 3. Pb2+ + 2OH- → Pb(OH)2 (white precipitate) 4. Cu2+ + 2OH- → Cu(OH)2 (blue precipitate) 5. Zn2+ + 2OH- → Zn(OH)2 (white precipitate) 6. Mg2+ + 2OH- → Mg(OH)2 (white precipitate)

By observing the color and appearance of the precipitate formed, you can identify the specific metal ions present in the wastewater.

Task 6: Reactions Describing Transformations

The given reactions describe a series of transformations involving calcium (Ca) and its compounds. Here are the reactions:

1. Ca → CaO (Calcium metal reacts with oxygen to form calcium oxide) 2. CaO + H2O → Ca(OH)2 (Calcium oxide reacts with water to form calcium hydroxide) 3. Ca(OH)2 + CO2 → CaCO3 + H2O (Calcium hydroxide reacts with carbon dioxide to form calcium carbonate and water) 4. CaCO3 → CaO + CO2 (Calcium carbonate decomposes upon heating to form calcium oxide and carbon dioxide) 5. CaO + HCl → CaCl2 + H2O (Calcium oxide reacts with hydrochloric acid to form calcium chloride and water) 6. CaCO3 + H2CO3 → Ca(HCO3)2 (Calcium carbonate reacts with carbonic acid to form calcium bicarbonate) 7. CaCO3 → CaO + CO2 (Calcium carbonate decomposes upon heating to form calcium oxide and carbon dioxide)

These reactions represent various chemical transformations involving calcium and its compounds.