Answer: G

To determine whether the protesters should prevail in their lawsuit, we must analyze the legal principles governing the First and Fourteenth Amendments in the context of private property.

1.  **State Action Doctrine**: The First and Fourteenth Amendments of the U.S. Constitution protect individuals from government interference with their rights, including freedom of speech. These protections only apply when there is "state action"—that is, when the government (federal, state, or local) is the actor involved.

2.  **Private Property vs. Public Forum**: The shopping mall in this scenario is explicitly described as "privately owned and operated." While it is "open to the public," it is not a government-owned space. Therefore, it does not constitute a "public forum" (like a public park or a city sidewalk) where the First Amendment provides a constitutional right to protest.

3.  **Private Property Rights**: Owners of private property, including shopping malls, generally have the right to control the activities that occur on their premises. This includes the right to set rules for conduct, remove individuals who violate those rules, and maintain the atmosphere for their customers and businesses.

4.  **Analysis of the Claims**:
    *   The protesters' claim is based on a violation of the First and Fourteenth Amendments.
    *   Because the mall is a private entity, its actions do not constitute state action.
    *   Without state action, the constitutional guarantees of the First and Fourteenth Amendments do not apply to the mall's actions.
    *   The fact that the protesters were peaceful and did not interfere with traffic is relevant in a public forum (where "time, place, and manner" restrictions are evaluated), but it does not grant them a constitutional right to protest on private property.
    *   The mall's marketing ("a small town with a big-town feel") has no legal bearing on the property's status as private.

5.  **Conclusion**: The protesters should not prevail because the mall is private property and there was no state action involved.

ANSWER: G


Answer: B

To determine which compound chlorine gas ($Cl_2$) reacts most readily with, we can examine the types of chemical reactions each compound undergoes with halogens.

1.  **Alkanes (Butane, Propane, Ethane):** These are saturated hydrocarbons. They react with chlorine only via a substitution reaction, which requires ultraviolet (UV) light to initiate the formation of chlorine radicals. This process is relatively slow compared to addition reactions.
2.  **Alkenes (Ethylene):** Ethylene ($C_2H_4$) is an unsaturated hydrocarbon containing a carbon-carbon double bond. It reacts with chlorine via an **addition reaction**. This reaction occurs readily at room temperature without the need for UV light or a catalyst, as the $\pi$ electrons of the double bond are easily accessible to the chlorine molecule.
3.  **Aromatic Hydrocarbons (Toluene, Benzene):** These react with chlorine via electrophilic aromatic substitution. This reaction requires a Lewis acid catalyst (such as $FeCl_3$ or $AlCl_3$) to proceed at a reasonable rate.
4.  **Alcohols (Ethanol) and Carboxylic Acids (Acetic Acid/Ethanoic Acid):** While these can react with chlorine, the reactions are typically slower and less "ready" than the rapid addition of chlorine across a double bond in an alkene.

**Comparison:**
-   **Ethylene (B):** Addition reaction (Very fast, no catalyst/UV needed).
-   **Butane (A), Propane (F), Ethane (G):** Substitution reaction (Requires UV light).
-   **Toluene (D), Benzene (I):** Substitution reaction (Requires catalyst).
-   **Acetic acid (C, H), Ethanol (E):** Other reactions (Slower than alkene addition).

Therefore, chlorine gas reacts most readily with ethylene.

ANSWER: B


Answer: B

To calculate the maximum non-expansion work per mole that can be obtained from a fuel cell, we need to find the Gibbs free energy change ($\Delta G$) for the combustion of methane at 298.15 K. The maximum non-expansion work ($w_{\text{max}}$) is equal to $-\Delta G$ for a process occurring at constant temperature and pressure.

The chemical reaction for the combustion of methane is:
$$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$

We use the standard Gibbs free energies of formation ($\Delta G_f^\circ$) at 298.15 K:
- $\Delta G_f^\circ(\text{CH}_4, g) = -50.7 \text{ kJ/mol}$
- $\Delta G_f^\circ(\text{CO}_2, g) = -394.3 \text{ kJ/mol}$
- $\Delta G_f^\circ(\text{H}_2\text{O}, l) = -237.1 \text{ kJ/mol}$
- $\Delta G_f^\circ(\text{O}_2, g) = 0 \text{ kJ/mol}$

