Title: Double Replacement Reactions (Data and Calculations) Objective: Classify the chemical reaction through observation‚ which each reagent produce when mixed with another reagent. After careful observation‚ be able to prove each observation using the net ionic equation. Background: First‚ a double-replacement reaction is when two cations in different compound switch anions‚ AX + BZ → BY. If either compounds are insoluble a precipitate occurs‚ and if there is no precipitate formed there is
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Double Replacement Precipitation Reactions Determine whether an ionic double replacement reaction has occurred.If I place different amounts of ionic solutions in a well plate‚ then I will be able to determine which of the solutions has had an ionic double reaction placement because I will use the solubility rules to decide which product produced the solid precipitate.Place five drops of silver nitrate into well A1 through A4. Place five drops of Iron (III) Nitrate into rows B1 through B4. Place
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Double Replacement Reactions Lab Pre-Lab 1. If a solid‚ a gas or a weakly ionizing compound such as H2O formed‚ then a reaction has occurred. 2. The formation of a solid‚ the formation of a gas and the formation of a weakly ionizing compound are the driving forces that control a double replacement reaction. 3. a. Balanced equation: CaCl2 (aq) + NaPO4 (aq) Ca3(PO4)2 (s) + 6NaCl (aq) Ionic equation: 3Ca2+(aq) + 6Cl-(aq) + 6Na+(aq) + 2PO43- (aq) Ca3(PO4)2 (s) + 6Na+ (aq)
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Information: A double replacement reaction is a chemical reaction that usually takes place between two aqueous ionic compounds. In the reaction‚ the cation of one compound replaces the other compound’s cation that yields two produce two new and different compounds. A precipitation reaction is a chemical reaction in which the formation of a solid‚ as one of the products‚ arises “when two solutions are mixed”(Zumdahl‚ Decoste 184). This solid is also known as the precipitate of the reaction. One type of
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Micro-Reactions: Predicting the Products of Double Replacement Reactions Introduction – A double replacement reaction is a chemical reaction between two compounds where the positive ion of one compound is exchanged with the positive ion of another compound. If you have the reactants of two reaction solution that you can determine the products. All you need to do is pair the positive parts of the compounds with the other compounds negative part. Once you find the products you can determine their
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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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Chemistry Lab Report Jeffrey Kenneth Bangero Introduction Firstly we calculated the mass of the beaker and then we put the assigned grams of sodium carbonate and calcium chloride. Then we add 50 ml of distilled water to each substance‚ sodium carbonate dissolved faster than calcium carbonate. After we mixed both sodium carbonate and calcium chloride and they form a solid precipitate. Then we poured it a funnel with a filter paper to get the solid precipitate. Purpose ● The purposed
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A double displacement reaction is a type of chemical reaction in which two compounds react in order to create two new chemicals. The two cations and anions switch places with each other which forms two new compounds. As an example AB+CC can switch to AD+CB. Both sides of these equations must balance out in order to be neutral. So any pairs of compounds (such as A+B) must have a total balanced charge. In this example A could have a charge of positive 2 (+2) and B could have a charge of negative
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|Zn |Mg |Cu |Pb | | |(+) clear liquid but the |(+) a black spot appeared on|(-) no reaction occurred |(-) no reaction | |Pb(NO3)2 |Zinc itself turned glittery |the sliver of magnesium. | | | | |(+) the liquid looks |(+)
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purpose of this lab was to find the theoretical and experimental percentage yields of the double displacement reaction between the solutions Lead (II) Nitrate (PbNO3) and Potassium Iodide (KI). It is important to obtain amounts of Lead (II) Nitrate and Potassium Iodide as close to 1.44g as possible. This reaction creates Lead (II) Iodide and Potassium Nitrate. The precipitate during this reaction is Lead (II) Iodide. The balanced equation is Pb(NO3)2 + 2KI= PbI2 + 2KNO3. In this lab the Lead (II) Iodide
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