Molar mass of CaCO3 = 100.0869 g/mol
Molar mass of CaNO32 = 566.0655 g/mol
For this you need the atomic (molecular) mass of CaCO3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CaCO3= 100.1 grams2.50 moles CaCO3 × (100.1 grams) = 250.25 grams CaCO3
Atomic weight of carbon: [12,0096; 12,0116]. Atomic weight of chlorine: [35,446; 35,457]. The molar mass of chlorine is greater; rounded 35,45 g/mol.
3 mol CaCO3 will produce 3 mol CaO (and 3 mol CO2) on dry-heating. So 3 times the molar mass of CaO (56,1 g/mol) which is 168 g CaO
The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol. To find the mass of 0.250 mol of calcium carbonate, you would multiply the number of moles by the molar mass: 0.250 mol x 100.09 g/mol = 25.02 grams of calcium carbonate.
The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CO2 is 44.01 g/mol. By using stoichiometry, you can calculate that 15.2 grams of CaCO3 would produce 6.51 grams of CO2. Using the ideal gas law, you can then convert the mass of CO2 to volume using its molar volume at STP (22.4 L/mol). The volume of CO2 produced would be around 3.32 liters.
Molar mass of CaCO3 = 100.0869 g/mol
Both C7H16 and CaCO3 have a molar mass of approximately 116 grams per mole.
To find the number of moles in 73.4 kg of CaCO3, we first need to calculate the molar mass of CaCO3. The molar mass of CaCO3 is 100.09 g/mol. Converting 73.4 kg to grams gives 73,400 g. Dividing 73,400 g by the molar mass of CaCO3 gives approximately 733 moles.
To convert alkalinity (HCO3) to CaCO3, you need to use the molar mass ratio. For every mole of bicarbonate (HCO3), you have one mole of carbonate (CO3) in CaCO3. So, to convert, you can multiply the HCO3 concentration by a factor of 50.04 (molar mass of CaCO3/molar mass of HCO3).
Molar mass of CaCO3 = 66.1221g CaCO3/mole CaCO3. This means that 1 mole CaCO3 = 66.1221g CaCO3. To find the mass of 4.5 mole CaCO3, complete the following calculation: 4.5g CaCO3 X 1mol CaCO3/66.1221g CaCO3 = 0.068 mole CaCO3.
250 grams CaCO3 (1 mole CaCO3/100.09 grams) = 2.50 moles of calcium carbonate
The molar mass of CaCO3 is approximately 100.09 g/mol. Therefore, the mass of 0.5 moles of CaCO3 would be 50.045 g.
To calculate the mass of 5.66 mol of CaCO3, you first need to find the molar mass of CaCO3. The molar mass of CaCO3 is 100.09 g/mol (40.08 g/mol for Ca + 12.01 g/mol for C + 3(16.00 g/mol for O)). Then, multiply the molar mass by the number of moles to find the mass: 5.66 mol x 100.09 g/mol = 566.3 grams.
1.15 (g CaCO3) / 100.1 (g/mol CaCO3) =1.149*10-2 (mol Ca)1.149*10-2 (mol Ca) = 1.149*10-2 (mol Ca) * 40.08 (g/mol Ca) = 0.4604 g Ca0.4604 g Ca = 0.4604 g Ca / 2.70 g Supplement = 0.1705 * 100% = 17.1% Calcium (m%)
The molar mass of calcium carbonate (CaCO3) is approximately 100.09 g/mol. To find the mass of 0.5 moles of CaCO3, you would multiply the molar mass by the number of moles: 0.5 moles x 100.09 g/mol = 50.045 g. Therefore, the mass of 0.5 moles of calcium carbonate is 50.045 grams.
For this you need the atomic (molecular) mass of CaCO3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CaCO3= 100.1 grams2.50 moles CaCO3 × (100.1 grams) = 250.25 grams CaCO3
Atomic weight of carbon: [12,0096; 12,0116]. Atomic weight of chlorine: [35,446; 35,457]. The molar mass of chlorine is greater; rounded 35,45 g/mol.