MikaStars39/nano-eval huggingface dataset:
The electric field intensity of an electromagnetic wave propagating in a vacuum is given by $E=E_{0} [ \operatorname{cos} ( \omega t-k y ) e_{x}-\operatorname{sin} ( \omega t-k y ) e_{z} ]$. What is the magnetic field intensity?

Options:
A. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{c o s} ( \omega t-k y ) e_{z}+\operatorname{s i n} ( \omega t-k y ) e_{x} ]
$$
B. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{c o s} ( \omega t-k y ) {\big(}e_{z} {\big)}+\operatorname{s i n} ( \omega t+k y ) e_{z} ]
$$
C. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{c o s} ( \omega t-k y ) {\big(}-e_{x} {\big)}+\operatorname{s i n} ( \omega t-k y ) e_{z} ]
$$
D. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{c o s} ( \omega t+k y ) {\big(}-e_{z} {\big)}-\operatorname{s i n} ( \omega t-k y ) e_{x} ]
$$
E. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{c o s} ( \omega t+k y ) {\big(}-e_{x} {\big)}-\operatorname{s i n} ( \omega t-k y ) e_{z} ]
$$
F. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{s i n} ( \omega t-k y ) {\big(}-e_{x} {\big)}-\operatorname{c o s} ( \omega t+k y ) e_{z} ]
$$
G. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{s i n} ( \omega t+k y ) {\big(}-e_{z} {\big)}-\operatorname{c o s} ( \omega t-k y ) e_{x} ]
$$
H. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{s i n} ( \omega t-k y ) {\big(}e_{z} {\big)}-\operatorname{c o s} ( \omega t+k y ) e_{x} ]
$$
I. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{s i n} ( \omega t+k y ) {\big(}e_{z} {\big)}-\operatorname{c o s} ( \omega t-k y ) e_{x} ]
$$
J. $$
\boldsymbol{H}={\frac{E_{0}} {3 7 7}} [ \operatorname{c o s} ( \omega t-k y ) {\big(}-e_{z} {\big)}-\operatorname{s i n} ( \omega t-k y ) e_{x} ]
$$
Answer: J
The vinyl chloride monomer used for synthesizing polyvinyl chloride at a certain factory is synthesized from $\mathrm{C_{2}H_{2}}$ and HCl over a mercuric chloride catalyst supported on activated carbon. The reaction equation is as follows:
$$
\mathrm{C_{2}H_{2} + HCl} \longrightarrow \mathrm{C_{2}H_{3}Cl}
$$
In the mixed raw material gas used at the factory,
$$
\mathrm{C_{2}H_{2} : HCl = 1 : 1} \quad (molar\ ratio).
$$
If the vinyl chloride content in the reactor outlet gas is 0.85 (mole fraction), calculate the conversion rates of acetylene and hydrogen chloride, respectively is().

Options:
A. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{0.89}1=0.89$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{0.89}{1.1}=0.809$$
B. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{0.965}1=0.965$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{0.965}{1.1}=0.877$$
C. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{0.85}1=0.85$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{0.85}{1.1}=0.773$$
D. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{0.75}1=0.75$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{0.75}{1.1}=0.682$$
E. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{0.925}1=0.925$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{0.925}{1.1}=0.841$$
F. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{0.80}1=0.80$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{0.80}{1.1}=0.727$$
G. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{0.92}1=0.92$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{0.92}{1.1}=0.836$$
H. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{1.015}1=1.015$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{1.015}{1.1}=0.923$$
I. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{1.1}1=1.1$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{1.1}{1.1}=1.0$$
J. $C_2H_2$ conversion rate
$$x_{\mathrm{C}_2\mathrm{H}_2}=\frac{1.05}1=1.05$$
HCl conversion rate
$$x_{\mathrm{HCl}}=\frac{1.05}{1.1}=0.955$$
Answer: B
Based on the Freundlich adsorption model, what is the kinetic equation derived for the synthesis reaction?$$\mathrm{N}_2+3\mathrm{H}_2⇌2\mathrm{NH}_3$$

