Gloria is writing thechemical formula for a compound using its chemical name. She has justidentified the names of the elements in the compound. The tool that she willneed to use next is a textbook to learn the IUPAC naming of compounds or ahandbook of chemical compounds.
Answer:
the periodic table
Explanation:
verified correct
Answer:
710.g
Explanation:
An oxidation reaction involves the loss of electrons or an increase in the oxidation state of an element. Options B, C, and D represent oxidation reactions, while option A does not.
Let's analyze the given options:
A. AuCl₄⁻ → AuCl₂⁻
In this reaction, the gold atom goes from a +3 oxidation state (AuCl₄⁻) to a +1 oxidation state (AuCl₂⁻). This means the gold atom gains two electrons, indicating a reduction rather than oxidation. Therefore, option A is not an oxidation reaction.
B. Mn⁷⁺ → Mn²⁺
In this reaction, the manganese atom goes from a +7 oxidation state to a +2 oxidation state. This indicates a decrease in the oxidation state, which means the manganese atom gains electrons. Therefore, option B represents an oxidation reaction.
C. Co³⁺ → Co²⁺
In this reaction, the cobalt atom goes from a +3 oxidation state to a +2 oxidation state. Similar to option B, this indicates a decrease in the oxidation state, representing an oxidation reaction.
D. Cl₂ → ClO₃⁻
In this reaction, the chlorine molecule (Cl₂) is converted to a chlorate ion (ClO₃⁻). Here, the chlorine atom undergoes an increase in oxidation state, changing from an oxidation state of 0 in Cl₂ to an oxidation state of +5 in ClO₃⁻. This indicates the loss of electrons by chlorine, making option D an oxidation reaction.
Therefore, options B, C, and D represent oxidation reactions, while option A does not.
Learn more about oxidation reaction from the link given below.
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B) The bonds strongly hold ions together, reducing the boiling point.
C) The bonds prevent ions from moving throughout the crystal, so a solid ionic compound is a poor conductor.
D) The bonds prevent electrons from moving throughout the crystal, so a solid ionic compound is a poor conductor.
Ionic bonds in ionic compounds result in a crystal lattice structure that prevents free movement of ions, making solid ionic compounds poor conductors of electricity. However, dissolved or melted ionic compounds can conduct electricity. Additionally, these strong bonds lead to high melting and boiling points for ionic compounds.
Ionic bonds greatly affect the properties of ionic compounds. Ionic bonds are formed when one atom donates an electron to another, resulting in charged ions that attract each other. Option C) and D) are both partly correct. Ionic bonds lock these ions in a crystal lattice structure which prevents free movement. This makes solid ionic compounds poor conductors of electricity. However, if these compounds are dissolved in water or melted (thus allowing ions to move freely), they can conduct electricity.
Moreover, ionic bonds are strong. As a result, it requires a lot of energy to break these bonds, leading to high melting and boiling points for ionic compounds, which eliminates options A) and B).
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Answer:
Ka = 1.5 -10^-5
Explanation:
Step 1: Data given
Molarity of the solution = 0.25 M
pH = 2.71
Step 2 The equation
C3H7COOH + NaOH ⇆ C3H7COONa + H2O
Step 3: Calculate pKa
pH = (pKa-log[acid])/2
2pH = pKa -log[acid]
pKa = 2pH +log[acid]
⇒with pH = 2.71
⇒with log[acid] = -0.60
pKa=2*2.71 -0.60
pKa = 5.42 - 0.60
pKa = 4.82
Step 4: Calculate Ka
pKa = -log(Ka)
Ka = 10^-4.82
Ka = 1.5 -10^-5