\(a,m_{H_2SO_4}=\dfrac{500.49}{100}=245g\\ n_{H_2SO_4}=\dfrac{245}{98}=2,5\left(mol\right)\\ b,m_{NaOH}=\dfrac{60.20}{100}=12g\\ n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\\ b,m_{CuSO_4}=\dfrac{50.10}{100}=5g\\ n_{CuSO_4}=\dfrac{5}{160}=\dfrac{1}{32}\left(mol\right)\\ d,m_{BaCl_2}=\dfrac{200.5}{100}=10g\\ n_{BaCl_2}=\dfrac{10}{208}=\dfrac{5}{104}\left(mol\right)\)
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