Qualitative Observations of Double Displacement Reactions Lab Table 1.0 Qualitative Observation of Products Formed |Balanced Chemical Equations |Qualitative Observations | |BaCl2 (aq) + 2NaOH (aq)( BaOH2(aq) + 2NaCl(s) |An aqueous solution formed | | |Precipitate
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Zn + CuSo4 mol ratio | Observations | .01/.09 | Clear‚ light gray zinc | .02/.08 | Clear‚ medium gray zinc | .03/.07 | Milky‚ dark gray zinc | .04/.06 | Light blue liquid‚ black zinc | .05/.05 | Light/medium blue‚ brown zinc | .06/.04 | Medium blue‚ dark brown zinc | .07/.03 | Dark blue‚ brownish black zinc | .08/.02 | Bright darker blue‚ medium brown zinc | .09/.01 | Bright darker blue‚ lighter brown zinc | Quantitative Data: Group # | Moles Zn | Moles CuSo4 | 3rd Hour
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opposite ions that attract to one another. In this lab research‚ I learned that the following reaction will happen: lead nitrate + potassium chromate à potassium nitrate + lead chromate‚ the balanced chemical reaction is: Pb (NO3)2 + K2CrO4 --> 2 KNO3 + PbCrO4 Procedure Materials: - (1) 100ml Beaker - (1) 50 ml(s) of potassium chromate 1M solution - (1) 50 ml of lead (II) nitrate 1M - (1) 250 mL Erlenmeyer flask - (1) Buchner funnel - (1) Test tube Procedure: 1. Gather required
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measured and weighed. It was found in the experiment that the yield of copper hydroxide in 40%. Introduction: The copper (II) sulphate is then placed in 100 mL of distilled water. Then 20 mL of CuSO4 is measured and placed 100 mL of distilled water. This can later be weighed to determine the mol of CuSO4 and the mol/L concentration. Then this was used to find out how many mL of 0.5 NaOH solution is needed to react completely with all the copper (II) sulphate in solution. Then titrate ¼ of the estimated
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A: Ba(NO3)2 B: AgNO3 C: CuSO4 D) CuCl2 E) KCl Description how to identify solution: _ We have two blue solution which are CuSO4 and CuCl2 or C and D‚ according to chemical reaction experiment‚ C didn’t have any reaction with other solution like B and D beside A‚ so if we look at the solubility chart‚ SO2- have only one precipitation with Ba2-. So we come to conclusion that C is CuSO4 and A is Ba(NO3)2 _ So now that we know C is CuSO4‚ other blue solution must be CuCl2
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single-displacement reaction | Fe + CuSO4(aq) --> FeSO4 + Cu | 5. PbSO4 (milk white) | double-displacement reaction | Pb(NO3 )2(aq) + H2SO4(aq) --> PbSO4 + 2HNO3 | 6. CaCO3 (s) appears to dissolve and lot of bubbles comes off out of container. | double-displacement reaction | CaCO3 (s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l) | 7. When copper sulfate hydrate was heated‚ we can see some white precipitate. | decomposition reactions | CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g) | 8. Red
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Formula of a Hydrate Lab Wednesday October 29‚ 2014 Chemistry Honors Purpose This lab was done to determine the percentage of water in a hydrate‚ which was CuSO4 ?H20. Not only the percentage of water can be found‚ the moles of water can be found per one mole of anhydrous salt. An anhydrous salt is a hydrate that lost its water. Using various lab equipment such as burners‚ crucible‚ and balance‚ and techniques such as the mass-to-mole ratio and mass to percentage‚ the percentage of water in a hydrate
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|the MgCl was added | | | | |(+) There is a black tint to|(+) there was black spotting|(-) no reaction |(-) no reaction | |CuSO4 |the Zinc and the edges are |on the magnesium and the | |
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ranking the pH of these solutions‚ you will then test your predictions in the laboratory. 1. Arrange the following 0.1 M solutions in order of increasing pH and state why you placed each solution in that position: NaCH3COO‚ HCl‚ HCN‚ NaOH‚ NH3‚ NaCN‚ KNO3‚ H2SO4‚ NH4Cl‚ H2SO3‚ NaHCO3‚ Na3PO4 and CH3COOH. In order of increasing pH: H2SO4: This is because the first hydrogen is strong and completely ionizes and the second is weak and ionizes very minutely. It is also a
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=50.480% Actual mass= (Paper + residue) – (filter paper) = 3.96 – 1.32 =2.64 Atom economy AE (%) = (Desired)/ (Total Products) Desired mass ∑ (desire product)/ mass ∑ (all products) Two ways to make copper sulphate CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l) 159.6 + 18= 177.6 159/177.6 x 100 =89.86486% Copper
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