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There are 0.07871604895385 moles of CaC12 in 14.5g of CaC12.
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What is the mass of 2.430 mole of CaCl2?
2.430 moles CaCl2 x 110.98 g CaCl2/mole CaCl2 = 269.7 grams (4 sig figs)
How many moles are in 1 g of Cl2?
# of Moles = Mass in grams divided by Molar Mass =5o divided by (cl x 2) =50 divided by 71 =0.704 moles use: 1 mol = Mr in grams that is 35.5x2 g of Cl2 = 1 mol 71g of Cl2 = 1 mol therefore 50g of Cl2 = (1/71) x 50 =0.704 mol
How many moles are in 23.34 g of NaCl?
The answer is 0,4 moles.
How many moles are in 978 g Ca?
978 g calcium contain 24,4 moles.
How many moles are in 14.84 g Mg?
14,84 g magnesium are equivalent to 0,61 moles.
How many moles are present in 140.g of Cacl2?
1.26 moles
What is the mass of 2.430 mole of CaCl2?
2.430 moles CaCl2 x 110.98 g CaCl2/mole CaCl2 = 269.7 grams (4 sig figs)
How many grams are in 0.74 moles of CaCl2?
Atomic Weight of Calcium = 40 Atomic Weight of Chlorine = 35.5 Therefore, 1 mole of CaCl2 => 40 + 2 (35.5) = 111 g 0.74 moles of CaCl2 => 0.74 (111) = 82.14 g
How many moles are in 1 g of Cl2?
# of Moles = Mass in grams divided by Molar Mass =5o divided by (cl x 2) =50 divided by 71 =0.704 moles use: 1 mol = Mr in grams that is 35.5x2 g of Cl2 = 1 mol 71g of Cl2 = 1 mol therefore 50g of Cl2 = (1/71) x 50 =0.704 mol
Determine the number of moles of Cl in 1.9 mol CaCl2?
Ca-1(40.08)g=40.8g/mol of Ca Cl-2(35.45)g=70.9g/mol of Cl2 40.08g+70.9g=110.98g/mol CaCl2 110.98g CaCl2/1mol * 1.9mol =210.86 g of CaCl2
What mass of CaCL2 is needed to prepare 2.850 L of a 1.56 M solution?
2.850 L x (1.56 moles CaCl2 / 1 L) x (111.0 g CaCl2 / mole CaCl2) = 493g
How many grams of calcium chloride will you add to 250mL of water to make a 0.236 M solution?
Molarity is moles of solute / liters of solvent. Plugging in the data: 0.236M = x / 0.250L; x = (0.236M)(0.250L) = 0.0590 moles of CaCl2. The molecular weight of CaCl2 is 40.1 + 2(35.5) = 111.1 g / mole. The mass of CaCl2 = (MW)(moles) = (111.1g/mole)(0.0590moles) = 6.55g
How many grams of CaCl2 are needed to prepare 125 ml of a 1.50 M solution?
Molarity = moles of solute/volume of solution 1.50 M = X Moles/125ml = 187.5 millimoles, which is 0.1875 moles The molar mass of CaCl2 = 110.98 grams 0.1875 moles CaCl2 (110.98g/1mol) = 20.8 grams needed
How many grams of cacl2 are in 35.00 ml of a 0.237 m solution of cacl2 a 0.307 g b 0.460 g c 0.920 g d 1.84 g e 3.75 g?
Assuming that "m" was supposed to mean "molar" ("M" is usually used), the answer to this question can be found from the following calculation: The moles of calcium chloride present are 0.237(35/1000) = 0.00829. CaCl2 has a "molecular" mass of about 110; the product of this number and the number of grams is 0.920.
How many grams of CaCl2 can be obtained if 14.6 g HCl is allowed to react with excess CaCO3?
22.2g CaCl2
How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
If you have g of beryllium how many moles is this equivalent to?
The formula is: number of moles = g Be/9,012.
