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H(2) S + HNO(3) = NO + S + H(2) O...

`H_(2) S + HNO_(3) = NO + S + H_(2) O`

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To balance the redox reaction \( H_2S + HNO_3 \rightarrow NO + S + H_2O \), we will follow these steps: ### Step 1: Assign Oxidation Numbers - In \( H_2S \), sulfur (S) has an oxidation number of -2. - In \( HNO_3 \), nitrogen (N) has an oxidation number of +5. - In \( NO \), nitrogen (N) has an oxidation number of +2. - In elemental sulfur (S), the oxidation number is 0. - In \( H_2O \), oxygen (O) has an oxidation number of -2. ...
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The stoichiometric coefficient of S in the following reaction H_(2)S + HNO_(3) to NO + S+ H_(2)O is balanced (in acidic medium):

Number of acids having central atom in +3 oxidation state among the following is (a) HNO_(2) (b) HNO_(3) (c) H_(3)PO_(2) (d) H_(3)PO_(3) (e) H_(3)PO_(4) (f) H_(4)P_(2)O_(5) (g) H_(4)P_(2)O_(7) (h) H_(2)SO_(3) (i) H_(2)S_(2)O_(7) (j) H_(2)S_(2)O_(8) (k) H_(2)SO_(4)