Use the uncertainty principle to estimate the binding energy of the $\mathrm{H}_2$ molecule by calculating the difference in kinetic energy of the electrons between when they are in separate atoms and when they are in the molecule. Take $\Delta x$ for the electrons in the separated atoms to be the radius of the first Bohr orbit, $0.053 \mathrm{~nm}$, and for the molecule take $\Delta x$ to be the separation of the nuclei, $0.074 \mathrm{~nm}$. [Hint: Let $\left.\Delta p \approx \Delta p_x.\right]$

Suppose that a firm's capital stock $K(t)$ satisfies the differential equation

$$\dot{K}(t)=I-\delta K(t)$$

where investment $I$ is constant, and $\delta K(t)$ denotes depreciation, with $\delta$ a positive constant.

(a) Find the solution of the equation if the capital stock at time $t=0$ is $K_0$.

(b) Let $\delta=0.05$ and $I=10$. Explain what happens as $t \rightarrow \infty$ when:

(i) $K_0=150$;

(ii) $K_0=250$.

Predict the signs of $\Delta H, \Delta S$, and $\Delta G$ of the system for the following processes at 1 bar:
ammonia melts at $100^{\circ} \mathrm{C}$. (The normal melting point of ammonia is $77.7^{\circ} \mathrm{C}$.)

For reactions in condensed phases (liquids and solids), the difference between $\Delta H$ and $\Delta E$ is usually quite small. This statement holds for reactions carried out under atmospheric conditions. For certain geochemical processes, however, the external pressure may be so great that $\Delta H$ and $\Delta E$ can differ by a significant amount. A well-known example is the slow conversion of graphite to diamond under Earth’s surface. Calculate ($\Delta H - \Delta E$) for the conversion of $1$ mole of graphite to $1$ mole of diamond at a pressure of $50,000 atm$. The densities of graphite and diamond are $2.25 \mathrm{~g/cm^3}$ and $3.52 \mathrm{~g/cm^3}$, respectively.

For the gas-phase reaction $\mathrm{A}+\mathrm{A} \rightarrow \mathrm{A}_{2},$ the experimental rate constant, $k_{a},$ has been fitted to the Arrhenius equation with a frequency factor $A=4.07 \times 10^{8}\ \mathrm{dm}^{3}\ \mathrm{mol}^{-1}\ \mathrm{s}^{-1}$ at 300 K and an activation energy of $65.4\ \mathrm{kJ}\ \mathrm{mol}^{-1}.$ Calculate $\Delta^{3} S, \Delta^{3} H, \Delta^{3} U,$ and $\Delta^{3} G$ for the reaction.

The following table gives data, in Btu, for a system under. going a thermodynamic cycle consisting of four processes in series. Determine

(a) the missing table entries, each in Btu.

(b) whether the cycle is a power cycle or a refrigeration cycle.

Show that, in principle, $\mathrm{Na}_2 \mathrm{CO}_3(\mathrm{aq})$ can be converted almost completely to $\mathrm{NaOH}(\mathrm{aq})$ by the reaction

$$\begin{aligned} \mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s})+\mathrm{Na}_2 \mathrm{CO}_3(\mathrm{aq}) & \longrightarrow \\ & \mathrm{CaCO}_3(\mathrm{~s})+2 \mathrm{NaOH}(\mathrm{aq}) \end{aligned}$$

Draw any angle $\angle A$. Then construct $\angle B$ congruent to $\angle A$.

Draw a sketch of: a. a reflex angle b. an acute angle c. a right angle d. an obtuse angle.

For each value of $\epsilon$, find a positive value of $\delta$ such that the graph of the function leaves the window $a−\delta <x<a + \delta, b − \epsilon<y<b + \epsilon$ by the sides and not through the top or bottom.

$k(x) = cos x, a = 0, b = 1, \epsilon = 0.1, 0.001, 0.00001.$.

Sirius $\mathrm{B}$ has a mass and radius of $2.1 \times 10^{50} \mathrm{~kg}$ and $5.5 \times$ $10^5 \mathrm{~m}$, respectively. Assuming Sirius $B$ is perfectly spherical in shape, compute the ratio of the $\Delta t_{\mathrm{r}} / \Delta t_{\mathrm{d}}$. Using this result, confirm that the redshift quoted in Section $12.2$ for this star if $z-\Delta t_{\mathrm{r}} / \Delta t_{\mathrm{d}}-1$.

use Archimedes’ Principle, which states that when a sphere of radius r with density

$$d_s$$

is placed in a liquid of density

$$d_L=62.5lb/ft^3$$

,it will sink to a depth h where .

$$\dfrac{π}{3}(3rh^2-h^3)d_L=\dfrac{4}{3}πr^3d_s$$

Find an approximate value for h if:

$$r=5 ft$$

and

$$d_s=20lb/ft^3$$

$$\Delta A B C \cong \Delta H I J.$$

Name three pairs of congruent angles and three pairs of congruent sides.

Refer to the figure and use the principle of superposition to Find the component of the current $i$ through $R_{3}$ that is due to $V_{S 2}$.

$$\begin{aligned} & V_{S 1}=V_{S 2}=450 \mathrm{~V} \\ & R_{1}=7 \Omega \quad R_{2}=5 \Omega \\ & R_{3}=10 \Omega \quad R_{4}=R_{5}=1 \Omega \end{aligned}$$