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Indicate the number of equivalents in each of the following: a. 1 mole of \(\mathrm{Mg}^{2+}\) b. \(0.5\) mole of \(\mathrm{H}^{+}\) c. 4 moles of \(\mathrm{Cl}\) d. 2 moles of \(\mathrm{Fe}^{3+}\)

Short Answer

Expert verified
2 equivalents, 0.5 equivalents, 4 equivalents, 6 equivalents.

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01

Understanding Equivalents

The number of equivalents of an ion is calculated as the product of the moles of the ion and its valency (charge).
02

Calculate Equivalents for \(\rightarrow \text{Mg}^{2+}\)

For 1 mole of \( \text{Mg}^{2+} \), the valency is 2. Therefore, the number of equivalents is \( 1 \text{ mole} \times 2 = 2 \text{ equivalents} \).
03

Calculate Equivalents for \(0.5 \rightarrow \text{H}^{+} \)

For 0.5 moles of \( \text{H}^{+} \), the valency is 1. Thus, the number of equivalents is \( 0.5 \text{ moles} \times 1 = 0.5 \text{ equivalents} \).
04

Calculate Equivalents for 4 moles of \( \text{Cl} \)

For 4 moles of \( \text{Cl} \) with a valency of 1, the number of equivalents is \( 4 \text{ moles} \times 1 = 4 \text{ equivalents} \).
05

Calculate Equivalents for 2 moles of \( \text{Fe}^{3+} \)

For 2 moles of \( \text{Fe}^{3+} \) with a valency of 3, the number of equivalents is \( 2 \text{ moles} \times 3 = 6 \text{ equivalents} \).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

moles calculation
The concept of moles is essential in chemistry. A mole represents a specific number, known as Avogadro's number, which is approximately 6.022 x 10^{23} particles. These particles can be atoms, molecules, ions, or other chemical entities. Moles help simplify the measurement of substances involved in chemical reactions. For example, instead of saying you have 6.022 x 10^{23} hydrogen atoms, you simply say you have 1 mole of hydrogen atoms. This simplification makes calculations more manageable when dealing with chemical equations and reactions.
valency
Valency is the combining capacity of an element. It tells you how many electrons an atom can lose, gain, or share to form a stable chemical bond. For metals, valency often corresponds to the number of electrons lost, while for non-metals, it can correspond to the number of electrons gained or shared. For instance, magnesium (Mg) has a valency of 2, meaning it can lose two electrons, forming a Mg^{2+} ion. Similarly, chlorine (Cl) has a valency of 1 because it gains one electron to form a Cl^{-} ion. Knowing the valency is crucial for determining the equivalents in a chemical reaction.
equivalents calculation
Calculating equivalents involves determining the effective number of reacting particles. An equivalent relates to the amount of a substance that reacts with or supplies one mole of hydrogen ions (H+). To find the number of equivalents, you multiply the moles of a substance by its valency (charge). For example, 1 mole of Mg^{2+} (with a valency of 2) equates to 2 equivalents (1 mole x 2). Similarly, 0.5 moles of H+ (with a valency of 1) equates to 0.5 equivalents (0.5 moles x 1). This concept is key when balancing chemical equations and understanding reaction stoichiometry.
ionic charge
Ions are atoms or molecules that have gained or lost electrons, resulting in a net charge. The charge of an ion is crucial in understanding its chemical behavior and forming compounds. Positive ions are called cations (e.g., Na^{+}, Mg^{2+}), and negative ions are called anions (e.g., Cl^{-}, SO_{4}^{2-}). The ionic charge affects how ions interact, bond, and react with each other. For example, in a solution, Mg^{2+} ions will attract two Cl^- ions to form magnesium chloride (MgCl_{2}). Understanding ionic charges also helps in calculating the number of equivalents, which depends on both the moles and the charge of the ions involved.

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Most popular questions from this chapter

Calculate the concentration of each of the following diluted solutions: a. \(2.0 \mathrm{~L}\) of a \(6.0 \mathrm{M} \mathrm{HCl}\) solution is added to water so that the final volume is \(6.0 \mathrm{~L}\). b. Water is added to \(0.50 \mathrm{~L}\) of a \(12 \mathrm{M} \mathrm{NaOH}\) solution to make \(3.0 \mathrm{~L}\) of a diluted \(\mathrm{NaOH}\) solution. c. A \(10.0-\mathrm{mL}\) sample of a \(25 \%(\mathrm{~m} / \mathrm{v})\) KOH solution is diluted with water so that the final volume is \(100.0 \mathrm{~mL}\). d. A \(50.0\) -mL sample of a \(15 \%(\mathrm{~m} / \mathrm{v}) \mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to water to give a final volume of \(250 \mathrm{~mL}\).

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Calculate the grams or milliliters of solute needed to prepare the following solutions: a. \(50.0 \mathrm{~mL}\) of a \(5.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{KCl}\) solution b. \(1250 \mathrm{~mL}\) of a \(4.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{NH}_{4} \mathrm{Cl}\) solution c. \(250 . \mathrm{mL}\) of a \(10.0 \%(\mathrm{v} / \mathrm{v})\) acetic acid solution

Determine the volume, in milliliters, required to prepare each of the following diluted solutions: a. \(255 \mathrm{~mL}\) of a \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) solution from a \(4.00 \mathrm{M} \mathrm{HNO}_{3}\) solution b. \(715 \mathrm{~mL}\) of a \(0.100 \mathrm{M} \mathrm{MgCl}_{2}\) solution using a \(6.00 \mathrm{M}\) \(\mathrm{Mg} \mathrm{Cl}_{2}\) solution c. \(0.100 \mathrm{~L}\) of a \(0.150 \mathrm{M} \mathrm{KCl}\) solution using an \(8.00 \mathrm{M} \mathrm{KCl}\) solution

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