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How many moles of ions are in 2 moles of CaOH2?
2 moles of Ca and 4 moles of OH
How many moles of calcium are in 425 g calcium (Ca)?
425 g calcium (Ca) is equal to 10,604 moles.
How many moles of ca would be in 3.8G?
3.8 grams calcium (1 mole Ca/40.08 grams) = 0.09 moles calcium ==============
How many moles are in 978 g Ca?
978 g calcium contain 24,4 moles.
How many moles of atoms are present in 4 moles of Ca(HCO3)2?
In Ca(HCO3)2, there are 2 moles of carbonate ions (CO3^2-), each containing 3 atoms. So there are 6 atoms in 1 molecule of Ca(HCO3)2. Therefore, in 4 moles of Ca(HCO3)2, there are 4 x 6 = 24 moles of atoms.
How many grams are there in .761 moles of Ca?
For this you need the atomic mass of Ca. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. Ca= 40.1 grams.761 moles Ca × (40.1 grams) = 30.5 grams Ca
How many moles of Ca are there in 0.18moles of Ca F2?
There are 0.18 moles of Ca2+ ions in 0.18 moles of CaF2.
How many atoms would 3 moles of Ca have?
3 x 6.02E23 molecules.
How many moles of calcium is 126g of ca?
126 grams calcium (1 mole Ca/40.08 grams) = 3.14 moles of calcium ------------------------------
How many moles of atom contained in 0.688 g Ca?
0,688 g calcium is equivalent to 0,017 moles.
How many moles of calcium atoms are in 45.8 g of Ca?
To determine the number of moles of calcium atoms in 45.8 g of Ca, first find the molar mass of calcium (Ca) from the periodic table (40.08 g/mol). Then, divide the given mass by the molar mass of Ca to get the number of moles. In this case, 45.8 g of Ca is equal to 1.14 moles of Ca atoms.
A compound was analyzed and found to contain 13.5 g Ca 10.8g O and 0.675 g H what is the empirical formula of the compound?
First, determine the moles of each element: moles of Ca = 13.5 g / 40.08 g/mol = 0.336 moles moles of O = 10.8 g / 16.00 g/mol = 0.675 moles moles of H = 0.675 g / 1.01 g/mol = 0.668 moles Next, divide each mole value by the smallest mole value to get the ratio of atoms. The ratio is roughly 0.5 Ca to 1 O to 1 H. Therefore, the empirical formula of the compound is Ca(OH)2.
