Lấy 5ml dd hồ tinh bột vào ông nghiệm, thêm 4 giọt dd H2SO4 10%. Lắc kỹ, đun sôi 20 phút. cứ mỗi 5 phút lấy ra 5 giọt dd đnag đun cho vào ống nghiệm khác nhau. Để nguội, nhỏ 1-2 giọt dd iod. Quan sát, nhận xét, giải thích. Đến khi dd kong có màu với dd iod nữa thì ngừng đun. Để nguội, lấy 5 giọt dd thủy phân vào ống nghiệm. Trung hòa bằng NaOH , kểm tra bàng giấy quỳ. Thêm 5 giọt thuốc thử Fehling, lăc kỹ, đun cách thủy sôi nhẹ 5-7 phút. Quan sát, nhận xét, giải thích

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.