Analyze and calculate the percentage of nickel in a 1.201 grams sample of nickel ore which was analyzed by the volumetric method: solution of 48 ml containing 1.40 x 10-2 gram of KCN/ml and the excess cyanide required half a milliliter of silver nitrate solution with 1.25 x 10-2 gram AgNO3 per milliliter.

1. Determine the mass of CaCl2·2H2O (MW 147.02 g/mole) to weigh out using a four- place analytical balance to prepare 100 mL of a 0.1005 M solution of Ca2+. 2. Determine the mass of KCI to weigh out using a four-place analytical balance to prepare 100 mL of a 1 M solution of KCI (MW 74.5513 g/mole). 3. Record the solubility of calcium carbonate in water (mg/L) in your notebook. 4. Calculate the amount of calcium carbonate (mg) not in solution if a tablet is dissolved in 250 mL of water and the tablet's label claim is 600 mg calcium / tablet.

Identify the two mass balance equations for a 0.10 M solution of sodium hypochlorite, NaOCI. 0.10 M = 0.10 M = [Na] + [OCI] [OCIT] [OCIT] + [HOCI] [Na+] + [HOCI] [Na+]

The zinc in a 0.7162-g sample of foot powder was titrated with 21.27 mL of 0.01645 M EDTA.  What was the percent zinc in the sample (4 sf)?

The Kjeldahl method was used to determine the nitrogen content of a protein. A 275 µL aliquot of a 47.0 mg/mL protein solution was digested in boiling sulfuric acid. The solution was made basic and the liberated NH, was collected in 5.00 mL of 0.0512 M HCl. A volume of 7.00 mL of 0.0174 M NaOH was required to react with the excess HCl. Calculate the weight percent of nitrogen in the protein. weight percent: wt% N

25.00 mL of a solution containing KIO4 is treated with excess potassium iodide. IO4- + 8H+ + 7I- → 4I2 + 4H2O The resulting solution is titrated with 0.15 M sodium thiosulfate pentahydrate. The average titre was 21.76 mL.  I2 + 2S2O32- → 2I- + S4O62- What is the concentration of the sodium thiosulfate solution as a percentage (w/v)

A solution of malonic acid (H2C3H2O4) with a known concentration of 0.500 M H2C3H2O4 is tirated with a 0.255 M KOH solution. How many mL of KOH are required to reach the second equivalence point with a starting volume of 60.0 mL H2C3H2O4, according to the following balanced chemical equation: H2C3H2O4 + 2 KOH---> K2C3H2O4 + 2 H2O