Question

10.27. Соль А представляет собой лиловые кристаллы, которые при нагревании до 47,2 °C плавятся в своей кристаллизационной воде, обра-зуя красную жидкость. Вещество А получают действием разбавленнойазотной кислоты на железные стружки с последующим утпариваниеми кристаллизацией.а) Установите состав кристаллогидрата А, если известно, что приохлаждении 500 г насыщенного при 40 °С раствора на 20 градусов выпа-дает 204 г кристаллов. Растворимость безводной соли при 20 °С и 40 °Cсоставляет соответственно 82,5ги 105 г на 100 г воды.​

Answer 1

Ответ:

ответ: m(p-pa) = 40*100/25=160

m(h2o) = 160-40=120

Объяснение:

Answer 2

Composition of Hydrate A

The composition of the hydrate A can be determined by analyzing the given information. We know that when 500 g of a saturated solution at 40 °C is cooled by 20 degrees, it yields 204 g of crystals. Additionally, the solubility of the anhydrous salt at 20 °C and 40 °C is 82.5 g and 105 g per 100 g of water, respectively.

To find the composition of the hydrate A, we need to calculate the amount of water and the amount of the anhydrous salt in the 204 g of crystals obtained.

Let's assume the mass of the anhydrous salt in the crystals is x g, and the mass of water is y g.

According to the given information, the mass of the anhydrous salt is equal to the mass of the crystals minus the mass of water: x = 204 g - y

We also know that the solubility of the anhydrous salt at 40 °C is 105 g per 100 g of water. Therefore, the mass of the anhydrous salt in the saturated solution at 40 °C is: (105 g / 100 g) * (500 g - y) = (105/100) * (500 - y) g

Since the solution is saturated, the mass of the anhydrous salt in the crystals should be equal to the mass of the anhydrous salt in the saturated solution: x = (105/100) * (500 - y)

Now, we can solve these two equations simultaneously to find the values of x and y.

To do this, we can substitute the value of x from the first equation into the second equation: 204 g - y = (105/100) * (500 - y)

Simplifying the equation: 20400 - 100y = 52500 - 105y 5y = 32100 y = 6420 g

Substituting the value of y back into the first equation to find x: x = 204 g - 6420 g x = -6216 g

Since the mass of a substance cannot be negative, we can conclude that there is an error in the given information or calculations.

Unfortunately, based on the given information, it is not possible to determine the composition of hydrate A accurately.