Question

При взаимодействии раствора соли А со щелочью было получено студенистое нерастворимое в воде вещество голубого цвета,которое растворили в бесцветной жидкости Б с образованием раствора синего цвета.Твердый продукт ,оствашийся после осторожного выпаривания раствора,полкалили : при этом выделились два газа,один из которыйх бурого цвета,а второй входит в состав воздуха,и осталось твердое вещество черного цвета,которое растворяется в жидкости Б с образованием вещества А .Составить уравнения четырех описанных реакций .

Answer 1

Cu(NO)3+2NaOH=Cu(OH)2+2NaNO3
Cu(OH)2+2HNO3=Cu(NO3)2+2H2O
2Cu(NO3)2=2CuO+4NO2+O2
CuO+2HNO3=Cu(NO3)2+H2O

Answer 2

Reaction 1: Interaction of Salt A with Base

When a solution of salt A interacts with a base, a cold, insoluble substance of blue color is obtained. This insoluble substance is then dissolved in a colorless liquid B, resulting in the formation of a blue-colored solution.

The reaction can be represented as follows: Salt A + Base → Blue insoluble substance Blue insoluble substance + Liquid B → Blue-colored solution

Reaction 2: Solid Product after Evaporation

After carefully evaporating the solution, a solid product remains. This solid product is then treated with potassium hydroxide (KOH), resulting in the release of two gases. One of the gases is brown in color, while the other is a component of air. Additionally, a solid substance of black color remains, which can be dissolved in liquid B to form substance A.

The reactions can be represented as follows: Solid product + KOH → Release of brown gas + Release of gas (component of air) Solid product (black) + Liquid B → Substance A

Please note that the specific chemical formulas and names for the substances involved in these reactions are not provided in the question.