A sample of Caco3 has ca=40% if the law of constant proportion is true then the mass of Ca in 4 gram of a sample of CaCo3 from another source will be :
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As it is also given that the law of constant proportion holds true, then this means that even the sample of 4g of calcium carbonate will have the percentage composition of the constituent elements as: Ca = 40%; C = 12%; O = 48%. Therefore, this means that 100g of $CaC{O}_{3}$, 40 g of Ca is present.
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