What is the oxidation number of each atom in sodium hydrogen carbonate nahco3

The oxidation state of carbon in NaHCO3 is +4. Sodium has a +1 oxidation state. Explanation:Hydrogen has a +1 oxidation state. Oxygen has a -2 oxidation state

1. P₂O₅      → oxidation number of phosphorous is +5 and Oxygen is -2.2. (SO₄)²⁻ → oxidation number of sulfur is +6 and Oxygen is -2. 3. KClO₃   → oxidation number of Potassium is +1, Chlorine is +5, and Oxygen is -2.4. NH₄Cl  → oxidation number of Nitrogen is -3, Hydrogen is +1, and Chlorine is -15. (NH₄)₂S  → oxidation number of Nitrogen is -3, Hydrogen is +1, and Sulfur is -2Explanation:General Rules for assigning oxidation numbers The oxidation number of a free element is always 0.The oxidation number of a mono-atomic ion equals the charge of the ion.The alkali metals (group I) always have an oxidation number of +1.The alkaline earth metals (group II) are always assigned an oxidation number of +2.Oxygen almost always has an oxidation number of -2, except in peroxides (H₂O₂) where it is -1 and in compounds with fluorine (OF₂) where it is +2.Hydrogen has an oxidation number of +1 when combined with non-metals, but it has an oxidation number of -1 when combined with metals.The algebraic sum of the oxidation numbers of elements in a compound is zero.The algebraic sum of the oxidation states in an ion is equal to the charge on the ion.Using the above rules: 1. P₂O₅∵ it is a neutral compound its total charge is 0.Also, we know that oxygen has an oxidation number of -2.Let oxidation number of P be x∴ 2(x)+5(-2)=0   →   2x=+10   →   x=+5  ∴oxidation number of phosphorous is +5. 2. SO₄²⁻:∵ it is a charged ion its total charge is -2.Also, we know that oxygen has an oxidation number of -2.Let oxidation number of S be x∴ (x)+4(-2)= -2   →   x=+6   ∴oxidation number of sulfur is +6. 3. KClO₃:∵ it is a neutral compound its total charge is 0.Also, we know that oxygen has an oxidation number of -2 and the  oxidation number of K (group I) is +1 Let oxidation number of Cl be x ∴ (+1) + (x) + 3(-2) = 0   →   x=+5  ∴oxidation number of Chlorine is +5. 4. NH₄Cl: ∵ it is a neutral compound its total charge is 0.Also, we know that chloride has an oxidation number of -1Hydrogen has an oxidation number of +1 when combined with non-metals  Let oxidation number of N be x∴ (x) + 4(+1) + (-1) = 0   →   x=-3  ∴oxidation number of Nitrogen is -3.5. (NH₄)₂S: ∵ it is a neutral compound its total charge is 0.Also, we know that chloride has an oxidation number of -1Ammonium ion (NH₄⁺) has an oxidation number of +1  Let oxidation number of N be x∴ 2(+1) + (x) = 0   →   x= -2   ∴oxidation number of sulfur is -2.

EExplanation:To solve this question, we need to know the oxidation number in the compound in the question. The name of the compound is tetraoxosulphate(vi) acid. We can obviously see that the oxidation number of sulphur in this particular compound is 6.A. Is wrongThe name of the compound is trioxosulphate(iv) ionB. Is wrongS2O32-2S + 3(-2) = -2S = +2C.is wrongOxidation state there is -2D is wrong The oxidation number here is zeroE is correctOxidation number of oxygen is -2 and that of chlorine is -1Hence:S + 2(-2) + 2(-1) = 0S -4 -2 = 0S = 4+2 = +6

Let the oxidation number of P be x.  We know that oxidation number of Na= +1, O= -2  therefore,  3*1 + x+ 4* -2 = 0 (net charge on the molecule)  3 + x - 8 =0  x = +5  therefore, the answer is C)+5

No element decreases its oxidation number.Explanation:

Oxidation number of each atom in sodium hydrogen carbonate ( NaHCO3) areNa = +1H = +1C = +4O = -2Explanation:Let's pick the first atomNaHCO3find oxidation number of NaNa + 1 +4 + 3(-2) = 0( because there is no charge) Na + 5 - 6 = 0Na = 6 -5Na = +1Always put the signFind the oxidation number of H in NaHCO3+1 + H + 4 + 3(-2) = 0.H + 1 + 4 + 3(-2) = 0H + 5 -6 = 0H = 6 -5 H = +1Find the oxidation number of C In NaHCO3+1 +1 +C + 3(-2) = 0C + 1+1-6 = 0C +2 - 6= 0C = 6 -2C = +4Find the oxidation number of O in NaHCO3+1 + 1 + 4 + 3(O) = 06 + 3 (O) = 03(O) = -6Divide through by 3 to get O3(O) / 3 = -6 / 3O = -6/3O = -2

The answer to your question is: letter b SulphurExplanation:Remember that in neutral molecules the sum of the oxidation numbers equals zero.                                Mg⁺²S⁺⁶O⁻²₄Magnesium = +2Sulphur = +6Oxygen = -2 x 4 = -8

You may find bellow the assignment of the oxidation number for each atom.Explanation:Li₃PO₄Li have the oxidation number +1P have the oxidation number +5O have the oxidation number -2SO₃²⁻ S have the oxidation number +4O have the oxidation number -2 Cr₂S₃Cr have the oxidation number +3S have the oxidation number -2NO₃⁻N have the oxidation number +5O have the oxidation number -2Learn more about:oxidation number#learnwithBrainly

Na⁺¹₃P⁺⁵O⁻²₄+1*3 + (+5) + (-2*4) = 0

The balanced reaction is as follows;BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄this is a double displacement reaction the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄oxidation number of Cl is -1 in both BiCl₂ and NaCl oxidation number of Na is +1 in both Na₂SO₄ and NaCloxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .Answer for this question is no element decreases its oxidation number.