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What is the total number of grams of NaI(s) needed to make 1 L of a 0.010M solution?
To make a 0.010M NaI solution, you'll need 0.010 moles of NaI per liter. The molar mass of NaI is 149.89 g/mol. Therefore, to calculate the grams needed, you would multiply the molar mass by the number of moles, which gives you 1.499 g of NaI needed per liter of solution.
How do you make a Molar solution from 32 percent hydrochloric acid?
To make a molar solution from a 32% hydrochloric acid solution, you would need to first calculate the molarity of the 32% solution. Molarity is calculated by multiplying the percent concentration by the density of the solution and dividing by the molar mass of the solute. Once you determine the molarity, you can then dilute the solution to the desired molar concentration by adding the appropriate amount of solvent (usually water).
Potassium fluoride (KF) a salt has a molecular weight of 58.10 grams. How many grams would be needed to mix 1.0 liter of 0.10 M salt solution?
To calculate the grams needed to mix a 0.10 M potassium fluoride solution, you would first find the molar mass of KF (58.10 g/mol). Then use the formula Molarity = moles/Liter to find the moles of KF needed (moles = Molarity x Volume). Finally, convert moles to grams using the molar mass: grams = moles x molar mass. In this case, you would need 5.81 grams of KF to make a 0.10 M solution in 1.0 liter.
How much calcium carbonate in grams can be dissolved by 5.00 grams of hydrochloric acid?
The molar mass of calcium carbonate is 100.09 g/mol and the molar mass of hydrochloric acid is 36.46 g/mol. Using the mole ratio of the balanced chemical equation, you can calculate the amount of calcium carbonate that can be dissolved by 5.00 grams of hydrochloric acid, which is approximately 12.43 grams.
How many grams of CoCl2 6H2O are needed to prepare 50.0ml of a 0.30M CoCl2 solution?
To determine the mass of CoCl2 6H2O to prepare a 0.30M solution in 50.0ml, we first calculate the molar mass of CoCl2 6H2O. The molar mass is 237.93 g/mol, so for a 0.30M solution in 50.0ml, we can use the formula: mass = molarity x volume (L) x molar mass. Therefore, the mass of CoCl2 6H2O needed is 3.57 grams.
How many millimeters of a 2 Molar solution of hydrochloric acid are needed to completely react with 75 grams of iron sulfide?
FeS + 2HCl >> FeCl2 + H2S 75 grams FeS (1mole FeS/87.92 grams)(2 mole HCl/1 mole FeS) = 1.71 moles HCl 2 Molar HCl = 1.71 moles HCl/Liters = 0.855 Liters HCl, or as asked for; 855 milliliters of hydrochloric acid needed
What is the molar mass of a solution containing 2000 grams?
Only a compound has a molar mass not a solution.
How many grams of sodium fluoride are needed to make 6.3 liters of a 3.6 molar solution?
To find the grams of sodium fluoride needed, use the formula: grams = moles x molar mass. First calculate the moles by multiplying the volume (6.3 L) by the molarity (3.6 mol/L). Then multiply the moles by the molar mass of sodium fluoride (41.99 g/mol) to find the grams required. In this case, approximately 920 grams of sodium fluoride are needed.
Potassium fluoride (KF a salt has a molecular weight of 58.10 grams. How many grams would be needed to make 1.0 liter of 5.0 molar salt solution?
290 grams
What is the total number of grams of NaI(s) needed to make 1 L of a 0.010M solution?
To make a 0.010M NaI solution, you'll need 0.010 moles of NaI per liter. The molar mass of NaI is 149.89 g/mol. Therefore, to calculate the grams needed, you would multiply the molar mass by the number of moles, which gives you 1.499 g of NaI needed per liter of solution.
How do you make a Molar solution from 32 percent hydrochloric acid?
To make a molar solution from a 32% hydrochloric acid solution, you would need to first calculate the molarity of the 32% solution. Molarity is calculated by multiplying the percent concentration by the density of the solution and dividing by the molar mass of the solute. Once you determine the molarity, you can then dilute the solution to the desired molar concentration by adding the appropriate amount of solvent (usually water).
Potassium fluoride (KF) a salt has a molecular weight of 58.10 grams. How many grams would be needed to mix 1.0 liter of 0.10 M salt solution?
To calculate the grams needed to mix a 0.10 M potassium fluoride solution, you would first find the molar mass of KF (58.10 g/mol). Then use the formula Molarity = moles/Liter to find the moles of KF needed (moles = Molarity x Volume). Finally, convert moles to grams using the molar mass: grams = moles x molar mass. In this case, you would need 5.81 grams of KF to make a 0.10 M solution in 1.0 liter.
How much calcium carbonate in grams can be dissolved by 5.00 grams of hydrochloric acid?
The molar mass of calcium carbonate is 100.09 g/mol and the molar mass of hydrochloric acid is 36.46 g/mol. Using the mole ratio of the balanced chemical equation, you can calculate the amount of calcium carbonate that can be dissolved by 5.00 grams of hydrochloric acid, which is approximately 12.43 grams.
How many grams of CoCl2 6H2O are needed to prepare 50.0ml of a 0.30M CoCl2 solution?
To determine the mass of CoCl2 6H2O to prepare a 0.30M solution in 50.0ml, we first calculate the molar mass of CoCl2 6H2O. The molar mass is 237.93 g/mol, so for a 0.30M solution in 50.0ml, we can use the formula: mass = molarity x volume (L) x molar mass. Therefore, the mass of CoCl2 6H2O needed is 3.57 grams.
How many grams of KCl is needed to make 2 molar solution in 1 liter of water?
To prepare a 2 M solution of KCl in 1 liter of water, you would need to dissolve 149.5 grams of KCl. This is because the molar mass of KCl is approximately 74.5 g/mol, and 2 moles of KCl are needed to prepare a 2 M solution in 1 liter of water.
What is the morality of a solution that has 120 grams of NaOH in 500 ML of solution?
6 molar
How many grams of sodium chloride are required to make a 1.0 L solution with a concentration of 3.5 M?
The molecular weight of NaCl is 58.44; sodium =22.99; Chlorine=35.45. A 1 molar solution is the molecular weight in grams in 1 litre of water, so a 3.5 molar solution would be 58.44g multiplied by 3.5, which is 204.54g in 1L.
