Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
Fe2O3 + 3C → 2Fe + 3CO :)
Answer:
The answer to your question is: 69.6 %
Explanation:
Freon -112 (C₂Cl₄F₂)
MW = (12 x 2) + (35.5 x 4) + (19 x 2)
= 24 + 142 + 38
= 204 g
204 g of C₂Cl₄F₂ ----------------- 100%
142 g ----------------- x
x = (142 x 100 ) / 204
x = 69.6 %
The student's test average is : 0.92
Total score obtained = 369
Number of tests = 4
Since Each test is exactly 100 points ;
The total score obtainable is (100 × 4) = 400
Average = score obtained / total score obtainable
Average = 369 / 400
Average = 0.9225
Average = 0.92
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It will float, because it is almost weighs like a pen.