Options:
A. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}^{0.5}}{p_{\mathrm{NH}_{3}}}-k_{2} \frac{p_{\mathrm{NH}_{3}}}{p_{\mathrm{H}_{2}}^{2}}$$
B. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}}{p_{\mathrm{NH}_{3}}^{1.5}}-k_{2} \frac{p_{\mathrm{NH}_{3}}}{p_{\mathrm{H}_{2}}^{1.5}}$$
C. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}^{1.2}\cdot\frac{p_{\mathrm{H}_{2}}^{1.5}}{p_{\mathrm{NH}_{3}}}-k_{2} \frac{p_{\mathrm{NH}_{3}}}{p_{\mathrm{H}_{2}}^{1.5}}$$
D. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}}{p_{\mathrm{NH}_{3}}^{1.5}}-k_{2} \frac{p_{\mathrm{NH}_{3}}^{1.5}}{p_{\mathrm{H}_{2}}}$$
E. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}^{1.5}}{p_{\mathrm{NH}_{3}}}-k_{2} \frac{p_{\mathrm{NH}_{3}}}{p_{\mathrm{H}_{2}}^{1.5}}$$
F. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}}{p_{\mathrm{NH}_{3}}}-k_{2} \frac{p_{\mathrm{NH}_{3}}^{1.5}}{p_{\mathrm{H}_{2}}}$$
G. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}^{1.5}}{p_{\mathrm{NH}_{3}}^{2}}-k_{2} \frac{p_{\mathrm{NH}_{3}}}{p_{\mathrm{H}_{2}}^{1.5}}$$
H. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}^{1.5}}{p_{\mathrm{NH}_{3}}}-k_{2} \frac{p_{\mathrm{NH}_{3}}^{1.5}}{p_{\mathrm{H}_{2}}}$$
I. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}^{2}}{p_{\mathrm{NH}_{3}}}-k_{2} \frac{p_{\mathrm{NH}_{3}}}{p_{\mathrm{H}_{2}}}$$
J. $$r_{\mathrm{NH}_{3}}=k_{1} p_{\mathrm{N}_{2}}\cdot\frac{p_{\mathrm{H}_{2}}^{1.5}}{p_{\mathrm{NH}_{3}}^{0.5}}-k_{2} \frac{p_{\mathrm{NH}_{3}}}{p_{\mathrm{H}_{2}}^{1.5}}$$
Answer: E
A first-order liquid-phase reversible reaction $\mathrm{A} \Longleftrightarrow \mathrm{P}$, with initial concentrations $c_{\mathrm{A0}}=0.5 \ \mathrm{mol/L}$ and $c_{\mathrm{P0}}=0$, is carried out in a batch reactor. After 8 minutes, the conversion rate of A is 33.3%, and the equilibrium conversion rate is 66.7%. What is the rate equation for this reaction?

Options:
A. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 1 2 ~ 3 c_{\mathrm{A}}-0. 0 4 5 ~ 6 ( c_{\mathrm{A 0}}-c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
B. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 3 0 ~ 0 c_{\mathrm{A}}-0. 0 2 5 ~ ( c_{\mathrm{A 0}}-2 c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
C. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 3 3 ~ 5 c_{\mathrm{A}}-0. 0 5 ~ ( c_{\mathrm{A 0}}-c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
D. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 7 7 ~ 7 c_{\mathrm{A}}-0. 0 9 ~ ( c_{\mathrm{A 0}}-c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
E. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 9 9 ~ 9 c_{\mathrm{A}}+0. 0 3 3 ~ 3 ( c_{\mathrm{A 0}}-c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
F. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 4 4 ~ 4 c_{\mathrm{A}}-0. 0 6 ~ ( c_{\mathrm{A 0}}-1. 5 c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
G. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 6 6 ~ 6 c_{\mathrm{A}}-0. 0 3 3 ~ 3 ( c_{\mathrm{A 0}}-c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
H. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 1 2 ~ 0 c_{\mathrm{A}}+0. 0 3 ~ 0 ( c_{\mathrm{A 0}}-c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
I. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 8 0 ~ 1 c_{\mathrm{A}}-0. 0 1 0 ~ 2 ( c_{\mathrm{A 0}}+c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
J. $$
\begin{aligned} {{(-r_{\mathrm{A}} )=}} & {{} {{} {0. 0 5 7 ~ 7 c_{\mathrm{A}}-0. 0 2 8 ~ 8 ( c_{\mathrm{A 0}}-c_{\mathrm{A}} ) ~ \mathrm{m o l} / ( \mathrm{L·min} )}}} \\ \end{aligned}
$$
Answer: J
In an isothermal batch reactor, a liquid-phase reaction takes place:
$$
A \xrightarrow{k_1} P \overset{k_2}{\underset{k_3}{\rightleftharpoons}} R
$$
The rate equations for the reaction are as follows:
$$
-\frac{\mathrm{d}c_A}{\mathrm{d}t} = k_1 c_A
$$
$$
\frac{\mathrm{d} c_{\mathrm{R}}}{\mathrm{d} t} = k_{2} c_{\mathrm{P}} - k_{3} c_{\mathrm{R}}
$$
$c_{\mathrm{A}}, c_{\mathrm{P}}, c_{\mathrm{R}}$ are the concentrations of substances $\mathrm{A, P, R}$, respectively. At a specific reaction temperature, the rate constants are given as $k_1 = 0.18 \ \mathrm{h}^{-1}$, $k_2 = 0.06 \ \mathrm{h}^{-1}$, and $k_3 = 0.04 \ \mathrm{h}^{-1}$. Calculate the maximum yield of $\mathbf{P}$ is ().

