Calcium has Wt 40, carbon is 12 and oxygen 16, so the MWt of calcium carbonate is 40+12+3x16=100. As this contains 40 calcium, calcium carbonate is 40% calcium. 40% of 418 is 167.2grams
The chemical formula for calcium carbonate is CaCO3, not calcium carbon monoxide. Carbon monoxide is a different compound consisting of one carbon atom and one oxygen atom with the chemical formula CO.
To find the number of molecules of carbon monoxide in 3.69 grams, first calculate the number of moles using the molar mass of carbon monoxide (28.01 g/mol). Next, use Avogadro's number to determine the number of molecules in those moles of carbon monoxide.
Calcium carbide is prepared from quick lime (calcium oxide) and coke in an electric arc furnace at temperatures above 2000°C. The process involves the reaction between calcium oxide and coke to produce calcium carbide and carbon monoxide gas. This reaction is represented by the equation: CaO + 3C -> CaC2 + CO.
The reaction between calcium carbonate and sodium metal would likely produce calcium oxide, sodium oxide, and carbon. The calcium oxide and sodium oxide would be the main products, with carbon formation as a byproduct.
Cobalt = Co transition metal and has a charge of 3+ Bromide = Br halogen and has a charge of 1- Co of 3+ and Br of 1- Cross method Metal First Formula: CoBr3
13.2g
To calculate the amount of calcium in 34.5 g of CaCO3, you need to consider the molar mass of CaCO3 which is 100.09 g/mol. Calcium accounts for approximately 40.08 g in every 100.09 g of CaCO3, which means there are (40.08/100.09) * 34.5 g = 13.82 g of calcium in 34.5 g of CaCO3.
160...cant quite grasp HOW though
0.87 g
Fe2O3 (s) + 3 CO (g) 🡪 2 Fe (s) + 3 CO2 (g) Calculate the number of grams of CO that can react with 250 g of Fe2O3. Calculate the number of grams of Fe and the number of grams of CO2 formed when 250 grams of Fe2O3 reacts.
To calculate the grams of CO needed to react with Fe2O3, you need to write a balanced chemical equation for the reaction, determine the moles of Fe2O3 given the mass provided, and use the stoichiometry of the balanced equation to find the moles of CO needed. Finally, convert the moles of CO to grams using the molar mass of CO.
26,3 g cobalt is equivalent to 0,446 moles.
The answer is 445,6 g carbon dioxide.
To calculate the amount of CO formed from 35.0 grams of oxygen, you need to determine the limiting reactant. First, convert 35.0 grams of O2 to moles. Then, use the balanced equation to calculate the moles of CO that can be formed from the moles of O2. Finally, convert the moles of CO to grams using the molar mass of CO.
5.34 grams W(CO)6 (1 mole W(CO)6/351.86 grams)(6 moles C/1 mole W(CO)6)(6.022 X 1023/1 mole C) = 5.48 X 1022 atoms of carbon ========================
To determine the mass of carbon monoxide in 2.55 moles, we first find the molar mass of CO, which is 28.01 g/mol. Then, we multiply the molar mass by the number of moles: 28.01 g/mol * 2.55 mol = 71.53 grams of CO in 2.55 moles of the compound.
There is no compound with the formula CaCO the here is howver CaCO3 which is calcium carbonate. If the question was chemical names of Ca and CO then the answer is calcium and carbon monoxide