有NaCl和NaI的混合物共26.7 g,溶于水,通入足量氯气后,蒸干、灼烧,固体质量变为17.55 g。求原混合物中氯化钠的质量分数。 答案:解析:NaCl和NaI的混合物中通入氯气后,只有NaI发生反应,NaCl不反应,所以,固体的质量变化是由NaI变为NaCl引起的,且生成的I2蒸干、灼烧时会升华,最后的固体只有NaCl。 2NaI+Cl2══2NaCl+I2 Δm 300 183 x When he grows up, he's going to do _________.
[ ]A. what he wants to do B. what does he want to do C. what he want to do D. what will he like