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How many moles of calcium are in 425 g calcium (Ca)?
425 g calcium (Ca) is equal to 10,604 moles.
How many grams is contained in .27 moles of calcium?
(2.7 mol Ca / 1) * (40.078 g Ca / 1 mol Ca) = 108.21 g Ca. After significant figures, the answer is 110 g Ca.
How many moles are in 16.4 g of Ca(NO3)2?
To find the number of moles in 16.4 g of calcium nitrate, Ca(NO3)2, you first need to calculate its molar mass. The molar mass of Ca(NO3)2 is approximately 164.1 g/mol. Using the formula for moles (moles = mass / molar mass), you can calculate the moles: ( \text{moles} = \frac{16.4 , \text{g}}{164.1 , \text{g/mol}} \approx 0.100 , \text{moles} ). Therefore, there are about 0.100 moles of Ca(NO3)2 in 16.4 g.
How many moles are represented 40.1 g Ca?
To find the number of moles of calcium (Ca) in 40.1 grams, you can use the formula: moles = mass (g) / molar mass (g/mol). The molar mass of calcium is approximately 40.08 g/mol. Therefore, the calculation is 40.1 g / 40.08 g/mol, which is about 1.0005 moles of calcium.
How many moles are in 12.0 g of Ca(OH)2?
To find the number of moles in 12.0 g of calcium hydroxide, Ca(OH)₂, you first need to calculate its molar mass. The molar mass of Ca(OH)₂ is approximately 74.09 g/mol (calcium: 40.08 g/mol, oxygen: 16.00 g/mol, hydrogen: 1.01 g/mol). Using the formula ( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} ), you can calculate the moles: ( \frac{12.0 , \text{g}}{74.09 , \text{g/mol}} \approx 0.162 , \text{moles} ). Thus, there are approximately 0.162 moles of Ca(OH)₂ in 12.0 g.
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 are in 156 g of the element?
156 g calcium is equivalent to 3,89 moles.
How many moles of atom contained in 0.688 g Ca?
0,688 g calcium is equivalent to 0,017 moles.
How many grams is contained in .27 moles of calcium?
(2.7 mol Ca / 1) * (40.078 g Ca / 1 mol Ca) = 108.21 g Ca. After significant figures, the answer is 110 g Ca.
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.
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 are present in 808g of calcium?
40 g of Ca = 1 mole So 808 g of Ca = 20.2 mole
How many moles of calcium are there in 77.4 grams of Ca?
To find the number of moles, you need to divide the given mass of calcium by its molar mass. The molar mass of calcium (Ca) is approximately 40.08 g/mol. Number of moles = 77.4 g / 40.08 g/mol ≈ 1.93 moles.
Which value most closely represents the number of grams in 9.0 moles of calcium metal 200 g Ca 6.02 x 1023 g Ca 400 g Ca 360 g Ca?
9,0 moles of calcium is equal to 360,702 g.
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
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.
How many moles of calcium ca atoms are in 98.5 g of ca?
To determine the number of moles of calcium atoms in 98.5 g of Ca, you need to first calculate the molar mass of Ca (40.08 g/mol). Then, divide the given mass by the molar mass to get the number of moles. So, 98.5 g of Ca is equal to 98.5 g / 40.08 g/mol ≈ 2.46 moles of Ca atoms.
