Chemistry High School

Which metal can be prepared by electrolysis of an aqueous solution of one of its salts?
a. magnesium
b. potassium
c. sodium
d. aluminum
e. copper

Answers

Answer 1

Out of the options provided, aluminum (d) can be prepared by electrolysis of an aqueous solution of one of its salts

Aluminum can be prepared by electrolysis of an aqueous solution of one of its salts, specifically aluminum oxide (Al2O3) dissolved in molten cryolite (Na3AlF6). This process is known as the Hall-Héroult process.

In the Hall-Héroult process, a carbon anode and a graphite cathode are placed in a cell containing molten cryolite. The aluminum oxide is dissolved in the molten cryolite, and when an electric current is passed through the solution, electrolysis occurs.

At the cathode, aluminum ions (Al3+) are reduced to form molten aluminum metal, which collects at the bottom of the cell. At the anode, oxygen ions (O2-) are oxidized, producing oxygen gas.

The overall reaction can be represented as follows:

2 Al2O3(l) → 4 Al(l) + 3 O2(g)

Out of the options provided, aluminum (d) can be prepared by electrolysis of an aqueous solution of one of its salts. Magnesium (a), potassium (b), sodium (c), and copper (e) cannot be obtained through electrolysis of their aqueous salt solutions.

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Related Questions

If 15.00 mL of a 1.00 M HCl solution is mixed with 25.00 mL of a 0.250 M NaOH solution in a process that produces 273 mg of NaCl, what is the percentage yield of NaCl?

Answers

The percentage yeild of the NaCl from the calculation is 73.8 %

What is the percentage yield?

The percentage yield is a measure of the efficiency of a chemical reaction or process, indicating the proportion of the theoretical yield that is actually obtained in practice.

Number of moles of HCl = 1 * 15/1000

= 0.015 moles

Number of moles of NaOH = 25/1000 * 0.250

= 0.00625 moles

Since the reaction is 1:1, we can see that NaOH is the limiting reactant

Theoretical yeild = 0.00625 moles * 58.5 g/mol

= 0.37 g or 370 mg

We have the percentage yeild is;

273/370 * 100/1

= 73.8 %

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is sio2 most likely a molecular, metallic, ionic, or covalent-network solid?

Answers

Sio2, also known as silicon dioxide or silica, is most likely a covalent-network solid.

To determine the type of solid, we need to analyze the nature of bonding in SiO2. Silicon (Si) and oxygen (O) are both nonmetals, and their bonding is typically covalent.

In SiO2, silicon forms four covalent bonds with four oxygen atoms, and each oxygen atom forms two covalent bonds with two silicon atoms. This results in a three-dimensional network structure of alternating silicon and oxygen atoms.

Covalent-network solids have a high melting point, are hard and brittle, and do not conduct electricity because the electrons are localized within the covalent bonds. In the case of SiO2, the strong covalent bonds between silicon and oxygen atoms give rise to its characteristic properties.

Based on the nature of bonding in SiO2, it is most likely a covalent-network solid. The three-dimensional network structure formed by covalent bonds between silicon and oxygen atoms is responsible for its high melting point and other properties associated with covalent-network solids.

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write the complete electron configuration for bromine using the periodic table

Answers

Bromine is a non-metal element with the chemical symbol Br and atomic number 35. To write the complete electron configuration of bromine, we first need to determine the number of electrons in its neutral state. Since bromine has an atomic number of 35, it means that it has 35 electrons in its neutral state.

The electron configuration of bromine can be written by using the Aufbau principle, which states that electrons fill the lowest energy level orbitals first before moving to higher energy levels. The electron configuration for bromine can be written as follows:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

The first number represents the principal quantum number, which determines the energy level of the electrons. The letters represent the subshells, where s, p, d, and f are the different subshells. The superscript numbers represent the number of electrons in each subshell.

In the case of bromine, the first two electrons are in the 1s orbital, followed by two electrons in the 2s orbital and six electrons in the 2p orbital. After that, there are two electrons in the 3s orbital and six electrons in the 3p orbital. The remaining ten electrons are in the 4s, 3d, and 4p orbitals, with five electrons in the 4p orbital.

Thus, the complete electron configuration for bromine using the periodic table is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5.

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The Ksp of Al(OH)3 is 2.0 x 10-31 at 298 K. What is \DeltaGo (at 298 K) for the precipitation of Al(OH)3 according to the equation below? Al3+(aq) + 3OH- (aq) -> Al(OH)3 (s) The answer is -175 kJ mol-1 , I do not understand why it is negative and not positive because, every time I plug the equation in my calculator, the answer is positive and not negative.

Answers

The negative value of ΔG° (-175 kJ/mol) indicates that the precipitation of Al(OH)3 is thermodynamically favorable and spontaneous at 298 K. The negative sign signifies that the reaction will proceed in the forward direction without the need for an external energy input.

The sign of ΔG° determines the spontaneity of a reaction. A negative ΔG° indicates a thermodynamically favorable process, where the reaction will occur spontaneously in the forward direction. In the case of the precipitation of Al(OH)3, the given value of -175 kJ/mol suggests that the reaction is energetically favorable. The calculation of ΔG° involves the equilibrium constant (K) of the reaction. For this reaction, K corresponds to the solubility product constant (Ksp) of Al(OH)3, which is 2.0 x 10^(-31). When plugging this value into the equation ΔG° = -RTln(K), the natural logarithm of a very small number (Ksp) yields a large positive value. The negative sign outside the equation makes the overall ΔG° negative, indicating thermodynamic favorability.

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