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well take sodium, which the molar mass is 23.0 g/mol and multiply by # of moles (4) and you get 92g
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How would you prepare 400ml of a 0.24M NaCl solution?
We first calculate the amount, in moles, of NaCl that we will need.Amount of NaCl needed = 0.24 x 400/100 = 0.096mol. Mass of NaCl needed = (23.0 + 35.5) x 0.096 = 5.616g So to produce 400ml of 0.24M NaCl solution, accurately add 5.616 grams of NaCl to 400ml of deionised water.
How many grams of sodium fluoride are needed to prepare a 0.400m NaF solution that contains 750g of water?
The answer to your question is 12.6 because you multiply the .400 times .750 and then you take the answer (0.3) and multiply this times the molar mass of NaF (42) and get 12.6.
If 400.0 mL of 0.420 M NaCl are mixed with 110.0 mL of 0.240 M NaCl what is the molarity of the final solution?
In 400 ml of the 0.420 M NaCl there are 0.1680 moles of NaCl.In 110 ml of the 0.240 M NaCl there are 0.0264 moles of NaCl.There are 0.1944 moles of NaCl in 510 ml of solution. Dividing 0.1944 moles by 0.510 l gives 0.381 mole/l.So You get a 0.381 M NaCl.
The sterile solution used to rinse contact lenses can be made by dissolving 400 mg of NaCl in sterile water and diluting to 100 mL. what is the molarity of this solution?
First, find the moles of NaCl. Because you have 400 milligrams of NaCl convert to grams by dividing 1000mg. 400mg/1000mg = 0.400 grams of NaCl. Now we need to convert to moles. For every mole of NaCl there is 58.44 grams. (22.99g Na + 35.45 g Cl) so using stoichiometry (grams cancel) I get .0068 moles of NaCl. Conver 100 mL to L by dividing by 1000mL. Therefore there is .001 L in the solution. Molarity is moles per Liter. So 0.0068 mols of NaCl / 0.001 L = 6.84 M
How many grams of calcium nitrate needs to be added to 400g of water to make a solution that is 12.5 percent by mass of calcium nitrate?
Call the unknown mass m. Then, from the problem statement, m/(400 + m) = 12.5 % = 0.125. Applying standard algebra techniques, multiply both sides of this equation by (400 + m) to result in m = 0.125(400 + m); m(1 - 0.125) = 400 X 0.125; 0.875m = 50; m = 57.1 grams, to the justified number of significant digits.
How many moles of sodium chloride are in 400 mL of a 6 M NaCl solution?
The answer is 2,4 moles.
How do you calculate the number of moles of NaI in 50.0mL of a 0.400M solution?
First convert mL to liters. So .05L. We know that we have .400 moles/liter, so multiply .05L*.400 mol/L and the L cancels out, leaving us with .02 moles.
How many mg in 2 g?
For this you need the atomic mass of Na. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.2.1 moles Na × (23.0 grams) = 48.3 grams Na
How do you calculate the volume in milliliters of a 0.61M sodium nitrate solution that contains 400mmol of sodium nitrate?
By definition, a 0.61M sodium nitrate solution contains 0.61 moles of sodium nitrate per liter, which is equivalent to 0.61 mmol/ml. Therefore, the volume of this solution required to contain 400mmol is 400/0.61 or 6.6 X 102 ml, to the justified number of significant digits.
What is 400 subtracted from 2000?
-1600
How many moles are in 400 grams of nickel sulphate?
400 grams of nickel sulphate (anhydrous) is equivalent to 2,58 moles.
How many moles of aluminum oxide are present in a sample having a mass of 178.43g?
For this you need the atomic (molecular) mass of Al2O3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. Al2O3= 102 grams408 grams Al / (102 grams) = 4.00 moles Al
What is the number of moles of solute in 400 ml of a 0.4 M solute?
The answer is 0,16 moles.
Suppose you have 20.8 g of Cr What number of moles of Cr is present?
.400 moles
How many grams of solid sodium hydroxide are required to prepare 400 milliliters of a 4 hundredth molar solution?
So you want 0.04M but you have 400ml, not a litre. 0.04/1000*400 is 0.016 moles wanted. 0.016*40 (molecular weight) is 0.64g
How many moles are there in 400g of calcium cabonate?
400 g of calcium cabonate is equal to 3,996 moles.
300.0 Ml of water is added to .40 L of a .400 M Na2CrO4 solution what is the molarity of the resulting solution?
300.0 ml of water is added to .40 L of a .400 M Na2CrO4 solution what is the molarity of the resulting solution? Na2CrO4 = 2 Cr +Cr + 4 O's Molar mass = (2*23 + 52 + (4*16) = 162 A .400 M N Na2CrO4 solution has .400 moles of Na2CrO4 in a liter of water. .400 moles of Na2CrO4 = 0.400 * 162 = 64.8 grams of Na2CrO4 in a liter of water. Since you only have .40 L, you have 64.8 grams/liter * 0.4L = 25.92 grams of Na2CrO4 in 0.4 liter of solution. When you add 300.0 ml of water, you have total of 700 ml of solution. You still have 25.92 grams of Na2CrO4, but now you have 700 ml of solution. Molarity = moles of solute per liter of solution. Moles of solute = grams of solute ÷ Molar mass of solute Moles of solute = 25.92 ÷ 162 = 0.16 moles of Na2CrO4. Molarity = 0.16 moles of Na2CrO4 ÷ 0.700 L of solution. Molarity = 0.23 M
