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Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be called a "2 normal" solution.
For example, in a solution with a 1:5 dilution ratio, entails combining 1 unit volume of solute (the material to be diluted) with 5 unit volumes of the solvent to give 6 total units of total volume. In photographic development, dilutions are normally given in a '1+x' format.
Conversion and its related terms yield and selectivity are important terms in chemical reaction engineering.They are described as ratios of how much of a reactant has reacted (X — conversion, normally between zero and one), how much of a desired product was formed (Y — yield, normally also between zero and one) and how much desired product was formed in ratio to the undesired product(s) (S ...
Triton X-100 is a commonly used detergent in laboratories. [5] Triton X-100 is widely used to lyse cells to extract protein or organelles, or to permeabilize the membranes of living cells. [6] Some applications include: Inactivation of lipid-enveloped viruses (e.g. HIV, HBV, HCV) in manufacturing of biopharmaceuticals; Industrial purpose ...
An earlier definition, used especially for chemical elements, holds that an equivalent is the amount of a substance that will react with 1 g (0.035 oz) of hydrogen, 8 g (0.28 oz) of oxygen, or 35.5 g (1.25 oz) of chlorine—or that will displace any of the three. [8]
Ethanol is used in a 2.5-fold excess (5.0 mol ÷ 2.0 mol). The theoretical molar yield is 2.0 mol (the molar amount of the limiting compound, acetic acid). The molar yield of the product is calculated from its weight (132 g ÷ 88 g/mol = 1.5 mol). The % yield is calculated from the actual molar yield and the theoretical molar yield (1.5 mol ÷ ...
Free cash flow margin expanded appropriately 8 percentage points to reach a record high of 11.1%. Net income per share was $0.09, using $362.1 million diluted weighted average shares outstanding.
11.6 g of NaCl is dissolved in 100 g of water. The final mass concentration ρ(NaCl) is ρ(NaCl) = 11.6 g / 11.6 g + 100 g = 0.104 g/g = 10.4 %. The volume of such a solution is 104.3mL (volume is directly observable); its density is calculated to be 1.07 (111.6g/104.3mL) The molar concentration of NaCl in the solution is therefore