The principal oxidation state of NA in NA2O2 and K in KO2 is +1
+1 for each Na -1 for oxygen (as it is peroxide)
4 Na + O2 → 2 Na2O2 Na + O2 → 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.
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
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.
+1 for each Na -1 for oxygen (as it is peroxide)
4 Na + O2 → 2 Na2O2 Na + O2 → 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.
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
Na2O2, or Na-O-O-Na.
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.
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.
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.
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.
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.
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.
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.