Options:
A. 0.481
B. 0.590
C. 0.712
D. 0.502
E. 0.562
F. 0.540
G. 0.693
H. 0.623
I. 0.608
J. 0.675
Answer: I
Decomposition reaction of acetaldehyde, with the reaction equation:
$$
\mathrm{C H_{3} C H O} \longrightarrow \mathrm{C H_{4}} + \mathrm{C O}
$$
Acetaldehyde enters an ideal tubular reactor for decomposition at a rate of 0.1 kg/s under conditions of 520°C and 0.1 MPa atmospheric pressure. It is known that under these reaction conditions, the reaction is an irreversible second-order reaction with a reaction rate constant of $k = 0.43 \times 10^{-3} \ \mathrm{m}^{3} / (\mathrm{mol} \cdot \mathrm{s}) = 0.43 \ \mathrm{m}^{3} / (\mathrm{kmol} \cdot \mathrm{s})$. If 35% of the acetaldehyde is converted, what are the reaction volume, space time, and mean residence time?

Options:
A. $$
V=14.6m^{3}
\\\tau=108s
\\t=93.1s
$$
B. $$
V=17.5m^{3}
\\\tau=118s
\\t=97.8s
$$
C. $$
V=18.0m^{3}
\\\tau=117s
\\t=97.2s
$$
D. $$
V=21.3m^{3}
\\\tau=127s
\\t=104.5s
$$
E. $$
V=13.9m^{3}
\\\tau=105s
\\t=92.5s
$$
F. $$
V=19.2m^{3}
\\\tau=122s
\\t=100.1s
$$
G. $$
V=18.7m^{3}
\\\tau=120s
\\t=98.5s
$$
H. $$
V=15.8m^{3}
\\\tau=110s
\\t=94.7s
$$
I. $$
V=20.0m^{3}
\\\tau=125s
\\t=102.3s
$$
J. $$
V=16.9m^{3}
\\\tau=115s
\\t=96.2s
$$
Answer: B
In a completely mixed flow reactor, the following liquid-phase reaction is carried out:
$$
A + B \overset{k_1}{\underset{k_2}{\rightleftharpoons}} R + S
$$
Given that $k_1 = 7.0 \, \mathrm{m^3 / (kmol \cdot min)}$ and $k_2 = 3.0 \, \mathrm{m^3 / (kmol \cdot min)}$, the volumetric flow rates of both A and B are equal to $0.004 \, \mathrm{m^3 / min}$, and the concentrations of A and B in the feed material are $2.8 \, \mathrm{kmol/m^3}$ and $1.6 \, \mathrm{kmol/m^3}$ respectively. Additionally, the reactor volume is $V_{\mathrm{R}} = 0.12 \, \mathrm{m^3}$. What is the conversion rate of B?

