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- Titration of a 25.00 mL sample of acid rain required 12.4 mL of 0.009879 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration (in mol/l and %) of sulfuric acid in this sample of rain?Titration of a 25.00 mL sample of acid rain required 13.23 mL of 0.009567 M KOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration (in mol/l and %) of sulfuric acid in this sample of rain?What volume of 0.171 M KOH,0.171 M KOH, ?,V, can be neutralized with 58.5 mL58.5 mL of 1.21 M HNO3? V=
- In an impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgN03 formed 2.044 g of AgCl. Listing the percentage of NaCl is in the impure sample. What is the list of three sources of error in this experiment?Sample: Saline 0.900% (m/v), in sodium chloride (NaCl). Data: M.A. (g/mol): Na = 22.9898; K = 39.0983; Cr=51.9961; Ag = 107.8682; Cl = 35.453; N=14.0067; O = 15.9994. Material available in the laboratory's warehouse: Reagents: distilled water; standardized solution of silver nitrate (AgNO₃), at a concentration of 0.09980 mols/L; Potassium chromate solution (K₂CrO₄) 1%; ammoniacal ferric alum solution [Fe(NH₄)(SO₄)₂] and; nitrobenzene (or cooking oil, alternatively); 0.1000 mols/L potassium thiocyanate (KSCN) standard solution. Glassware: beakers of all sizes available on the market; 50, 100, 150, 200 and 250 ml Erlenmeyers; 10.00 and 25.00 mL volumetric pipettes; 15.00 mL burette; 30.00, 50.00 and 100.00 mL volumetric flasks Calculate the molarity of saline, in terms of the NaCl concentration (concentration in mols/L). (a) 0.900 mols/L (b) 5.84 mols/L (c) 0.154 mols/L (d) 0.0154 mols/LWhat volume of 0.441 M KOH,0.441 M KOH, ?,V, can be neutralized with 50.0 mL50.0 mL of 1.64 M HNO3? ?= mL
- To prepare 50 mL solutions of 0.01 M KMnO4 in water (calculation to prepare chemicals in spectrophotometer)What volume in mL of 0.1354 M NaOH is required to nitrate 15.00 ML of 0.1002 M acid (HAc)?What volume of water should be added to 100.0 mL of a 0.500 N Cr2(SO4)3·18H2O solution to obtain a final concentration of 0.075 M? Show complete solution *Also find the number of replaceable ions (n)
- Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811MNaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain?Use the following atomic masses (in g/mol): Mg = 24.31; H = 1; S = 32.06; O = 16; Na = 23; Cl = 35. 45; Ca = 40.08; C = 12.01; N = 14.01; Mg = 24.31 A Kjeldahl procedure was used to analyze 356 μL of a solution containing 48.36 mg protein/mL. The liberated NH3 was collected in 5 mL of 0.0336 M HCl and the remaining acid required 6.34 mL of 0.010 M NaOH for complete titration. How many % nitrogen is in the protein sample?To what volume (in mL) must 27.4 mL of 17.2 M H2SO4 be diluted to produce 1.43 M H2SO4?
