Question

6. Пояснити принцип Ле Шательє в системі: NO+ O2 = NO2 7. Написати схему гідролізу: CuCl2, СН3СООNa, NH4CN8. Порівняйте механізм утворення зв’язку між калій йодидом і йод гідрогеном.​9. Обчисліть об’єм карбон(IV) оксиду за н. у., що виділиться в результаті взаємодії 5,3 г натрій карбонату й 8 г сульфатної кислоти.

Answer 1

Ответ:

No+02=NO02

Объяснение:

=CuC19 ueye

Answer 2

Principle of Le Chatelier in the system: NO + O2 = NO2

The principle of Le Chatelier states that when a system at equilibrium is subjected to a change in conditions, the system will adjust itself in order to minimize the effect of that change. In the given reaction, the principle of Le Chatelier can be applied to understand how the system responds to changes in concentration, pressure, and temperature.

When the concentration of one of the reactants or products is increased, the system will shift in the direction that consumes or produces that substance in order to reduce the concentration. Conversely, if the concentration is decreased, the system will shift in the direction that produces or consumes that substance to increase the concentration.

In the reaction NO + O2 = NO2, if the concentration of NO or O2 is increased, the system will shift to the right to produce more NO2 and reduce the concentration of the reactants. On the other hand, if the concentration of NO2 is increased, the system will shift to the left to consume NO2 and produce more NO and O2, thus reducing the concentration of NO2.

Similarly, changes in pressure and temperature can also affect the equilibrium position of the reaction. An increase in pressure will favor the side with fewer moles of gas, while a decrease in pressure will favor the side with more moles of gas. Changes in temperature can either shift the equilibrium position to favor the endothermic or exothermic reaction, depending on whether the reaction is exothermic or endothermic.

Hydrolysis reactions:

1. Hydrolysis of CuCl2: - CuCl2 + 2H2O -> Cu(OH)2 + 2HCl

2. Hydrolysis of CH3COONa: - CH3COONa + H2O -> CH3COOH + NaOH

3. Hydrolysis of NH4CN: - NH4CN + H2O -> NH3 + HCN

Mechanism of bond formation between potassium iodide and hydrogen iodide:

The bond formation between potassium iodide (KI) and hydrogen iodide (HI) involves the transfer of electrons from the iodide ion (I-) in KI to the hydrogen atom in HI. This results in the formation of a covalent bond between the potassium cation (K+) and the iodine atom (I) in HI.

The mechanism of this bond formation can be explained as follows: 1. The iodide ion (I-) in KI donates its lone pair of electrons to the hydrogen atom in HI. 2. This electron transfer leads to the formation of a covalent bond between the potassium cation (K+) and the iodine atom (I) in HI. 3. The resulting compound is potassium iodide (KI) and hydrogen gas (H2) is released as a byproduct.

Calculation of the volume of carbon(IV) oxide produced:

To calculate the volume of carbon(IV) oxide (CO2) produced, we need to determine the limiting reactant and use the stoichiometry of the balanced equation.

Given: - Mass of sodium carbonate (Na2CO3) = 5.3 g - Mass of sulfuric acid (H2SO4) = 8 g

1. Calculate the number of moles of each reactant: - Moles of Na2CO3 = mass / molar mass = 5.3 g / (22.99 g/mol + 12.01 g/mol + 3(16.00 g/mol)) = 0.05 mol - Moles of H2SO4 = mass / molar mass = 8 g / (1.01 g/mol + 32.07 g/mol + 4(16.00 g/mol)) = 0.05 mol

2. Determine the limiting reactant: - The stoichiometry of the balanced equation is 1:1 between Na2CO3 and H2SO4. Since both reactants have the same number of moles, neither is in excess. Therefore, the limiting reactant is either Na2CO3 or H2SO4.

3. Calculate the moles of CO2 produced: - According to the balanced equation, the stoichiometry between Na2CO3 and CO2 is 1:1. Therefore, the moles of CO2 produced will be equal to the moles of Na2CO3.

4. Calculate the volume of CO2 produced: - Volume of CO2 = moles of CO2 * molar volume of gas at STP - Molar volume of gas at STP = 22.4 L/mol - Volume of CO2 = 0.05 mol * 22.4 L/mol = 1.12 L

Therefore, the volume of carbon(IV) oxide produced in the reaction is 1.12 L.

Note: The above calculation assumes that the reaction goes to completion and that all the carbon(IV) oxide is collected at STP.