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required to react with 4.35 g of solid containing BaSO4 (s)? Atomic mass: Ba 137.33, N 18. How many milliliters of 3.00 M H2S
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Firstly mass of barium nitrate is calculated from the given percent and then molecular weight is also calculated, then using stoichiometry moles of sulfuric acid is calculated and then volume is calculated.

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Answees (H25043=3.00M mass of sample= 4.35g. Ba(NO3), mass = 4.35 x 2 3.2 =1,00929. Ba2++5042--- BaSO465) Molar mass of Ba(N

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