2 moles of calcium is 80,156 g; 1 part per million is 1 mg/kg (or L).
The number of moles of calcium carbonate are 3.5 moles. , there are 1 mole of calcium (Ca) atom, 1 mole of carbon (C) atom and 3 moles of oxygen (O) atoms.
To determine the volume of carbon dioxide needed, you would need to know the stoichiometry of the reaction between carbon dioxide and calcium carbonate. In this case, since 20 grams of calcium carbonate is given, you would convert that to moles using the molar mass of calcium carbonate. Then, using the balanced equation, you can determine the mole ratio between carbon dioxide and calcium carbonate. Finally, using the molar volume of carbon dioxide gas at the given conditions (usually 22.4 L/mol at standard temperature and pressure), you can calculate the volume of carbon dioxide needed.
0,27 moles of calcium contain 10,82 g calcium.
425 g calcium (Ca) is equal to 10,604 moles.
5000 grams calcium (1 mole Ca/40.08 grams) = 124.75 moles Calcium
2,8 moles of calcium carbonate have 240,208 g.
The number of moles of calcium carbonate are 3.5 moles. , there are 1 mole of calcium (Ca) atom, 1 mole of carbon (C) atom and 3 moles of oxygen (O) atoms.
Calcium carbonate, CaCO3 has formula mass of 40.1+12.0+3(16.0) = 100.1Amount of CaCO3 = 1.719/100.1 = 0.0172molThere are 0.0172 moles of calcium carbonate in a 1.719 gram pure sample.
1400 grams
The formula given shows that each formula unit or mole contains one calcium atom; therefore, 2.5 moles of calcium chloride contains 2.5 moles of calcium atoms.
.06 moles
3.5 moles CaCO3 (1 mole carbon/1 mole CaCO3) = 3.5 moles
1. Calculate formula massCalcium carbonate has chemical formula CaCO3.Its formula mass is 40.1 + 12.0 + 3(16.0) = 100.12. Apply formula to calculate number of moles of CaCO3Amount of CaCO3= mass/formula mass= 50/100.1= 0.50mol
Calcium Carbonate (CaCO3)I mol CaCO3 contains 3 mol Oxygen atomsso 4.25 mol CaCO3 will have 12.75 mol Oxygen Atoms.
i am asking the same thing
This question is solved with the help of mole concept . 1 mole of anhydrous calcium carbonate weighs 40+12+48=100 gm . 1.25 mole of similar anhydrous calcium carbonate will be 100* 1.25 = 125 gm
250 grams CaCO3 (1 mole CaCO3/100.09 grams) = 2.50 moles of calcium carbonate