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Question: how would you expect solubility of product of potassium bitartrate to vary between distilled water and a KNO3 solution?
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- Please fast solveA solution consists of 0.020 mol.kg-1 KCl(aq) and 0.035 mol.kg-1 Ca(NO3)2(aq). For Cd(NO3)2(aq), calculate (i) the average activity coefficient, and (ii) the activities of Cd2+ and NO31) How many milliliters of 70% ethanol and 20% ethanol must be mixed to prepare 500ml of 30% ethanol? 2) A gaseous blend of nitric oxide and nitrogen (NOmax) contains 0.09% v/v nitric oxide (NO) and 99.92% v/v nitrogen. Express the strength of nitric oxide in PPM (round to whole number) 3) The ratio strength of a 10mL vial of neostigmine methylsulfate injection USP is 1:1500.Calculate the amount of neostigmine methylsulfate in milligrams contained in the 10mL vial. (Round to the whole number).1) The ratio strength of a 10mL vial of neostigmine methylsulfate injection USP is 1:1500. Calculate the amount of neostigmine methylsulfate in milligrams contained in the 10mL vial. (Round to the whole number).
- A 18 g of unknown organic sample was dissolve in 756 mL of benzene. The boiling point of benzene was increased by 3.36oC. As the first step of analysis, determine the moecular weight of the unknow sample? Kb of benzene= 2.64oC/m Bb of benzene = 80.09 oC density of benzene = 0.874 g/mL at 25 °C Answer in whole number, no units required.(a) During the analysis of water sample by argentometric titration, results obtained are as follows: Experiment1 2 3 4 Volume 15.5, 15.2 ,15.1 ,15.4 A)Calculate average deviation and relative average deviation for the given data. (b)Calculate the molecular weight of an unknown acid if 8.5 g of it is dissolved in 200.0 ml of water and requires 50.0 ml of 1.5 M sodium hydroxide for complete neutralization .EX 2.Given-> Volume of Na+ = 500 ml Molarity of Na+= 0.0100M Molar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volume Number of millimole = 0.0100 × 500 = 5 millimole Na2CO3 ---> 2Na+ + CO32- Millimole of Na2CO3 = millimole of Na+/2 Millimole of Na2CO3 =5/2 = 2.5 millimole Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole) Weight of Na2CO3 required = mole × molar mass = 2.5 × 10-3 × 105.99 =0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.
- Density of solution:Trial 1: 1.2 g/mLTrial 2: 1.2 g/mLTrial 3: 1.2 g/mL Average density = 1.2 g/mL What is the relative average deviaion, %?Calculate the solubility of lead(II) sulfate (Ksp = 2.53x10-8) in a 0.0034 M solution of sodium sulfate. Give your answer to three sig. figs. and in exponential form (e. g. 1.23E-3).2.3-3 Calculate the Ksp of the following aqueous solutions: a) 3.60x10-5M Ag2CO3 b) 4.30x10-3M of Ag+ in Ag2C2O4 solution c) 4.80 mol of PbCrO4 dissolved in 2.00 L container.