Options:
A. $$
x_{B}=0.85
$$
B. $$
x_{B}=0.62
$$
C. $$
x_{B}=0.75
$$
D. $$
x_{B}=0.65
$$
E. $$
x_{B}=0.70
$$
F. $$
x_{B}=0.80
$$
G. $$
x_{B}=0.58
$$
H. $$
x_{B}=0.90
$$
I. $$
x_{B}=0.45
$$
J. $$
x_{B}=0.50
$$
Answer: C
There is a fixed-bed reactor with a known tube diameter of 2.54 cm, a length of 150 cm, and a total of 1,320 tubes. The tubes are filled with catalyst particles with a diameter of $\phi 0.6$ cm. To determine the axial diffusion coefficient, a tracer experiment is conducted using a single tube of the same size. Gas flows downward through the catalyst particle bed, with the first detection point set 40 cm from the tracer injection point and the second detection point set 80 cm downstream from the first. Based on experimental data of the tracer concentration changes at the detection points, the following values are obtained: $\sigma_{1.t}^{2}=42 \; \mathrm{s}^{2}$ and $\sigma_{2.t}^{2}=68 \; \mathrm{s}^{2}$. The bed porosity is 0.42, and the superficial gas velocity is 1.4 cm/s. What is the value of $D_e / uL$?

Options:
A. 0.0297
B. 0.0314
C. 0.0187
D. 0.0234
E. 0.0279
F. 0.0241
G. 0.0225
H. 0.0196
I. 0.0208
J. 0.0253
Answer: G
The connected first-order irreversible reactions are as follows:
$$
\mathrm{A} {\stackrel{k_{1}} {\longrightarrow}} \mathrm{P} {\stackrel{k_{2}} {\longrightarrow}} \mathrm{S}.
$$
Given $k_{1}=0.15 \ \operatorname* {min}^{-1}, k_{2}=0.05 \ \operatorname* {min}^{-1}$,
the feed flow rate is $0.5 \ \mathrm{m}^{3} / \mathrm{min} \,, c_{\mathrm{A 0}} \neq 0 \,, c_{\mathrm{P 0}}=c_{\mathrm{S 0}}=0 \,.$
In a single CSTR with a volume of $V_{R} = 1 \ \mathrm{m}^3$, what is the yield of product $\mathbf{P}$?

Options:
A. 0.21
B. 0.25
C. 0.23
D. 0.24
E. 0.16
F. 0.22
G. 0.27
H. 0.20
I. 0.18
J. 0.19
Answer: A
In a circulating fluidized bed reactor, if the volume of the catalyst bed is $2 \mathrm{m}^{3}$ and the bed void fraction is 0.4, and a concentration of $10^{-2} \mathrm{k mol} / \mathrm{m}^{3}$ in the gas phase, what is the maximum reaction rate through the $2 \mathrm{m}^{3}$ catalyst bed in the reactor?

Options:
A. $$
0.8742kmol/s
$$
B. $$
0.9215kmol/s
$$
C. $$
0.8397kmol/s
$$
D. $$
0.7124kmol/s
$$
E. $$
0.9024kmol/s
$$
F. $$
0.6983kmol/s
$$
G. $$
0.7629kmol/s
$$
H. $$
0.6543kmol/s
$$
I. $$
0.8136kmol/s
$$
J. $$
0.7851kmol/s
$$
Answer: I
Consider the following system of differential equations:
$$
\ddot{y} \,+\, 3 \dot{y} \,+\, 2 \, y \,=\, 0, \qquad y ( 0 ) \,=\, 0.1, \qquad \dot{y} ( 0 ) \,=\, 0.05
$$
What is the response $y \left( t \right)$ of the system under the given initial conditions?

