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Find [H] and the pH of 0.05 M LiNO3.
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- 4- Calculate the pH of a solution that is 0.0400 M in a) H3PO4. b) NaH2PO4. c) Na3PO4.4. Determine the pH of 0.05 M CH3COONa. 5. Determine the pH of 0.05 M NH4NO3. 6. Determine the pH of a solution in which 1.00 mol H2CO3 (Ka = 4.2 x 10-7) and 1.00 mole NaHCO3 is dissolved in enough water to form 1.00 L of solution.1. A silver nitrate solution contains 14.77 g of primary standard AgNO3 in 1.00 L. What volume of this solution will be needed to react with a) 0.2631 g of NaCl? b) 0.1799 g of Na CrO4? c) 64.13 mg of NazAsO4?
- Complete the balanced dissociation equation for the compound below in aqueous solution. If the compound does not dissociate, write NR after the reaction arrow. HCIO,(aq) – 4- 2- n2+ 3+ D 4+ + 1 3 4 6. 7 8 Os O8 (s) (1) (g) (aq) NR CI Reset • x H,O Delete x H̟O 2 11 3. 2. +4. What is the pH of 39.74 mL of a 0.1888 M solution of a weak monobasic base (B:, Kb = 9.56 x 10-6) prior to its titration with 0.1765 M HClO3? Your answer must have 2 decimal places.Help me please
- What is the pH of a river water sample in which [HCO3-] = 1.0 x 10-4 M and the concentration of dissolved CO2 in equilibrium with atmospheric CO2 is 1.4 x 10-5 M? (The pKa of H2CO3 is 6.37)7. What is the pH of a 0.35 M solution of KBrO₂? Ka of HBrO2 = 1.2 x 10-5Calculate the amount of 0.300M H2SO4 required to neutralize a 0.600g sample that is 50% FeO.
- 1. Titration of 0.485g sample by the Mohr method required 38.8mL of standard 0.1060 M AgNO3 solution. Calculate the percentage of chloride in the sample.What mass of NaCN must be added to 1 L of 0.010 M Mg(NO3)2 in order to produce the first trace of Mg(OH)2?Buffer solutions with the component concentrations shown below were prepared. Which of them should have the lowest pH? O [CH3COOH] = 0.25 M, [CH3COO]= 0.25 M %3D %3D [CH3COOH] = 0.75 M, [CH3COO"] = 0.75 M %3D [CH3COOH] = 0.75 M, [CH3CO0] = 0.25 M [CH3COOH] = 0.25 M, [CH3COO"] = 0.75 M %3D [CH3COOH] = 1.00 M, [CH3COO"] = 1.00 M %3D Page 31 of Next Page Previous Page 42 of 70 questions saved
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