(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) = h2seo3+no+NO2 =, so the answer is: Se+2HNO3 (concentrated) =

(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) According to the reaction equation, the consumption of SEO 2 ~ 2I2 ~ 4Na2S2O3 is n(Na2S2O3)=0.2000? Mol/L×0.025L=0.005mol, and according to the relation, n (SEO 2) = 0.005mol×14 = 0.00125mol in the sample, so the mass of seo2 is 0.00 125mol× 65438+. Therefore, the mass fraction of SeO2 in the sample is 0.13875g 0.2500g×100% = 55.5%, so the answer is 55.5%.