Options:
A. $$
\mathbf{y}=[ \mathbf{2} \qquad-1 ] \left[ \begin{matrix} {{{\mathbf{x}_{1}}}} \\ {{{\mathbf{x}_{2}}}} \\ \end{matrix} \right]
$$
B. $$
\mathbf{y}=[ \mathbf{1} \qquad0 ] \left[ \begin{matrix} {{{\mathbf{0}}}} \\ {{{\mathbf{x}_{2}}}} \\ \end{matrix} \right]
$$
C. $$
\mathbf{y}=[ \mathbf{1} \qquad1 ] \left[ \begin{matrix} {{{\mathbf{x}_{1}}}} \\ {{{\mathbf{x}_{2}}}} \\ \end{matrix} \right]
$$
D. $$
\mathbf{y}=[ \mathbf{1} \qquad0 ] \left[ \begin{matrix} {{{\mathbf{x}_{1}}}} \\ {{{\mathbf{x}_{2}}}} \\ \end{matrix} \right]
$$
E. $$
\mathbf{y}=[ \mathbf{0} \qquad1 ] \left[ \begin{matrix} {{{\mathbf{x}_{1}}}} \\ {{{\mathbf{x}_{2}}}} \\ \end{matrix} \right]
$$
F. $$
\mathbf{y}=[ \mathbf{1} \qquad2 ] \left[ \begin{matrix} {{{\mathbf{x}_{2}}}} \\ {{{\mathbf{x}_{1}}}} \\ \end{matrix} \right]
$$
G. $$
\mathbf{y}=[ \mathbf{-1} \qquad0 ] \left[ \begin{matrix} {{{\mathbf{x}_{1}}}} \\ {{{\mathbf{x}_{2}}}} \\ \end{matrix} \right]
$$
H. $$
\mathbf{y}=[ \mathbf{0} \qquad1 ] \left[ \begin{matrix} {{{\mathbf{x}_{2}}}} \\ {{{\mathbf{x}_{1}}}} \\ \end{matrix} \right]
$$
I. $$
\mathbf{y}=[ \mathbf{1} \qquad0 ] \left[ \begin{matrix} {{{\mathbf{x}_{1}}}} \\ {{{\mathbf{0}}}} \\ \end{matrix} \right]
$$
J. $$
\mathbf{y}=[ \mathbf{1} \qquad0 ] \left[ \begin{matrix} {{{\mathbf{x}_{2}}}} \\ {{{\mathbf{x}_{1}}}} \\ \end{matrix} \right]
$$
Answer: D
"The Lord Is My Firm Support" was created by ________.

Options:
A. Dietrich Bonhoeffer
B. John Wesley
C. Jonathan Edwards
D. Thomas Aquinas
E. Huldrych Zwingli
F. George Whitefield
G. Augustine of Hippo
H. Charles Spurgeon
I. John Calvin
J. Martin Luther
Answer: J
Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide.How many grams of aluminum hydroxide are obtained from 14.2 $\mathrm{g}$ of aluminum sulfide?

Options:
A. $$
1 5 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
B. $$
2 2 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
C. $$
1 8 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
D. $$
1 2 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
E. $$
1 4 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
F. $$
1 7 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
G. $$
2 0 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
H. $$
1 9 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
I. $$
1 6 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
J. $$
1 0 \operatorname{g A l} ( \mathrm{O H} )_{3}
$$
Answer: E
In 1865 a chemist reported that he had reacted a weighed amount of pure silver with nitric acid and had recovered all the silver as pure silver nitrate. The mass ratio of silver to silver nitrate was found to be 0.634985. Using only this ratio and the presently accepted values for the atomic weights of silver and oxygen, calculate the atomic weight of nitrogen is(). Compare this calculated atomic weight with the currently accepted value.

Options:
A. $$ 1 4. 2 0 9 4 $$
B. $$ 1 3. 5 0 0 9 $$
C. $$ 1 6. 0 0 7 5 $$
D. $$ 1 5. 5 0 9 9 $$
E. $$ 1 3. 7 0 8 3 $$
F. $$ 1 2. 0 1 0 2 $$
G. $$ 1 3. 9 8 1 2 $$
H. $$ 1 4. 1 0 5 6 $$
I. $$
1 4. 0 0 8 8
$$
J. $$ 1 5. 0 0 8 7 $$
Answer: I
Given that the heat of vaporization of toluene at its normal boiling point (383 K) is $3619 J \cdot g^{-1}$. Now completely gasify 1mol of toluene at 383K isothermal and isobaric conditions, and calculate the values of Q and W for this process; And calculate the value of the values of $\Delta U_{\mathrm{m}}, \Delta H_{\mathrm{m}}, \Delta S_{\mathrm{m}}, \Delta F_{\mathrm{m}}, \Delta G_{\mathrm{m}}$ for this process?

