Dissolve 29,22 g NaCl p.a. in 1 L demineralized water at 20 0C, in a volumetric flask.
fehling's solution is dark blue at room temperature "Fehling's solution" is prepared by dissolving separately 34'639 grammes of copper sulphate, 173 grammes of Rochelle salt, and 71 grammes of caustic soda in water, mixing and making up to l000 ccs.; 10 ccs. of this solution is completely reduced by o 05 grammes of hexose
The molar mass of the hydrated compound is 208 g/mol. To find the formula of the hydrate, we need to determine the molar mass of the anhydrous compound (XY) and subtract it from the total molar mass. With that information, we can calculate the molar mass of water in the hydrate and determine the ratio between XY and water molecules, giving us the formula of the hydrate.
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To determine the number of moles in 300 grams of sulfur, you need to know the molar mass of sulfur. The molar mass of sulfur is 32.06 g/mol. You can calculate the number of moles by dividing the given mass by the molar mass: 300 g / 32.06 g/mol ≈ 9.35 moles.
If you are asking if x=0, y=5 is a solution to 5x-3y=15, then no. 0, -5 would be as if you sub in 0 for x you get 5(0) -3y=15, i.e. -3y=15, i.e. y=-5
137.4 g First you need to calculate the molar mass of the molecule. Be = 9.01 g/mol C = 12.01 g/mol I = 126.9 g/mol Thus BeCI2 is equivalent to ( 9.01 + 12.01 + 2(126.9) ) g/mol or 274.82 g/mol. Then, using conversions, you multiply the amount you have by the molar mass so that: (.05 mol BeCI2)*(274.82 g/mol). The moles cancel out and you are left with a weight of 137.41 g.
50ml = .05L of HCL 1.0 M = 1mol / 1L of HCL simply multiply - .05 by 1.0, and get your answer!
(.05)X(grams of total solution) = grams of acetic acid (grams of acetic acid)/ (mol. wt. of acetic acid(=60g/mol)) = mol. acetic acid (mol. acetic acid)/ (Liters of total solution) = molarity(M)
The date 25-05-05 in Roman numerals is XXV.V.V and the date 25-05-2005 is XXV.V.MMV
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