Question

Помогите пожалуйста с химией!!! Задали, много! Я не знаю как делать, помогите плиз, подробное решение если можно) Сделать мне надо до понедельника!!!!!1. Найдите молекулярную формулу алкена, массовая доля водорода в котором составляет 14,3%, относительная плотность этого вещества по водороду равна 21.2. В воде массой 135 г растворили 15 г соли. Какова массовая доля соли в этом растворе.3. При соединении 18 г алюминия с кислородом выделяется 547 кДж теплоты. Составьте термохимическое уравнение этой реакции.4. Рассчитайте, какую массу оксида серы (IV) можно получить при взаимодействии 50 г серы с кислородом, если практический выход процесса равен 45%.5. К раствору, содержащему 40 г сульфата меди (II) прибавили 12 г железных опилок. Рассчитайте массу выделившейся меди.6. Найдите молекулярную формулу алкена, массовая доля водорода в котором составляет 14,3%, относительная плотность этого вещества по водороду равна 21.7. Сколько кислорода необходимо для сжигания метана количеством вещества 0,1 моль?Задача 1. Глюкозумассой 50 г растворили в 100 г воды. Вычислите массовую долю глюкозы в получившемся растворе.Задача 2.Вычислите массу уксусной кислоты, затраченную на реакцию с раствором гидроксида натрия массой 120 г с массовой долей щелочи 25%.Задача 3. Какой объем водорода (н.у.) выделится при взаимодействии уксусной кислоты с 10 г магния, содержащего 20% примесей?Задача 4. Какая масса фенолята натрия может быть получена при взаимодействии фенола массой 4,7 г с раствором гидроксида натрия, содержащего 2,4 г NаОН.Задача 5. Какой объем углекислого газа (н.у.) образуется при спиртовом брожении глюкозы массой 250г, если объемная доля выхода газа составляет 95%?

Answer 1

1. Массовая доля водорода во ВСЕХ алкенах одинаковая! 
Алкены имеют общую молекулярную формулу СnН2n, поэтому их молярная масса равна 14n. Массовая доля водорода равна w(H)=2n/14n=0,143 (14,3 %). 
Если плотность по водороду равна 21, следовательно, молярная масса алкена равна 2*21=42(г/моль) . Тогда 14n=42, n=42/14=3. ФОРМУЛА АЛКЕНА С3Н6 

Answer 2

I'd be happy to help you with your chemistry questions! Let's go through each question one by one and find the solutions.

1. To find the molecular formula of an alkene with a hydrogen mass fraction of 14.3% and a relative density of 21.2, we need to use the concept of empirical formula and molecular formula. The empirical formula gives the simplest whole number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms in a compound.

To find the molecular formula, we need the molar mass of the compound. The molar mass can be calculated using the relative density and the molar mass of hydrogen. The molar mass of hydrogen is approximately 1 g/mol.

Let's calculate the molar mass: Relative density = (molar mass of compound) / (molar mass of hydrogen) 21.2 = (molar mass of compound) / 1 molar mass of compound = 21.2 g/mol

Now, let's calculate the empirical formula: - Assume we have 100 g of the compound. - The mass of hydrogen in 100 g of the compound is 14.3 g. - The mass of carbon in 100 g of the compound is (100 - 14.3) g = 85.7 g.

Next, we need to find the moles of hydrogen and carbon: - Moles of hydrogen = (mass of hydrogen) / (molar mass of hydrogen) = 14.3 g / 1 g/mol = 14.3 mol - Moles of carbon = (mass of carbon) / (molar mass of carbon) = 85.7 g / 12 g/mol = 7.14 mol

Now, we need to find the simplest whole number ratio of moles of hydrogen to carbon. Divide both moles by the smallest number of moles (7.14): - Moles of hydrogen = 14.3 mol / 7.14 mol = 2 mol - Moles of carbon = 7.14 mol / 7.14 mol = 1 mol

Therefore, the empirical formula is CH2.

To find the molecular formula, we need to know the molar mass of the compound. Let's assume the molar mass is 42.2 g/mol (just for calculation purposes). The molar mass of the empirical formula (CH2) is 14 g/mol (1 carbon atom + 2 hydrogen atoms).

To find the molecular formula, divide the molar mass of the compound by the molar mass of the empirical formula: Molecular formula = (molar mass of compound) / (molar mass of empirical formula) Molecular formula = 42.2 g/mol / 14 g/mol = 3

Therefore, the molecular formula of the alkene is C3H6.

2. To find the mass fraction of salt in a solution, we need to divide the mass of the salt by the total mass of the solution.

Given: Mass of water = 135 g Mass of salt = 15 g

Total mass of the solution = Mass of water + Mass of salt = 135 g + 15 g = 150 g

Mass fraction of salt = (Mass of salt) / (Total mass of the solution) = 15 g / 150 g = 0.1 = 10%

Therefore, the mass fraction of salt in the solution is 10%.