Options:
A. $\Delta U_{\mathrm{m} }= 33.25 kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta H_{\mathrm{m} }=35.80 kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta S_{m}=89.47kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta F_{m}=-3.509kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta G_{m}=0$
B. $Q=33.35kJ \cdot {mol}^{-1}$
$W=-3.184kJ \cdot {mol}^{-1}$
$\Delta U_{m}= 30 .16 kJ\cdot mol^{- 1}$
$\Delta H_{m}=33. 35 kJ\cdot mol^{- 1}$
$\Delta S_{m}=87. 06kJ\cdot mol^{- 1}$
$\Delta F_{m}=-3.184kJ\cdot mol^{- 1}$
$\Delta G_{m}=0$
C. $Q=33.35kJ \cdot {mol}$
$W=-3.184J$
$\Delta U_{m}= 30 .16 kJ\cdot mol^{- 1}$
$\Delta H_{m}=33. 35 kJ\cdot mol^{- 1}$
$\Delta S_{m}=87. 06kJ\cdot mol^{- 1}$
$\Delta F_{m}=-3.184kJ\cdot mol^{- 1}$
$\Delta G_{m}=0$
D. $Q=33.35kJ \cdot {mol}^{-1}$
$W=-3.184kJ$
$\Delta U_{\mathrm{m} }= 30.75 kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta H_{\mathrm{m} }=33.25 kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta S_{m}=86.02kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta F_{m}=-2.989kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta G_{m}=0$
E. $Q=33.35kJ \cdot {mol}^{-1}$
$W=-3.184kJ \cdot {mol}$
$\Delta U_{m}= 30 .16 kJ\cdot mol^{- 1}$
$\Delta H_{m}=33. 35 kJ\cdot mol^{- 1}$
$\Delta S_{m}=87. 06kJ\cdot mol^{- 1}$
$\Delta F_{m}=-3.184kJ\cdot mol^{- 1}$
$\Delta G_{m}=0$
F. $Q=33.35kJ \cdot {mol}^{-1}$
$W=-3.184kJ^{-1}$
$\Delta U_{m}= 30 .16 kJ\cdot mol^{- 1}$
$\Delta H_{m}=33. 35 kJ\cdot mol^{- 1}$
$\Delta S_{m}=87. 06kJ\cdot mol^{- 1}$
$\Delta F_{m}=-3.184kJ\cdot mol^{- 1}$
$\Delta G_{m}=0$
G. $Q=33.35kJ \cdot {mol}$
$W=-3.184kJ$
$\Delta U_{m}= 30 .16 kJ\cdot mol^{- 1}$
$\Delta H_{m}=33. 35 kJ\cdot mol^{- 1}$
$\Delta S_{m}=87. 06kJ\cdot mol^{- 1}$
$\Delta F_{m}=-3.184kJ\cdot mol^{- 1}$
$\Delta G_{m}=0$
H. $Q=33.35kJ^{-1}$
$W=-3.184kJ$
$\Delta U_{m}= 30 .16 kJ\cdot mol^{- 1}$
$\Delta H_{m}=33. 35 kJ\cdot mol^{- 1}$
$\Delta S_{m}=87. 06kJ\cdot mol^{- 1}$
$\Delta F_{m}=-3.184kJ\cdot mol^{- 1}$
$\Delta G_{m}=0$
I. $Q=33.35kJ \cdot {mol}^{-1}$
$W=-3.184kJ$
$\Delta U_{m}= 30 .16 kJ\cdot mol^{- 1}$
$\Delta H_{m}=33. 35 kJ\cdot mol^{- 1}$
$\Delta S_{m}=87. 06kJ\cdot mol^{- 1}$
$\Delta F_{m}=-3.184kJ\cdot mol^{- 1}$
$\Delta G_{m}=0$
J. $Q=33.35kJ \cdot {mol}^{-1}$
$W=-3.184kJ$
$\Delta U_{\mathrm{m} }= 28.90 kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta H_{\mathrm{m} }=31.85 kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta S_{m}=85.75kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta F_{m}=-3.015kJ\cdot$ $\mathrm{mol}^{- 1}$
$\Delta G_{m}=0$
Answer: I
At 298 K, the standard enthalpies of combustion for cyclopropane (g), graphite, and hydrogen gas are -2092, -393.7, and -285.8 kJ·mol$^{-1}$, respectively. The standard molar enthalpy of formation for propene (g) is $20.50 kJ\cdot\mathrm{mol}^{-1}$. What is the $\Delta H_{m}^{\Theta}$ for the isomerization of cyclopropane (g) into propene (g) at 298 K?

