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The principal oxidation state of NA in NA2O2 and K in KO2 is +1

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What is the oxidation number of Na in Na2O2?

+1 for each Na -1 for oxygen (as it is peroxide)


What type of reaction occurs between Na plus O2?

4 Na + O2 → 2 Na2O2 Na + O2 → Na2O2


What is the oxidation number of Na2O2?

The oxidation number of Na in Na2O2 is +1, because it belongs to group 1 of the periodic table. The oxygen in O2 has an oxidation number of -1, so to balance the charges in the compound, the sodium must be +1 to give a total charge of 0.


What is the percent yield for the reaction of 6.92 g of K and 4.28 g of O2 if 7.36 g of KO2 is recovered in the laboratory?

First you need to find the theoretical yield for the reaction:2Na + O2 ---> Na2O2(1) find the theoretical yield of 3.74 g of Na, i.e. if all of the Na is reacted3.74 g Na *(1 mol Na/22.99 g Na)*(1 mol Na2O2 formed/2 mol Na reacted)*(77.98 g Na2O2/1 mol Na2O2) = 6.34 g Na2O2(2) % yield = (actual yield / theoretical yield)*(100%)% yield = (5.34 g / 6.34 g) *(100%) = 84.2% yield


What is chemical symbol for sodium peroxide?

Na2O2, or Na-O-O-Na.


What is the oxidation state of NaBrO3?

The oxidation states for Na, Br, and O are +1, +5, and -2, respectively. In NaBrO3, there is 1 Na, 1 Br, and 3 O atoms, leading to a total charge of 0. Thus, the oxidation state of NaBrO3 is +5 to balance the charge.


How does the oxidation state of Na change in the following reaction Li NaOH LiOH Na?

The oxidation state of Na changes from 0 in Na to +1 in NaOH and then back to 0 in Na in the reaction. Na starts as a neutral metal atom, gains an electron in NaOH to have an oxidation state of +1, and then loses that electron to return to its neutral state in the final product.


How does the oxidation state of Na change in the following reaction?

In the reaction where Na is oxidized to Na+ in a chemical reaction, the oxidation state of Na changes from 0 to +1. This means that Na loses one electron and is oxidized.


What is the oxidation state of na2so4?

In Na2SO4, the oxidation state of sodium (Na) is +1, the oxidation state of sulfur (S) is +6, and the oxidation state of oxygen (O) is -2. To calculate the oxidation state of the whole compound, you can use the rule that the sum of the oxidation states in a neutral compound is zero, so in this case it would be +1*2 + (-2)*4 = 0.


What is the oxidation number of H in the compound NaHSO4?

The oxidation number of H in NaHSO4 is +1. In this compound, Na has an oxidation state of +1, S has an oxidation state of +6, and O has an oxidation state of -2. By adding up the oxidation states and solving for H, it is determined to be +1.


What is the oxidation state of Sodium Nitrate?

Sodium Oxide, or NaOH, has no oxidation state. It has a charge, which is zero. The elements that make up NaOH, however, do have oxidation states. The oxidation state of sodium (Na) is +1, and it will forever be +1 because it is impossible for it to be anything else, no matter what situation. The same applies for Hydrogen. Oxygen has an oxidation state of -2, and almost always will have an oxidation state of -2. There is one notable exception: H2O2. In this case, since the total charge of the compound is neutral and the oxidation state of Hydrogen must be +1 and, seeing as there are two hydrogens, bringing the overall charge up to 2, the oxidation state of oxygen must be -1. If it was -2, then the molecule would have an overall charge of -2.


What is the oxidation number in NaNO3?

The oxidation number of Na in NaNO3 is +1, since Na typically has a +1 oxidation state in compounds. The oxidation number of N in NO3 is +5, since oxygen is usually assigned a -2 oxidation state and there are three oxygen atoms bonded to nitrogen in NO3.