The standard Gibbs free energy change for the reaction ($\Delta G^\circ$) is calculated as:
$$\Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})$$
$$\Delta G^\circ = [\Delta G_f^\circ(\text{CO}_2, g) + 2 \cdot \Delta G_f^\circ(\text{H}_2\text{O}, l)] - [\Delta G_f^\circ(\text{CH}_4, g) + 2 \cdot \Delta G_f^\circ(\text{O}_2, g)]$$
$$\Delta G^\circ = [-394.3 \text{ kJ/mol} + 2(-237.1 \text{ kJ/mol})] - [-50.7 \text{ kJ/mol} + 2(0 \text{ kJ/mol})]$$
$$\Delta G^\circ = [-394.3 - 474.2] - [-50.7]$$
$$\Delta G^\circ = -868.5 + 50.7 = -817.8 \text{ kJ/mol}$$

The maximum non-expansion work per mole is:
$$w_{\text{max}} = -\Delta G^\circ = 817.8 \text{ kJ/mol}$$

Comparing this to the given options, the value 817.90 $\text{kJ mol}^{-1}$ is the closest match.

ANSWER: B


Answer: G


Answer: F

To find the electric field $E$ at a radial distance $r = 4.0 \text{ cm}$, we use Gauss's Law for a cylindrical symmetry. The formula for the electric field at a distance $r$ from the axis of a long cylindrical charge distribution is:
$$E = \frac{\lambda_{encl}}{2\pi \varepsilon_0 r}$$
where $\lambda_{encl}$ is the charge per unit length enclosed by a Gaussian cylinder of radius $r$.

In this problem, we have two concentric cylindrical shells:
1.  Inner shell: radius $r_1 = 3.0 \text{ cm}$, charge per unit length $\lambda_1 = 5.0 \times 10^{-6} \text{ C/m}$.
2.  Outer shell: radius $r_2 = 6.0 \text{ cm}$, charge per unit length $\lambda_2 = -7.0 \times 10^{-6} \text{ C/m}$.

We are asked to find the electric field at $r = 4.0 \text{ cm}$. Since $r_1 < r < r_2$, the Gaussian cylinder of radius $r = 4.0 \text{ cm}$ only encloses the charge on the inner shell. Therefore, the enclosed charge per unit length is:
$$\lambda_{encl} = \lambda_1 = 5.0 \times 10^{-6} \text{ C/m}$$

Now, we substitute the values into the formula:
- $\lambda_{encl} = 5.0 \times 10^{-6} \text{ C/m}$
- $r = 4.0 \text{ cm} = 0.04 \text{ m}$
- $\varepsilon_0 \approx 8.854 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2)$

$$E = \frac{5.0 \times 10^{-6}}{2\pi(8.854 \times 10^{-12})(0.04)}$$
$$E = \frac{5.0 \times 10^{-6}}{5.563 \times 10^{-11} \cdot 0.04}$$
$$E = \frac{5.0 \times 10^{-6}}{2.225 \times 10^{-12}}$$
$$E \approx 2.247 \times 10^6 \text{ N/C}$$

Rounding to two significant figures, we get $E \approx 2.3 \times 10^6 \text{ N/C}$. Comparing this with the given options:
A) $3.0 \times 10^6 \text{ N/C}$
B) $4.5 \times 10^6 \text{ N/C}$
C) $0 \text{ N/C}$
D) $1.2 \times 10^6 \text{ N/C}$
E) $2.9 \times 10^6 \text{ N/C}$
F) $2.3 \times 10^6 \text{ N/C}$
G) $3.6 \times 10^6 \text{ N/C}$
H) $1.8 \times 10^6 \text{ N/C}$
I) $2.0 \times 10^6 \text{ N/C}$
J) $5.2 \times 10^6 \text{ N/C}$

The calculated value matches option F.

ANSWER: F


Answer: G

To determine whether the title insurance company is obligated to pay the friend's claim, we must examine the nature of a title insurance policy.

1.  **Contractual Nature of Title Insurance:** A title insurance policy is a contract between the title insurance company and the "named insured" (in this case, the woman). It is a personal contract that protects the insured's interest in the property against specific risks (like undisclosed liens).
2.  **Third-Party Status:** The friend is a third party to this contract. Because the friend did not purchase the policy and was not named as an insured, they have no direct contractual relationship with the title insurance company.
3.  **Transfer of Property vs. Transfer of Policy:** When the woman conveyed the property to the friend via a quitclaim deed, she transferred her interest in the property, but the title insurance policy did not automatically transfer to the friend. Standard title insurance policies do not extend coverage to subsequent owners unless the policy is specifically assigned to them.
4.  **Evaluating the Options:**
    *   A and B are incorrect because they focus on the friend's actions (not conducting a search or not buying insurance), which are irrelevant to the insurer's contractual obligation to the woman.
    *   C and E address whether the woman's policy is still valid. While some policies may end upon sale, even if the woman's policy were still valid, it would only cover *her* interest, not the friend's.
    *   D and J are incorrect because, while the insurer made an error, that error only creates a liability to the insured (the woman), not to third parties (the friend).
    *   F is incorrect because a quitclaim deed relates to the transfer of interest, not the transfer of insurance contracts.
    *   G is the most accurate legal reason: the friend is not a party to the contract (the named insured).
    *   I is incorrect because, although the woman might be liable to the friend for a breach of warranty, the friend cannot file a claim directly with the insurer; only the woman (the insured) could potentially file a claim for the loss of her interest or her liability to the friend.

