ions of SO42-‚ CO32-‚ Cl-‚ and I-‚identify them separately‚ and use the observation to identify the unknown. Procedure Part I - Qualitative Analysis of Group 2 Elements Mix 0.02M K2CrO4 with each Mg(NO3)2‚ Ca(NO3)2‚ Sr(NO3)2 and Ba(NO3)2 together. Secondly‚ mix 0.1M (NH4)2C2O4 instead of 0.02M K2CrO4 together with the same reactants used before. Thirdly‚ mix 0.1M Na2SO4 with those reactants. Then‚ mix 0.1M NaOH with the same reactants used before again. Some precipitates should forms
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“Qualitative Test for some Cations 1”. For this exercise we obtained 3 test tube‚ each test tube was treated with different reagent (6N HCl‚ hot H2O‚ K2CrO4‚ 6N NH4OH‚ 6N HNO3). TT 1‚2 and 3 with reagent treatment 6N form a white precipitate‚ TT 1 with hot H2O was dissolved and color orange while TT 2 and 3 was undissolved. For reagent treatment K2CrO4‚ TT 1 form yellow precipitate‚ TT 2 treated with 6N NH4OH was dissolved while on TT 3 it form black precipitate. Lastly‚ for TT 2 treated with 6N HNO3
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019 g of potassium reacts with O2‚ it forms 1.860 g of a rather strange compound. Find the empirical formula of this compound‚ and explain why it is strange. 5. How many grams of BaSO4 (s) are formed when an excess of BaCl2 (aq) is added to 635 mL of 0.314 M Na2SO4 (aq)? BaCl2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2 NaCl (aq) 6. Determine the empirical and molecular formulas for: a. 75.69% C; 8.80% H; 15.51% O; MW = 206 b. 59.0% C; 7.1% H; 26.2% O; 7.7% N; MW = 180 1. Write (a) complete-formula
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If 25.8 mL of AgNO3 solution is required to precipitate all of the Cl ions in a 0.785-g sample of KCl (forming AgCl)‚ what is the molarity of the AgNO3 solution? 10. If 55.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 0.544-g sample of Na2SO4 (forming BaSO4)‚ what is the molarity of the BaCl2 solution? 11. In the laboratory‚ 6.67 g of Sr(NO3)2 is dissolved in enough water to form a volume of 0.750 L. A 0.100-L sample of this stock solution is mixed with a 0.0460 M solution
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Chemical Equilibrium ------------------------------------------------- ------------------------------------------------- RESULTS AND DISCUSSION A. Iron-Silver Equilibrium In studying equilibrium between iron and silver‚ 0.10 M FeSO4 and 0.10 M AgNO3 were used. The balanced equation for the reaction is: FeSO4 (aq) + 2 AgNO3 (aq) ↔ Fe(NO3)2 (aq) + Ag2SO4 (s) It has a net equation of: Fe2+(aq) + Ag+(aq) ↔ Fe3+(aq) + Ag(s) This part of the experiment
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a color because each element has an exactly defined emission spectrum‚ which one can use to identify them. For example‚ NaCl was highlighter yellow‚ Sr(NO3)2 was sun orange‚ CuCl2 was turquoise‚ LiCl was neon red‚ KCl was solar flare yellow‚ and BaCl2 was Voldemort green. Introduction- In Bohr’s model of the atom‚ electrons travel around the nucleus in an orbit. The concentric circles in his model represent the energy levels. Electrons can jump from energy level to energy level and absorb or emit
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Determination of Sulfate by Gravimetry Jose Luis E. De Guzman University of Santo Tomas‚ Sampaloc‚ Manila The determination of the percentage of sulfate in a sample is done as an instrument of learning a quantitative method of analysis‚ i.e.‚ of the gravimetric analysis. This type of analysis‚ which makes use of the weight of the samples was used for the experiment and required the precipitation of our analyte‚ SO4-2‚ and its filtration as a BaSO4 precipitate so as for weighing it. The weight
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lead to the fact that unknown solution #2 is H2SO4. Unknown #3: Solution 3 is Na2SO4. While doing the precipitate reactions‚ I found that solution 3 only precipitated with solution 9 (BaCl2). The precipitate can be seen in the double replacement reaction below: Na2SO4(aq) + BaCl2(aq) ⇒ 2NaCl(aq) + BaSO4(s) The solid was formed with barium and sulfate.
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The experiment’s objectives were to synthesize alum‚ test for various ions in alum‚ and form crystals in aqueous alum solution. The alum was synthesized from aluminum foil and run through a filtration system to isolate the alum. 2 Al(s) + 2 KOH + 22 H2O + 4 H2SO4 → 2 KAl(SO4)2●12H2O(s) + 3 H2(g) Its composition was analyzed using two precipitation tests and two flame tests. Three methods for growing crystals were then set up for observation next lab. Theoretical yield and percent yield: The theoretical
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Slowly turned darker | Rapidly turned darker red | Slowly turned into Brown (yellowish) | [Part 2] Procedure: The solutions that we used were: * 5 ml of 0.1 mol dm-3 of K2CrO4 * 5 ml of 0.1 mol dm-3 of K2Cr2O7 1. We named 4 test tubes as #5‚ #6‚ #7 and #8. 2. We put 5 ml of 0.1 mol dm-3 of K2CrO4 into tubes #5 and #6 3. We put 5ml of 0.1 mol dm-3 of K2Cr2O7 into tubes #7 and #8 4. We added 1.0 mold
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