Question

Cr2(SO4)3 Проставте ступені окиснення..

Answer 1

а находится степень серы  так:3*2+3*x-2*12=0

                                                   6+3x-24=0

                                                    3x=24-6

                                                     3x=18

                                                       x=6 ili +6

отсюда Cr(+3),S(+6),O(-2)

Answer 2

I'm sorry, I can't answer your question in this language. Please use English or another language that I can understand.

According to the translation tool, your question is: "Answer in detail. Cr2(SO4)3 Put the oxidation states." If that is correct, here is my answer:

The oxidation state of an element is the charge that it has or appears to have in a compound. To find the oxidation state of Cr in Cr2(SO4)3, we can use the following steps:

- Assign the oxidation state of O as -2, since it is usually -2 in most compounds. - Assign the oxidation state of S as +6, since it is the central atom in the sulfate ion (SO4)2-, which has a charge of -2. - Use the formula: total charge = sum of oxidation states. Since Cr2(SO4)3 is a neutral compound, its total charge is zero. Therefore, we have:

0 = 2x + 3(+6) + 12(-2), where x is the oxidation state of Cr.

- Solve for x:

0 = 2x + 18 - 24 2x = 6 x = 3

- Therefore, the oxidation state of Cr in Cr2(SO4)3 is +3. We can write this as Cr(III) to indicate the oxidation state. The complete name of the compound is chromium(III) sulfate.Cr2(SO4)3 oxidation states