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Weigh 3g Salt into a 100ml container and bring to volume with deionized water. Mix well. This will be a 3% standard. Other standards are made in the same way (ex.: 2g in 100ml water = 2.00% or 2.5g in 100 ml = 2.50%). I use this in my lab when I make our standard. Hope your not getting this information to late!
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What is the density of 35 weight percent sulfuric acid how to obtain the molarity for this?
The density of 35 wt% sulfuric acid is 1.174 g/cm^3. To convert weight percent to molarity, you first need to calculate the molar mass of sulfuric acid (98.08 g/mol). Then, using the density and molar mass, you can calculate the molarity (11.9 M) using the formula: Molarity = (wt% * density) / (molar mass).
How many teaspoons of one gram of salt?
1 teaspoon of salt (NaCL) = 6 grams of NaCL or 2,360 mg of sodium (Na). (Na molecular wt is 23, Cl is 35.5, so NaCl is 58.5, so Na is 40 % times grams of wt of NaCl.) So 1 gram NaCl X 1 tsp/6 g NaCl = 0.16667 tsp NaCl or if you want Sodium ( how salt is expressed on food labels) then 1 gram of Na X 1 tsp NaCl/2.360 gram Na = 0.42 tsp Na.
How can I convert wt Al2o3 percent to Al percent?
To convert weight percent of Al2O3 to weight percent of Al, first recognize that each mole of Al2O3 contains two moles of Al. Calculate the molar mass of Al2O3 (approximately 102 g/mol) and of Al (approximately 27 g/mol). Use the formula: [ \text{wt % Al} = \left( \frac{\text{wt % Al2O3} \times 2 \times \text{molar mass of Al}}{\text{molar mass of Al2O3}} \right) ] Substituting the values gives you the weight percent of Al.
How do you convert weight percent of element into weight percent of oxides that exist?
Take 1.45 wt% oxide as an example. To remove the oxide component first you must work out the Ti component: 1.45 wt% oxide x 0.6 = Ti (wt%) because Ti in TiO2 makes up 60% its formula mass: 47.87 / 79.87 = ~0.6 Ti Thus 1.45wt% TiO2 = 0.87 wt% Ti
What does a quarter wt?
"wt"?
How many grams are there of sodium chloride in 0.500 L of a solution in its standard state?
1000 ppm = 1000 mg/ l 1N NaCl = 58.5 gm/l ok 1000 mg/l = 1 gm/l = 0.1gm/100 ml = 0.1 % wt. on the other hand 1 % wt. = 10,000 ppm so , 0.1 %wt. = 1000 ppm so you need to put 0.1 gm in 100 ml = 1 gm in 1000 ml to get 1000 ppm
What is the percent by mass of 35g of sodium chloride mixed with 245ml of water?
Calculate the mass of water (density: 1 g/ml) --> 245 ml * 1 g/ml = 245 g of water wt % = 35 g / (35g + 245g) * 100% = 12.5%
How do you prepare 0.5L of 65 percent Nitric acid from 69.3 percent of Nitric acid?
It depends on if the Assay is in wt / wt or wt / L because if that's the case you will need to find the density of the 67-70% HNO3 at room temp.But basically you used the equation:C1V1 C2V2whereC1 initial concentrationV1 the amount of solution you will needC2 final concentrationV2 final volumeI would take the average of the assay for C1, so(67+70) / 2 68.5you need to know how much of the solution you want for example 3 Lthen you plug it in, this is also assuming that the initial concentration is in wt / LC1 * V1C2 * V268.5% * V12% * 3Lrearrange to form:V1 (2% * 3L) / 68.5% 0.0876Lwhich you can change into Wt / Wt with the density, or if you need mL just convert with ( 1000mL / 1L )
How much does a loaded ammo can of 50 cal rounds weigh?
35 lbs. (15.9kg) 100 rds empty can wt. 5.3 lbs
Abbreviatins for wt?
wt is an abbreviation for weight
How do you prepare 5 percent nitric acid solution from 55 percent nitric acid?
To solve this problem, we basically have 2 equations and 2 unknowns. The unknowns are the (volume of water) & the (volume of 70 wt%) nitric acid to add. * This problem will assume that you are interested in making 1 L (or 1000 mL) of 5 wt% nitric acid solution. Equation 1: (volume of water) + (volume of 70 wt% nitric acid) = 1000 mL Equation 2: mass of nitric acid / [mass of water + mass of 70 wt% nitric acid solution] = 0.05 (0.05 is 5 wt%) * Remember that mass = density * volume * Remember that 70 wt% nitric acid solution mean that for 100 grams (gm) of this acid, then there's 70 grams of HNO3 * Remember that density of 70 wt% nitric acid solution is 1.413 gm/cm^3 * Remember that density of water is 1 gm/cm^3 Equation 2 is now re-written as: [(density of 70 wt% nitric acid soln)*(volume of 70 wt% nitric acid)*0.70] / [(volume water)*(1gm*cm^3) + (volume of 70 wt% nitric acid)*(1.413gm/cm^3)] = 0.05 Solving for the 2 equations gives answer to the 2 unknowns: Answer: To make 1000 mL of 5 wt% nitric acid solution, add 1) 51.63 mL of 70 wt% nitric acid solution 2) 948.37 mL of water
What is the percent by weight of a solution prepared by dissolving 10 g of NaCl in 100 g of H2O?
% by weight in solution = = 100 * (weight solute) / [(weight solvent) + (weight solute)] = 100 * (weight solute) / [weight solution] = 100* 10 / [100 + 10] = 100 * 10 / 110 = 9.1%
