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- Mass of empty beaker=16.26Volume of CuSO4 5H2O=30mLMolarity of CuSO4 5H2O used=0.542Moles of CuSO4 5H2O=0.016Mass of aluminum foil used=0.26gMass of beaker and copper product=52.12gMass of copper metal product=35.86gMoles of Al used = 0.009636 molMoles of CuSO4 5H2O used=0.016mol Find the moles of copper product based on moles of Al.Find the moles of copper product based on moles of CuSO4 5H2OFind the limiting reactant in gramsA mixture of methanol and water contains 60.0% water by mass. Assuming volume additivity: If the actual specific gravity of the mixture is 0.9345: How many liters is required to provide 150 mol of methanol?Stock iron(II) solution (200Ug mL-1 Fe) ferrous ammonium sulfate hexahydrate mass= 0.1437g, transfer it to a 100 ml beaker. add 15 ml approx of water and 15m1 'approx of dilute sulphuric acid (2M H2SO.). then transfer FeII to 100 ml flask makeup to the mark with water. calculate the moles of ferrous ammonium sulfate hexahydrate solution in unit ug/mL.
- 93.6g to ng and 93.6g to tons with the set up in the image possible show how to get the answer pleaseA mass of KHP is measured at 0.3051g and contains ______ moles. Group of answer choicesGiven the following data forMass of test tube, beaker and cyclohexane = 100.17 gMass of test tube and beaker = 84.07 gFreezing point of cyclohexane = 6.59 oCMass of weighing paper + naphthalene =1.080 gMass of weighing paper = 0.928 gFreezing point solution = 5.11oCKf = 20.8oC/mDetermine the followinga. mass of cyclohexane in g (2 decimal places); _____b. mass of naphthalene in g (4 decimal places); _____c. freezing point depression (2 decimal places); _____d. molality of solution (3 significant figures); _____e. moles of naphthalene (3 significant figures); _____f. molar mass of naphthalene, experimentally (3 significant figures); _____g. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); ____
- Consider a blend of mass 5g formed from 2.2 g of C12H24 and 2.8g of C998H1000 paraffin's:, What are Mn and Mw of the blend?Use your balanced chemical reaction from the above problem. How many kilograms of the barium product can be produced at 100% yield from the double replacement reaction, made in industry at a larger scale of 675.00 pounds of barium nitrate that is then dissolved in water and mixed with an excess of the potassium sulfate solution. The product is collected by gravimetric filtration and dried. Use 453.592 g = 1 lb. Ba(NO3)2 + K2SO4 ---> BaSO4 + 2KNO3Given the following data, what is the molarity? Include the relative error Error is 1.0 mg or 1.0e-4
- Component B is separated by dissolution in water, followed by filtration. the filtrate is collected in a beaker weighting 96.47g . After evaporation of the water from the filtrate , the beaker and solid residue weight 97.25g. What is the weight of of component B ? Calculate the weight percent of component B in the original mixture.Magnesium +Acetic Acid --> C2HG5+ O2 ---> Iron (III) carbonate -->Given Active Ingredient: precipitated sulfur (ointment) Raw Materials: 500 g calcium polysulphide and 1.5 kg hydrochloric acid Actual Yield: 343.4g precipitated sulfur Formulation: 250 mg per jar Dosage form: Ointment packaging:100 jars per box Synthesis and Packaging (Need answer)- Balanced Chemical Equation:- % composition by mass of each compound:- Mass to Mass Stoichiometry Calculation:- Limiting Reagent:- Excess Reagent:- Amount (g) in excess: % Yield:- Number of dosage form and packaging that can be produced from stoichiometric solution: