Using the equation: 2 NaCl + H2SO4 -> 2 HCl + Na2SO4, we can see that 1 mole of NaCl will produce 1 mole of HCl. First, calculate the moles of NaCl (131g / 58.44g/mol). Then, using the mole ratio from the equation, you can find the moles of HCl produced. Finally, using the ideal gas law, you can convert the moles of HCl to volume at STP.
2 moles or 117 gram of NaCl is precepitated
1.31*.3=.393 mols NaCl. This also equals .393(22.99+35.45)=22.97 g NaCl.
The formula unit of sodium chloride (NaCl) contain 60,33 % chlorine.
The recommended concentration is 20 +/- 5 mg iodine/kg NaCl.
0.95% * 500 g = 4.75 g NaCl
To determine the grams of NaCl in 0.058% NaCl solution, you first convert 0.058% to a decimal by dividing by 100 (0.00058). Then, you set up a proportion to find out how much of the solution contains 1.5g of NaCl. 1.5g of NaCl is equivalent to 0.00058x, where x is the total grams of solution. Solve for x to find how much of the solution is needed.
You need 58,44 mg of ultrapure NaCl; dissolve in demineralized water, at 20 0C, in a thermostat, using a class A volumetric flask of 1 L.
To determine the amount of NaCl in the solution, you first need to calculate the moles of NaCl present. Using the given molarity (2.48 M) and the volume of the solution (assumed to be 806 g = 806 ml for water), you can find the moles of NaCl. Then, you convert the moles of NaCl to grams using the molar mass of NaCl (58.44 g/mol) to find the amount of NaCl in the solution.
how much is keratine produced?
Since the balanced chemical equation states that 1 mole of sodium carbonate reacts with 2 moles of HCl to produce 2 moles of NaCl, the theoretical yield of NaCl would be double the initial moles of HCl used. Given the molarity of HCl, you can calculate the moles of NaCl produced and convert to grams if the volume of the solution is known. Remember to consider stoichiometry when doing the calculation.
To find the amount of NaCl in 1.84 L of 0.200 M NaCl, you use the formula: amount = concentration x volume. Thus, amount = 0.200 mol/L x 1.84 L = 0.368 mol of NaCl.