(1) According to the information in the question, Se reacts with concentrated HNO3, and Se is oxidized to +4 valence H2SeO3, and HNO3 is reduced to NO and NO2, and the ratio of substances producing NO and NO2 is 1: 1, that is, the ratio of their stoichiometric coefficients is1:/kloc-0. The coefficient of Se is1× 3+1×14 =1,so the reaction equation is: Se+2HNO3 (concentrated) = H2SO4+NO+NO2 ↑,

So the answer is: Se+2HNO3 (concentrated) = H2SO4 3+NO =+NO2 =;

(2) In the redox reaction, the oxidizability of oxidant is stronger than that of oxidation products, so according to the reaction equation, the order of the oxidizability of SeO2, H2SO4 (concentrated) and SO2 from strong to weak is H2SO4 (concentrated) > SEO 2 > SO2.

So, the answer is: H2SO4 (concentrated) > SEO 2 > SO2.

(3) In the reaction ①, I- loses electrons to generate I2, * * * increases the valence by 2, and the simple substance obtained by +4 valence Se in seO2 is reduced to simple substance Se, * * * decreases the valence by 4, so the lowest common multiple of valence increase or decrease is 4, so the coefficient of KI is 4, the coefficient of I2 is 2, and the coefficients of SeO2 and Se are 1, KNO3.

So the answer is:

(4) According to the reaction equation, the consumption of N (Na2S2O3) in SEO 2 ~ 2I2 ~ 4Na2S2O3 is 0.2000 mol/l× 0.025L = 0.005 mol, and the N (SEO 2) = 0.005 mol×14 = 0. Therefore, the mass of SeO2 is 0.00125mol×1kloc-0/g/mol = 0.13875g, so the mass fraction of SeO2 in the sample is 0.13875g 0.

So the answer is 92.50%.