Question

РАССТАВИТЬ СТЕПЕНИ ОКИСЛЕНИЯK2Sr2O7

Answer 1

K2(+)Sr2(+6)O7(-2)
Yflj dthyj gbcfnm ajhveks

Answer 2

Окисление и степени окисления в K2Sr2O7

K2Sr2O7 is a chemical compound composed of potassium (K), strontium (Sr), and oxygen (O). To determine the oxidation states of each element in K2Sr2O7, we can use the following guidelines:

1. The oxidation state of potassium (K) is typically +1 in compounds. 2. The oxidation state of oxygen (O) is typically -2 in compounds. 3. The sum of the oxidation states in a compound should equal zero.

Based on these guidelines, we can determine the oxidation states of each element in K2Sr2O7:

1. Let's assume the oxidation state of potassium (K) is +1. 2. Since there are two potassium atoms in K2Sr2O7, the total oxidation state contribution from potassium is +2.

Now, let's determine the oxidation state of strontium (Sr):

3. The sum of the oxidation states of potassium and strontium should equal zero. 4. Since the total oxidation state contribution from potassium is +2, the oxidation state of strontium (Sr) must be +2 to balance the equation.

Finally, let's determine the oxidation state of oxygen (O):

5. The sum of the oxidation states of all elements in K2Sr2O7 should equal zero. 6. Since there are seven oxygen atoms in K2Sr2O7, the total oxidation state contribution from oxygen is -14.

To balance the equation, the oxidation state of oxygen (O) must be +6.

Therefore, the oxidation states in K2Sr2O7 are as follows: - Potassium (K): +1 - Strontium (Sr): +2 - Oxygen (O): -2

Please note that the oxidation states mentioned above are based on general guidelines and may vary in certain compounds or under specific conditions.