Conclusion: The title insurance company is not obligated to pay the friend's claim because the friend was not a party to the contract.

ANSWER: G


Answer: B

To determine the correct answer, we must analyze the sequence of events and the legal principles governing deeds and property ownership.

1.  **Deed 1:** The professor, who owns the apartment building in fee simple, executed a deed conveying the building to his son and daughter as tenants in common. Crucially, the deed included a reservation of a life estate for the professor. This means the professor conveyed his **remainder interest** to his children.

2.  **Delivery of Deed 1:** The professor told his children of the conveyance and placed the deed in the family vault. In property law, telling the intended donees of the conveyance and placing the deed in a safe place (or delivering it to them) constitutes valid delivery. Thus, Deed 1 was a valid conveyance of the remainder interest.

3.  **Deed 2 ("Correction Deed"):** After the first conveyance, the professor executed a second deed (labeled "correction deed") attempting to convey the same building to his children as joint tenants with right of survivorship, again reserving a life estate for himself.

4.  **Validity of Deed 2:** A "correction deed" is typically used to fix clerical errors (like a misspelled name or a wrong legal description). It is not intended to change the *substance* of a conveyance (such as changing the type of ownership from tenants in common to joint tenants). Even if it were intended as a new conveyance, a grantor cannot convey an interest they no longer own. Since the professor had already conveyed the remainder interest in Deed 1, he had no remainder interest left to convey in Deed 2. Therefore, Deed 2 is ineffective and does not replace Deed 1.

5.  **Ownership after Son's Death:** Because Deed 1 is the only valid conveyance, the son and daughter own the remainder interest as **tenants in common**. Each holds an undivided 1/2 interest in the remainder, subject to the professor's life estate.

6.  **Inheritance:** When the son died intestate, his 1/2 interest in the remainder passed to his sole heir, the doctor. The daughter retains her 1/2 interest. Because they are tenants in common (and not joint tenants), the son's interest does not pass to the daughter by survivorship; it passes to his heir.

7.  **Final Ownership:** The daughter and the doctor now own the remainder interest as tenants in common, both subject to the professor's life estate.

Comparing this to the options:
A) Incorrect (son is dead).
B) Correct (daughter and doctor as tenants in common, subject to the professor's life estate).
C) Incorrect (would only be true if Deed 2 were valid).
D) Incorrect (omits the professor's life estate).
E) Incorrect (professor only has a life estate).
F) Incorrect (omits the doctor's share).
G) Incorrect (omits the daughter's share).
H) Incorrect (incorrect ownership form).
I) Incorrect (incorrect ownership form).

ANSWER: B


Answer: H

To find the $x$-coordinate of the positively charged particle, we use the formula for the electric field $\vec{E}$ produced by a charge $q$ at a point $(x, y)$:
$$\vec{E} = \frac{k q}{r^3} \vec{r}$$
where $\vec{r} = (x - x_0) \hat{\mathrm{i}} + (y - y_0) \hat{\mathrm{j}}$ is the vector from the particle's position $(x_0, y_0)$ to the point $(x, y)$, and $r = \sqrt{(x - x_0)^2 + (y - y_0)^2}$ is the distance between them.

Given:
1. At point $P_1 = (3.0, 3.0) \text{ cm}$, $\vec{E}_1 = 7.2(4.0 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}}) \text{ N/C} = (28.8 \hat{\mathrm{i}} + 21.6 \hat{\mathrm{j}}) \text{ N/C}$.
2. At point $P_2 = (2.0, 0) \text{ cm}$, $\vec{E}_2 = 100 \hat{\mathrm{i}} \text{ N/C}$.