3. To write the thermochemical equation for the reaction between aluminum and oxygen, we need to know the balanced chemical equation and the enthalpy change (heat released) for the reaction.

The balanced chemical equation for the reaction is: 4 Al + 3 O2 -> 2 Al2O3

Given: Heat released = 547 kJ

The thermochemical equation can be written as: 4 Al + 3 O2 -> 2 Al2O3 + 547 kJ

4. To calculate the mass of sulfur dioxide (SO2) produced when 50 g of sulfur reacts with oxygen, we need to consider the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation for the reaction is: S + O2 -> SO2

Given: Mass of sulfur = 50 g Practical yield = 45% (45 g)

Using the balanced chemical equation, we can see that the molar ratio between sulfur and sulfur dioxide is 1:1. This means that 1 mole of sulfur reacts to produce 1 mole of sulfur dioxide.

First, let's calculate the number of moles of sulfur: Molar mass of sulfur = 32 g/mol Moles of sulfur = (mass of sulfur) / (molar mass of sulfur) = 50 g / 32 g/mol = 1.5625 mol

Since the molar ratio between sulfur and sulfur dioxide is 1:1, the number of moles of sulfur dioxide produced is also 1.5625 mol.

Next, let's calculate the mass of sulfur dioxide produced: Molar mass of sulfur dioxide = 64 g/mol Mass of sulfur dioxide = (moles of sulfur dioxide) * (molar mass of sulfur dioxide) = 1.5625 mol * 64 g/mol = 100 g

Therefore, the mass of sulfur dioxide produced is 100 g.

5. To calculate the mass of copper formed when 40 g of copper sulfate (CuSO4) reacts with iron filings (Fe), we need to consider the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation for the reaction is: CuSO4 + Fe -> FeSO4 + Cu

Given: Mass of copper sulfate = 40 g Mass of iron filings = 12 g

Using the balanced chemical equation, we can see that the molar ratio between copper sulfate and copper is 1:1. This means that 1 mole of copper sulfate reacts to produce 1 mole of copper.

First, let's calculate the number of moles of copper sulfate: Molar mass of copper sulfate = 159.6 g/mol Moles of copper sulfate = (mass of copper sulfate) / (molar mass of copper sulfate) = 40 g / 159.6 g/mol = 0.250 mol

Since the molar ratio between copper sulfate and copper is 1:1, the number of moles of copper formed is also 0.250 mol.

Next, let's calculate the mass of copper formed: Molar mass of copper = 63.5 g/mol Mass of copper = (moles of copper) * (molar mass of copper) = 0.250 mol * 63.5 g/mol = 15.875 g

Therefore, the mass of copper formed is 15.875 g.

6. The question about the molecular formula of an alkene with a hydrogen mass fraction of 14.3% and a relative density of 21.2 is the same as question 1. The molecular formula of the alkene is C3H6.

7. To calculate the amount of oxygen needed to burn 0.1 mole of methane (CH4), we need to consider the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation for the combustion of methane is: CH4 + 2 O2 -> CO2 + 2 H2O

From the balanced chemical equation, we can see that the molar ratio between methane and oxygen is 1:2. This means that 1 mole of methane reacts with 2 moles of oxygen.

Given: Moles of methane = 0.1 mol

Using the molar ratio, we can calculate the moles of oxygen needed: Moles of oxygen = (moles of methane) * 2 = 0.1 mol * 2 = 0.2 mol

Therefore, 0.2 moles of oxygen are needed to burn 0.1 mole of methane.

Let's move on to the next set of questions.

1. To calculate the mass fraction of glucose in a solution, we need to divide the mass of glucose by the total mass of the solution.

Given: Mass of glucose = 50 g Mass of water = 100 g

Total mass of the solution = Mass of glucose + Mass of water = 50 g + 100 g = 150 g

Mass fraction of glucose = (Mass of glucose) / (Total mass of the solution) = 50 g / 150 g = 1/3 = 0.3333 = 33.33%

Therefore, the mass fraction of glucose in the solution is 33.33%.

2. To calculate the mass of acetic acid (CH3COOH) required to react with a solution of sodium hydroxide (NaOH), we need to consider the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation for the reaction between acetic acid and sodium hydroxide is: CH3COOH + NaOH -> CH3COONa + H2O

Given: Mass of sodium hydroxide = 120 g Mass fraction of sodium hydroxide = 25%

To find the mass of acetic acid, we need to calculate the moles of sodium hydroxide first: Molar mass of sodium hydroxide (NaOH) = 40 g/mol + 16 g/mol + 1 g/mol = 56 g/mol Moles of sodium hydroxide = (mass of sodium hydroxide) / (molar mass of sodium hydroxide) = 120 g / 56 g/mol = 2.143 mol

From the balanced chemical equation, we can see that the molar ratio between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide