% K =2 x 39.1 x 100/ 138.2 =56.6
% C = 12 x 100/ 138.2 =8.68
% O = 3 x 16 x 100/ 138.2 =34.7
Potassium = 56.6 %
The percentage by mass of carbon in potassium carbonate is 8,695 %.
(39x2)+12+(16x3)=138
The percentage of nitrogen is 29,16 %.
The percentage of carbon is 71,58 %.
125 g of Zinc Carbonate is 1 mole and when it decomposes it forms 1 mole of Carbon Dioxide ZnCO3 = ZnO + CO2 1 mole of carbon dioxide has a mass of 44g 44g is the answer
The equation given shows that each formula mass of calcium carbonate produces one formula mass of CO2. The gram formula masses of calcium carbonate and carbon dioxide are 100.09 and 44.01 respectively. Therefore, to produce 4.4 grams of carbon dioxide, 4.4(100.09/44.01), or 10 grams of calcium carbonate, to the justified number of significant digits, are needed.
(39x2)+12+(16x3)=138
12 12.01
The percentage by mass of sodium (Na) in a formula unit of sodium hydrogen carbonate (NaHCO3) is 27,38 %.
The percentage of nitrogen is 29,16 %.
The percentage of carbon is 71,58 %.
125 g of Zinc Carbonate is 1 mole and when it decomposes it forms 1 mole of Carbon Dioxide ZnCO3 = ZnO + CO2 1 mole of carbon dioxide has a mass of 44g 44g is the answer
I believe it is: %O=mass of 3 mol of O / mass of 1 mol of potassium chlorate *100% If you measured it: %O=mass of oxygen lost / mass of potassium chlorate *100%
to keep the pH opf the reaction mass towards basic.
The mass of the carbonate used.
The equation given shows that each formula mass of calcium carbonate produces one formula mass of CO2. The gram formula masses of calcium carbonate and carbon dioxide are 100.09 and 44.01 respectively. Therefore, to produce 4.4 grams of carbon dioxide, 4.4(100.09/44.01), or 10 grams of calcium carbonate, to the justified number of significant digits, are needed.
Important is the percentage of calcium carbonate (CaCO3) in the material. For an equivalent mass of pure chalk or pure marble the quantity of released carbon dioxide is the same - the chemical formula is the same.
84.09