Lab Chemistry 107 (Final ) Test

Centrifugation is the correct answer because it is the most effective method to separate PbCl2 from a mixture of AgCl and Hg2Cl2 after adding hot water. Centrifugation involves spinning the mixture at high speeds, causing the heavier PbCl2 to settle at the bottom of the tube while the lighter AgCl and Hg2Cl2 remain suspended. This allows for easy separation of the PbCl2 by pouring off the supernatant liquid. Filtration would not be effective as the particles are too small to be trapped by the filter. Sublimation and distillation are not applicable in this scenario.

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The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this question, we are given the mass of Cu produced (0.750 g) and the molar mass of Cu (63.5 g/mol). By dividing the mass of Cu by its molar mass, we can find the number of moles of Cu produced. Since the reaction is stated to be with excess Zn, we can assume that all the CuSO4 reacted. The volume of the solution is given as 15 ml, which can be converted to liters by dividing by 1000. By dividing the number of moles of Cu by the volume of the solution, we can find the molarity of the CuSO4 solution, which is approximately 0.79 M.

HCl is the correct answer because it is a strong acid that can dissolve excess Zn (unreacted) in a redox reaction. It can react with the Zn to form ZnCl2, which is soluble in water. This allows for the removal of the excess Zn from the reaction mixture, increasing the quantitative yield of the desired product.

PbCl2 is the most soluble in hot water because it is a chloride compound, and chloride compounds tend to be more soluble in water compared to other compounds. Additionally, the presence of heat increases the kinetic energy of the particles, allowing them to overcome intermolecular forces and dissolve more readily.

Cu is the reducing agent in this reaction because it undergoes oxidation, losing two electrons to form Cu+2. The Ag+ ions are the oxidizing agents because they gain electrons, being reduced to form Ag atoms.

The molarity of a solution can be calculated using the formula: Molarity = (moles of solute) / (volume of solution in liters). In this question, we know the volume of NaOH solution (25 mL) and its molarity (0.2 M). Since NaOH and H2SO4 react in a 1:1 ratio, the moles of H2SO4 can be calculated by multiplying the volume of NaOH solution (25 mL) by its molarity (0.2 M). The volume of the H2SO4 solution is given as 14 mL. To find the molarity of H2SO4, we divide the moles of H2SO4 by the volume of the H2SO4 solution in liters (14 mL = 0.014 L). Therefore, the molarity of H2SO4 is 0.2 M * 25 mL / 0.014 L = 0.357 M, which is equivalent to 0.179 when rounded to three decimal places.

The answer is 9.4 grams because the question states that there is an excess of Zn, meaning that Zn is not the limiting reactant. Therefore, the amount of Cu produced will be determined by the amount of CuSO4 present. To find the amount of CuSO4, we use the formula Molarity = moles/volume. Rearranging the formula to solve for moles, we get moles = Molarity * volume. Plugging in the values given, we get moles of CuSO4 = 0.94 mol/L * 0.157 L = 0.14758 mol. Since the molar ratio between CuSO4 and Cu is 1:1, the moles of Cu produced will also be 0.14758 mol. Finally, we can convert moles to grams using the molar mass of Cu (63.5 g/mol), giving us 9.36133 grams, which rounds to 9.4 grams.

The freezing point of a solution is calculated using the formula: ΔTf = Kf * m.

Given that the freezing point depression constant (Kf) is 4.68 and the mass of the solution (m) is the sum of the mass of CHCL3 (0.840 kg) and eucalyptol (57.0 g), we can calculate the freezing point depression (ΔTf).

ΔTf = 4.68 * (0.840 + 0.057) = 4.68 * 0.897 = 4.1832

To find the freezing point (C0) of the solution, we subtract the freezing point depression from the freezing point of the pure solvent (Tf).

C0 = Tf - ΔTf = -63.5 - 4.1832 = -67.6832

Rounding to the nearest tenth, the freezing point of the solution is -65.7°C.

The given statement "Au oxidized by Ag+" is false. Gold (Au) is a noble metal and is not easily oxidized by other substances, including Ag+ (silver ion). In fact, gold is known for its resistance to oxidation and is often used in jewelry and other applications due to its inertness. Therefore, it is unlikely for gold to be oxidized by silver ions.

The correct answer is "dark red" because copper sulfate (CuSO4) is typically a deep red or maroon color. This is due to the presence of copper ions in the compound, which absorb certain wavelengths of light and reflect others, giving it a distinct color. The intensity of the red color may vary depending on the concentration and purity of the compound.

When zinc (Zn) reacts with copper sulfate (CuSO4), zinc sulfate (ZnSO4) and copper (Cu) are produced. The color of copper sulfate is dark blue, but when zinc is added to it, the color changes to brownish yellow. Therefore, the correct answer is brownish yellow.