From $\vec{E}_2$, we have the $y$-component $E_{2y} = 0$:
$$E_{2y} = \frac{k q (y_2 - y_0)}{r_2^3} = 0$$
Since $k q \neq 0$ and $r_2$ is finite, we must have $y_2 - y_0 = 0$. Given $y_2 = 0$, it follows that $y_0 = 0$.

Next, we use the components of $\vec{E}_1$ to find $x_0$:
$$\frac{E_{1x}}{E_{1y}} = \frac{x_1 - x_0}{y_1 - y_0}$$
Substituting the known values:
$$\frac{28.8}{21.6} = \frac{3.0 - x_0}{3.0 - 0}$$
$$\frac{4}{3} = \frac{3.0 - x_0}{3.0}$$
Multiplying both sides by 3.0:
$$4.0 = 3.0 - x_0$$
$$x_0 = 3.0 - 4.0 = -1.0 \text{ cm}$$

To verify, we check the consistency of $kq$:
At $P_1$: $r_1 = \sqrt{(3.0 - (-1.0))^2 + (3.0 - 0)^2} = \sqrt{4^2 + 3^2} = 5.0 \text{ cm}$.
$E_{1x} = \frac{kq(x_1 - x_0)}{r_1^3} \Rightarrow 28.8 = \frac{kq(4.0)}{5.0^3} = \frac{4kq}{125} \Rightarrow kq = \frac{28.8 \times 125}{4} = 900 \text{ N}\cdot\text{cm}^2/\text{C}$.
At $P_2$: $r_2 = \sqrt{(2.0 - (-1.0))^2 + (0 - 0)^2} = 3.0 \text{ cm}$.
$E_{2x} = \frac{kq(x_2 - x_0)}{r_2^3} \Rightarrow 100 = \frac{kq(3.0)}{3.0^3} = \frac{kq}{9} \Rightarrow kq = 900 \text{ N}\cdot\text{cm}^2/\text{C}$.
The values are consistent. Thus, the $x$-coordinate of the particle is $-1.0 \text{ cm}$.

ANSWER: H


Answer: E

To find the limit $\lim_{n \to \infty} \int_E \cos^2(nx + \xi_n) dx$, we can use the trigonometric identity:
$$\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$$
Applying this to the integral, we have:
$$\int_E \cos^2(nx + \xi_n) dx = \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} dx$$
$$\int_E \cos^2(nx + \xi_n) dx = \frac{1}{2} \int_E 1 dx + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) dx$$
The first term is:
$$\frac{1}{2} \int_E 1 dx = \frac{1}{2} m(E)$$
Given that the Lebesgue measure $m(E) = 2$, the first term is:
$$\frac{1}{2} \cdot 2 = 1$$
The second term is:
$$\frac{1}{2} \int_E \cos(2nx + 2\xi_n) dx = \frac{1}{2} \int_0^{2\pi} \chi_E(x) \cos(2nx + 2\xi_n) dx$$
where $\chi_E(x)$ is the characteristic function of the set $E$. Since $E \subset (0, 2\pi)$ is a measurable set, $\chi_E$ is an integrable function ($\chi_E \in L^1(0, 2\pi)$). According to the Riemann-Lebesgue Lemma, for any $f \in L^1(a, b)$,
$$\lim_{\lambda \to \infty} \int_a^b f(x) \cos(\lambda x + \phi) dx = 0$$
for any constant $\phi$. In our case, $\lambda = 2n$ and $\phi = 2\xi_n$. Although $\xi_n$ depends on $n$, the lemma still applies because $\cos(2nx + 2\xi_n)$ can be expanded as:
$$\cos(2nx + 2\xi_n) = \cos(2nx) \cos(2\xi_n) - \sin(2nx) \sin(2\xi_n)$$
Thus, the second integral is:
$$\frac{1}{2} \left[ \cos(2\xi_n) \int_0^{2\pi} \chi_E(x) \cos(2nx) dx - \sin(2\xi_n) \int_0^{2\pi} \chi_E(x) \sin(2nx) dx \right]$$
By the Riemann-Lebesgue Lemma, both $\int_0^{2\pi} \chi_E(x) \cos(2nx) dx$ and $\int_0^{2\pi} \chi_E(x) \sin(2nx) dx$ approach 0 as $n \to \infty$. Since $\cos(2\xi_n)$ and $\sin(2\xi_n)$ are bounded between -1 and 1, the entire second term approaches 0.
Therefore, the limit is:
$$\lim_{n \to \infty} \int_E \cos^2(nx + \xi_n) dx = 1 + 0 = 1.0$$
Comparing this result to the given options, we find it matches option E.

ANSWER: E