题目

为了测定某稀硫酸的溶质质量分数，在10g稀硫酸中滴加过量的氯化钡溶液，得到沉淀2.33g，求该稀硫酸的溶质质量分数． 答案：解：设该稀硫酸的溶质质量为xBaCl2+H2SO4=BaSO4↓+2HCl 98 233 x 2.33g{#mathml#}98x{#/mathml#}={#mathml#}2332.33g{#/mathml#}解得x=0.98g该稀硫酸的溶质质量分数= {#mathml#}0.98g10g{#/mathml#}{#mathml#}×{#/mathml#}100%=9.8%答：该稀硫酸的溶质质量分数为9.8%． She has been the subject of_____ media coverage.A．massiveB．negativeC．expensiveD．active