A certain hexahydrate was heated to evaporate off all the water. The remaining material was found to be Nickel(II) nitrate, Ni(NO3)2. What is the % nickel in the compound? O 11% O 23% O 32% O 41%
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- In the distillation of the mixture of acetic acid (MW=60.05 g/mole; d=1.05g/ml) and water (MW= 18.00 g/mole; 1.00 g/ml), a total of 0.028 Liters of 1.0 M std NaOH was used to reach the endpoint of a 23 ml of distillate. What is the mole ratio of the CH3COOH in the mixture? final answer in 3 decimal places.1.A student masses an unknown hydrate sample and finds there are 5.00g of hydrate. After dehydrating the sample in an oven for 1h, the sample has a mass of 3.00 g. How many moles of water were removed from the hydrate. use dimensional analysis. 2. a. What is the mass of anhydride for the student sample? ___________ g anhydride b. If the formula weight (molar mass) of the anhydride is 89.54g/mol, how many moles of anhydride are in the student sample? c. What is the ratio of moles of water to moles of anhydride? mol H2O . = _____________ mol anhydride d. What is the formula for the hydrate? Fill in the blank: X • ___ H2OStock 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.
- At -28 C sodium bromide and water from a eutectic mixture which contains 48 % NaBr by weight. What is the maximum weight of pure NaBr that can be recovered from a mixture containing 80 g NaBr and 20 g water? a. 33.3 g b. 28.6 g c. 66.6 g d. 40 gVolume of an unknown used was 30 mL, Initial Buret volume was 0 and the Final Buret volume was 8.5 mL. What is the molarity of the unknown solution if the Net volume of NaOH being used was 8.5 mL and Millimoles (mmoles) NaOH reacted was 0.791? Then, what is the Mass (g) of Acetic Acid in unknown sample and thr average percent (%) Acetic Acid? (assume density = 1g/mL)A mixture of Al2O3(s) and CuO(s) weighing 18.371 mg was heated under H2(g) at 1 0008C to give 17.462 mg of Al2O3(s) 1 Cu(s). The other product is H2O(g). Find wt% Al2O3 in the original mixture.
- Analysis of product: Calculate the percent recovery based on the amount of tea originally used and a final mass of 0.20g of caffeine obtained. Compare this value against a literature search about the approximate caffeine content by percent in black tea. Given: the observed melting point of the product crystals is 230oC-235oC -10 tea bags (black tea, contains ~45mg caffeine/bag) -total mass is 21.10g of the loose black tea19 g of unknown organic sample was dissolve in 640 mL of Dicloromethane (DCM). The boiling point of benzene was increased by 3.78oC. Determine the molecular weight of the unknown sample? Kb of DCM = 2.42oC/m Bb of benzene = 39.6 oC density of benzene = 1.33 g/mL at 25 °C Round your answer to the nearest whole number, no units required.It is known that acid content has a major effect on theflavor of vinegars, but most cheaper vinegars are diluted similarly to 5% acidity Wt./vol. % is equivalent to gsolute per 100mL solution (so 5% is equivalent to 5 g acid/100 mL solution). a.) First, calculate the approximate molar concentration of acetic acid in the 5% wt./vol vinegar. b.) Next, calculate the expected molarity of acetic acid in the solution upon dilution by a factor of 5. Thank you!
- Beaker 0.00200 M Fe(NO3)3, mL 0.00200 M NaSCN, mL total volume, mL 1 3.000 2.000 10.00 2 3.000 3.000 10.00 3 3.000 4.000 10.00 4 3.000 5.000 10.00 5 (blank) 3.000 0.000 10.00 In Solutions 1-4 you are adding successively larger volumes of 0.00200M SCN- to the Fe3+ solution and diluting to 10.00 ml. Calculate the final diluted molarity of SCN- in solution #1 Your answer should have 3 sig figs =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.25 ppm of CaCO3 was found in the water sample obtained on the lake in the nearby town.Calculate the moles of CaCO3 having a molar mass of 100.09 g/mol in 10 liters of water samplesolution having a density of 997 kg/m3. (Ans.: 24.95 moles CaCO3)