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2 mol of AgCl to 1 mol BaCl2
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How many total atoms are there in 0.550 mol of barium chloride (BaCl2)?
0.550 mole BaCl2 x 6.02x10^23 formula units/mole x 3 atoms/formula unit = 9.93x10^23 atoms
How do you make BaCl2 solution?
by dissolving the amount you want in mole or grams in one dm3 of water
What volume of 0.131 M BaCl2 is required to react completely with 42.0 ml of .453 MNa2SO4. This is the net ionic equation for the reaction. Ba2 plus plus SO42---- BaSO4. Correct answer equals 145 ML?
first of all, you need to recognize that one mole of Na2SO4 is reacting with one mole of BaCL2. so find the moles of the NaSO4, then you automatically have the moles of the BaCL2. if you get the moles of the BaCL2, its easy to calculate the volume of it because you already have the MOLARITY. good luck
An impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgNO3 formed 2.044 g of Ag-Cl what is the percentage of Na-Cl in the impure sample?
2.044 g AgCl / 143.33 g per mole = 0.01426 moles of AgClNaCl + AgNO3 --> AgCl + NaNO31 mole of AgCl = 1 mole of NaClThe sample contained 0.01426 moles of NaCl0.01426 moles NaCl x 58.44 g per mole = 0.8334 g NaCl0.8334 / 0.8421 = 98.97% pureI'm pretty sure this is the correct answer.
How do you calculate the formula of hydrated Barium Chloride given that a sample of 10g of BaCl2.nH2O crystals lost 1.475g of of water on heating?
Mass of BaCl2 = crucible + lid + BaCl2.nH2O - empty crucible + lidMass of water contained in hydrated BaCl2: Initial mass - final massNo. of moles of H2O in hydrated salt = No. of moles = mass of H2O/formula mass of H2O Use this answer in ratioNo. of moles of BaCl2 in 3.399 g = No. of moles = mass of BaCl2/formula mass of BaCl2 Use this answer in ratioThe ratio of BaCl2 to H2O is 1 : 2
When agno3 reacts with bacl2 agcl and bano32 are formed how many grams of agcl are formed when 10.0g of silver nitrate reacts with 15.0g of bacl2?
Hit wrong button
How many grams of solid barium sulfate form when 22.6 mL of 0.160 M barium chloride reacts with 54.6 mL of 0.055 M sodium sulfate?
Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced
How does the experimental mole ratio compare with the theoretical mole ratio?
the experimental mole ratio has a bigger penis
How many total atoms are there in 0.550 mol of barium chloride (BaCl2)?
0.550 mole BaCl2 x 6.02x10^23 formula units/mole x 3 atoms/formula unit = 9.93x10^23 atoms
How do you make BaCl2 solution?
by dissolving the amount you want in mole or grams in one dm3 of water
What is the balanced equation of barium chloride and sodium chromate?
The balanced equation is: Ag+ (aq) + Cl- (aq) → AgCl(s)
How do you figure out how much NaCl will be necessary to precipitate silver?
One mole of chloride (Cl-) to one mole of Ag+ ions: Cl- + Ag+ --> AgCl(s)
What volume of 0.131 M BaCl2 is required to react completely with 42.0 ml of .453 MNa2SO4. This is the net ionic equation for the reaction. Ba2 plus plus SO42---- BaSO4. Correct answer equals 145 ML?
first of all, you need to recognize that one mole of Na2SO4 is reacting with one mole of BaCL2. so find the moles of the NaSO4, then you automatically have the moles of the BaCL2. if you get the moles of the BaCL2, its easy to calculate the volume of it because you already have the MOLARITY. good luck
An impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgNO3 formed 2.044 g of Ag-Cl what is the percentage of Na-Cl in the impure sample?
2.044 g AgCl / 143.33 g per mole = 0.01426 moles of AgClNaCl + AgNO3 --> AgCl + NaNO31 mole of AgCl = 1 mole of NaClThe sample contained 0.01426 moles of NaCl0.01426 moles NaCl x 58.44 g per mole = 0.8334 g NaCl0.8334 / 0.8421 = 98.97% pureI'm pretty sure this is the correct answer.
How many ml of 0.500 M BaCl2 is needed to obtain 25 mili moles of BaCl2?
calculate the molarity of 0.25 mol in 0.250 L solution
How do you calculate the formula of hydrated Barium Chloride given that a sample of 10g of BaCl2.nH2O crystals lost 1.475g of of water on heating?
Mass of BaCl2 = crucible + lid + BaCl2.nH2O - empty crucible + lidMass of water contained in hydrated BaCl2: Initial mass - final massNo. of moles of H2O in hydrated salt = No. of moles = mass of H2O/formula mass of H2O Use this answer in ratioNo. of moles of BaCl2 in 3.399 g = No. of moles = mass of BaCl2/formula mass of BaCl2 Use this answer in ratioThe ratio of BaCl2 to H2O is 1 : 2
What is the mole ratio of salt to water in NA2SO4.10H2O?
An anhydrous sals hasn't water.
