10. Strong Bases: NaOH‚ KOH‚ Na2CO3‚ NaHCO3‚ Na3PO4‚ NaHPO4‚ Strong Acids: Fe(NO3)3‚ HCL‚ HNO3‚ Al(NO3)3‚ NiCl2‚ H2SO4 11. Acidic Neutral Basic NaCl KNO3 NaOH HC2H3O2 Na2CO3 NaC2H3O2 NaHCO3 Fe(NO3)3 NaNO3 Na3PO4 HCl MgSO4 KOH HNO3 Na2SO4 NaHPO4 CuSO4 NaNO2 CoCl2 Al(NO3)3 NiCl2 H2SO4 KCl NH4Cl 12. CuSO4 CuCO3.Cu(OH)2(s) + 2 H2SO4(aq) ----> 2 CuSO4(aq) + CO2(g) + 3 H2O(l) Na3PO4 Na3PO4 (aq) + 3H2O (l) --> H3PO4 (aq) + 3NaOH (aq) 13
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Part I: Properties of Hydrates 1.Place about 0.1 g of the following compounds in each one test tube: CuSO45 H2O‚ CoCl26 H2O‚ NiCl26 H2O‚ and KAl(SO4)212 H2O. 2. Heat each test tube gently over a Bunsen burner flame and record your observations in your notebook. 3. After the sample has cooled‚ add a few drops of deionized water. What happens and what can be concluded? Part
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of iron into the solution of nickel chloride a nickel deposit formed on the bottom of the well in the well plate. Where does nickel fit into your activity series? The chemical reactions for the preceding question B are: Cu (s) + NiCl2 (aq) à no reaction Fe (s) + NiCl2 (aq) à 2 Ni (s) + Fe Cl2 (aq) C. Suppose you inserted a piece of an unknown metal into a solution of zinc (II) nitrate and observed no reaction. Then if you inserted the unknown piece of the metal into the solution of iron (III)
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Magnetic Susceptability Michael J. Horan II Abstract: The change in weight induced by a magnetic field for three solutions of complexes was recorded. The change in weight of a calibrating solution of 29.97% (W/W) of NiCl2 was recorded to calculate the apparatus constant as 5.7538. cv and cm for each solution was determined in order to calculate the number of unpaired electrons for each paramagnetic complex. Fe(NH4)2(SO4)26(H20) had 4 unpaired electrons‚ KMnO4 had zero unpaired electrons‚ and
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no residue when evaporated ○ almost neutral ph ■ ^my group got pH 9 tho?? ○ no conductivity there was conductivity for 2 contamination could have been present causing the conductivity Unknown 3 ● NiCl2 or CuCl2 ○ i dont think its copper cuz of color of solution ○ I think that it can be NiCl2 or NiSo4 cuz Ni turned blue in presence of ammonia (i think only periods 1 and 2 can use this) ○ Was solution 3 soluble or insoluble in Bacl2 and Agno3? That can help us
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your equations. 1A Co(NO3)2*6H2O when mixed with Na3PO4*12H2O the chemical combination turned Purple. 1B Cu (NO3)2 *3H2O when mixed with Na3PO4*12H2O the chemical combination turned light blue 1C Fe (NO3)3 * 9H2O when mixed with Na3PO4*12H2O the chemical combination turned a light yellow 1D Ba (NO3)2 when mixed with Na3PO4*12H2O the chemical combination turned white and the mixture sizzled when mixed. 1E Ni (NO3)2 * 6H2O when mixed with Na3PO4*12H2O the chemical combination turned a light
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PURPOSE: To see which solutions are soluble and which are not. We were able to see this by mixing certain solutions together and observing changes that occurred. PROCEDURE: 1 Cotton Swabs‚ 1 Sheet each of white and black paper‚ 1 Distilled water‚ Goggles-Safety‚ 1 Well-Plate-24‚1 Well-Plate-96‚ Bag-CK1 1 Pipet‚ Empty Short Stem‚ Experiment Bag Ionic Reactions 1 Barium Nitrate‚ 0.1 M - 2 mL in Pipet‚ 1 Cobalt (II) Nitrate‚ 0.1 M - 2 mL in Pipet1 Copper (II) Nitrate‚ 0.1 M - 2 mL in Pipet‚ 1 Iron
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Purpose The purpose of this lab is to identify the different features that come about by using oxidation reduction and recording the observations‚ these tests will help determine the specific qualities each solution has. Hypothesis I theorize that nothing will happen with Magnesium because it is such a soft metal. Zinc should turn a greener color‚ because that’s what happens to a lot of statues and sculptures. I theorize that iron will tarnish badly because in air it can get tarnished so oxidation
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Ocean County College Department of Chemistry Chem 180 Lab 5: Ionic Reactions Submitted by Abstract: The purpose of this experiment is to work with aqueous solutions of ionic substances. Aqueous solutions are those solutions in which water is the solvent. When ionic substances are dissolved in water‚ the ions separate and become surrounded by water molecules. The focus of this experiment is on precipitates. The goal of this experiment is to study the nature of ionic reactions‚ write balanced
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Millimole – 1mmol = 10-3 mol 3. Molar mass – mass in grams of one mole of a substance. Example 3-5 page 76 4.62 g Na3PO4 Molar Mass Na3PO4 = (22.9898 gNa X 3) + (30.9738 gP) + (15.9994 gO X4) = 163.9408 g per mol Na3PO4 Moles Na3PO4 = 4.62 g X 163.9408 g/ mol = 2.818 X 10-2 mol Na3PO4 Moles Na = 2.818 X 10-2 mol Na3PO4 X 3 mol Na / mol Na3PO4 = 8.45 X 10-2 mol Na Na+ ions = 8.45 X 10-2 mol Na X (6.022 X 1023) = 5.08 X 1022 ions C. Solutions and Their Concentrations
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