K2Cr2O7 + 6 KI + 7 H2So4 = Cr2(So4)3 + 4 K2So4 + 3 I2 + 7 H2O
orange. equilibrium shifts right.
what is the intermediate after adding H2SO4 on naphthol?
It's kind of hard to make copper sulphate without any sulphate being present.
yellowish
MW of H2SO4 is 98.08. 2M = 2 x 98.08 in 1 L of water (1 gram=1 ml). Take 500 ml water in a 1 L measuring cylinder. Add 196.16 ml slowly along the side into water in the measuring cylinder. Use 50 ml pipette with automated pipettor. If needed you may want to keep the cylinder in ice to take care of the heat generated. Then make up to volume to 1 L with water. Eq. wt for H2SO4 = 98.08/2 = 49.039. SO for 2N solution, 2 eq.wt in 1 L. 98.08 ml in 1 L water adopting the method cited above.
Start with 1 dm^3 (1 Liter) of pure H2O. Molarity is defined as Mol/dm^3, so this is generally an easy way to go. Measure out 3 mol of H2SO4 -> First find the molar weight of Sulfuric acid, which is, according to Google: 98.079g/mol. So 3 mol * 98.079g/mol = 294.24 grams of H2SO4. Add that to 1 Liter of water, and you'll have an aqueous 3 Molar solution!
what is the intermediate after adding H2SO4 on naphthol?
It's kind of hard to make copper sulphate without any sulphate being present.
yellowish
MW of H2SO4 is 98.08. 2M = 2 x 98.08 in 1 L of water (1 gram=1 ml). Take 500 ml water in a 1 L measuring cylinder. Add 196.16 ml slowly along the side into water in the measuring cylinder. Use 50 ml pipette with automated pipettor. If needed you may want to keep the cylinder in ice to take care of the heat generated. Then make up to volume to 1 L with water. Eq. wt for H2SO4 = 98.08/2 = 49.039. SO for 2N solution, 2 eq.wt in 1 L. 98.08 ml in 1 L water adopting the method cited above.
Start with 1 dm^3 (1 Liter) of pure H2O. Molarity is defined as Mol/dm^3, so this is generally an easy way to go. Measure out 3 mol of H2SO4 -> First find the molar weight of Sulfuric acid, which is, according to Google: 98.079g/mol. So 3 mol * 98.079g/mol = 294.24 grams of H2SO4. Add that to 1 Liter of water, and you'll have an aqueous 3 Molar solution!
For preparation of standard solution of Mohr salt {FeSO4.(NH4)2SO4.6H2O}, it's necessary to add dilute H2SO4 to prevent the Fe2+ ions of Mohr salt solution from undergoing oxidation (to Fe3+).
Since acidity is a concentration of H+ ions, we can simply add them (in the form of an acid, such as HCl or H2SO4) or we can remove OH-. The easiest way to remove OH is to just add H, however.
add lemon juice
You will get a pink to a pinkish-red solution depending on how concentrated the sodium hydroxide (NaOH) is.
40g urea+ 80ml of 40%H2SO4! add 2g stannous chloride, heat untill solution is clear and make the volume upto 100ml. it is called foulger's reagent!
the die is called a solute the water is the solevent and when you mix the two together it is called a solution.
Concentrated H2SO4 is 96 %.( In laboratory ) As density of concentrated H2SO4 is 1.84gm/ml we will need this number as well, and as the atomic mass of H2SO4 is 98.08,as it is dibasic for normality it is 49 hence, Calculation=((96/100)(1000)(1.84))/49=36.04 If H2SO4 concentrated is 36.04 M then for make a 1L solution of 1M H2SO4 (36.04)X (x) = 1X(1) x = 1 X(1) / (36.04) x=0.0277gm/ml of water x = 27.7 mL of 36M H2SO4 per liter Hence for 1N H2SO4 dissolve 27.7ml of it to 1000ml of solvent(Water) that means for 0.1 N H2SO4 2.77 ml of it to 1000mL of solvent.