Which process is used to manufacture alkenes, and what is the term for species formed and consumed during a reaction pathway but not in the overall equation?

• A. Combustion of alkanes produces carbon dioxide and water.
• B. Polymerization reactions use alkenes as monomers to form polymers.
• C. Cracking of hydrocarbons (larger alkanes) breaks them down into smaller alkanes and alkenes, which is a major industrial method for producing alkenes.
• D. Addition of hydrogen to unsaturated vegetable oils (hydrogenation) produces saturated fats (alkanes).

The cracking of hydrocarbons is a common industrial process used to produce alkenes.

The final answer is $\boxed{\text{C}}$

• Intermediates are chemical species that are formed in one step of a reaction mechanism and consumed in a subsequent step. They do not appear in the overall balanced chemical equation.
• Reactants are the starting materials that are consumed in the overall reaction.
• Inhibitors slow down reactions but are not necessarily formed and consumed within the pathway.
• Products are the substances formed at the end of the overall reaction.

The description perfectly fits the definition of an intermediate.

The final answer is $\boxed{\text{A}}$

• A. Pentane (C$_{5}$H$_{12}$) has 3 structural isomers: n-pentane, isopentane (2-methylbutane), and neopentane (2,2-dimethylpropane). So, the statement "Pentane has five isomers" is incorrect.
• B. Butane (C$_{4}$H$_{10}$) has a higher molecular weight than propane (C$_{3}$H$_{8}$). For alkanes, boiling points generally increase with increasing molecular weight due to stronger London dispersion forces. Therefore, butane has a higher boiling point than propane. This statement is incorrect.
• C. Alkanes undergo substitution reactions with halogens (like chlorine) in the presence of UV light, not addition reactions. Addition reactions are characteristic of unsaturated compounds like alkenes and alkynes. This statement is incorrect.
• D. The general formula for an alkane is C$_{n}$H$_{2n+2}$. For an alkane with 49 carbon atoms (n=49), the number of hydrogen atoms is $2(49) + 2 = 98 + 2 = 100$. This statement is correct.

The final answer is $\boxed{\text{D}}$

• Ethanoic acid (CH$_{3}$COOH) is a weak acid.
• Sodium hydroxide (NaOH) is a strong base.

The titration of a weak acid with a strong base results in an equivalence point that is in the basic region (pH > 7).

• Phenolphthalein changes color in the pH range of approximately 8.2-10.0, which is suitable for a basic equivalence point.
• Methyl orange (pH 3.1-4.4), screened methyl orange (pH ~3.2-4.2), and methyl red (pH 4.4-6.2) change color in the acidic range, making them unsuitable for this titration.

The final answer is $\boxed{\text{B}}$

In the reaction:

• Zinc (Zn) goes from an oxidation state of 0 in $\text{Zn}_{(\text{s})}$ to +2 in $\text{ZnSO}_{4(\text{aq})}$ ($\text{Zn}^{2+}$). This is oxidation (loss of electrons).
• Copper (Cu) goes from an oxidation state of +2 in $\text{CuSO}_{4(\text{aq})}$ ($\text{Cu}^{2+}$) to 0 in $\text{Cu}_{(\text{s})}$. This is reduction (gain of electrons).

Reduction involves the gain of electrons. The copper ion $\text{Cu}^{2+}$ gains electrons to become neutral copper metal $\text{Cu}_{(\text{s})}$. To balance the charge, two electrons are needed.

Option C matches the derived reduction half-equation.

• Option A is a dissolution equation.
• Option B shows only one electron gained, which is incorrect for $\text{Cu}^{2+}$ to $\text{Cu}$.
• Option D is the oxidation half-equation for zinc.

The final answer is $\boxed{\text{C}}$