When x molecules are removed from 200mg of N2O 2.89*10(-3)moles are le

1.	When x molecules are removed from 200mg of N2O 2.89*10(-3)moles are left then x will be
Answer» 1mg = 10^-3 g then , 200mg = 200×10^-3g _________________________ 1 MOLE of NO2 CONTAINS = (14×2)+(16) = 44g/mol then " y moles contains = 200mg = 200×10^-3g y = 200×10^3/44 y = 4.55 × 10^-3 moles then , 4.55 × 10^-3 - X = 2.89 × 10^-3 - X = (2.89 × 10^-3) - (4.55 × 10^-3) - X = -1.66 × 10^-3 X = 1.66 × 10^-3 ___________________________________________ molar mass:- mass of the one mol of the SUBSTANCE is called molar mass onother method mass = number of moles × molar mass 200mg = number of moles × 44g/mol number of moles = 200×10^-3/44 number of moles = 4.55 × 10^-3 moles so 200mg of N2O contains 4.55 × 10^-3 therefore , 4.55 × 10^-3 - X = 2.89 × 10^-3 - X = -1.66 × 10^-3 X = 1.66 × 10^-3 so your answer is 1.66 × 10^-3

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