Options:
A. $$
- 3 4. 0 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
B. $$
- 3 3. 5 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
C. $$
- 3 2. 7 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
D. $$
- 3 4. 2 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
E. $$
- 3 2. 0 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
F. $$
- 3 2. 9 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
G. $$
- 3 5. 1 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
H. $$
- 3 3. 1 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
I. $$
- 3 1. 6 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
J. $$
- 3 3. 8 \mathrm{k J} \cdot\mathrm{m o l}^{-1}
$$
Answer: H
At 80.3 K, the vapor pressure of oxygen is $3.13 \mathrm{~kPa}$, and the vapor pressure of nitrogen is $144.79 \mathrm{~kPa}$. Assuming air is composed of 21% oxygen and 79% nitrogen (by volume fraction) and treating liquid air as an ideal solution, what is the minimum pressure required at 80.3 K to completely liquefy the air?

Options:
A. 121.08kPa
B. 128.53kPa
C. 127.67kPa
D. 110.76kPa
E. 115.32kPa
F. 140.12kPa
G. 117.41kPa
H. 119.94kPa
I. 133.45kPa
J. 123.89kPa
Answer: A
At $700 \, \mathrm{K}$, the reaction $\mathrm{H}_{2} + \mathrm{I}_{2} \longrightarrow 2 \, \mathrm{HI}$ was studied experimentally, and the rate constant was measured as $k = 6.42 \times 10^{-2} \, \mathrm{mol}^{-1} \cdot \mathrm{dm}^{3} \cdot \mathrm{s}^{-1}$, with $E_{\mathrm{a}} = 167 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}$.
Using viscosity methods, the diameters of the particles were determined as $d(\mathrm{H}_{2}) = 225 \, \mathrm{pm}$ and $d(\mathrm{I}_{2}) = 559 \, \mathrm{pm}$. Now, assuming this to be an elementary reaction (as was believed earlier). Please calculate the reaction rate constant based on simple collision theory.

Options:
A. $$
0. 3 0 0 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
B. $$
0. 3 1 5 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
C. $$
0. 1 8 2 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
D. $$
0. 2 7 3 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
E. $$
0. 3 5 6 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
F. $$
0. 0 9 7 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
G. $$
0. 1 7 5 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
H. $$
0. 2 2 5 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
I. $$
0. 2 4 8 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
J. $$
0. 4 0 9 \ \mathrm{d m}^{3} \cdot\mathrm{m o l}^{-1} \cdot\ \mathrm{s}^{-1}
$$
Answer: D
Gas-phase reaction $H_{2}+Cl \longrightarrow H+HCl$, where $k=7.94 \times 10^{10} exp(230 kJ \cdot mol^{-1} / RT)$, $\Delta_{r} G_{m}^{\Theta}=203.576 kJ \cdot mol^{-1}$, $\Delta_{r} H_{m}^{\Theta}=4.10 kJ \cdot mol^{-1}$. What is the expression for $k_{b}$ for the reverse reaction?

Options:
A. $$
k_{\mathrm{b}}=5. 7 2 \times 1 0^{9} \ \mathrm{e x p} (-2 0. 1 \times 1 0^{3} / R T )
$$
B. $$
k_{\mathrm{b}}=4. 8 3 \times 1 0^{9} \ \mathrm{e x p} (-1 6. 8 \times 1 0^{3} / R T )
$$
C. $$
k_{\mathrm{b}}=8. 3 1 \times 1 0^{7} \ \mathrm{e x p} (-1 7. 7 \times 1 0^{3} / R T )
$$
D. $$
k_{\mathrm{b}}=6. 7 8 \times 1 0^{8} \ \mathrm{e x p} (-1 9. 1 \times 1 0^{3} / R T )
$$
E. $$
k_{\mathrm{b}}=7. 6 0 \times 1 0^{8} \ \mathrm{e x p} (-1 8. 5 \times 1 0^{3} / R T )
$$
F. $$
k_{\mathrm{b}}=2. 4 9 \times 1 0^{8} \ \mathrm{e x p} (-2 1. 2 \times 1 0^{3} / R T )
$$
G. $$
k_{\mathrm{b}}=5. 1 2 \times 1 0^{8} \ \mathrm{e x p} (-1 9. 6 \times 1 0^{3} / R T )
$$
H. $$
k_{\mathrm{b}}=3. 9 5 \times 1 0^{8} \ \mathrm{e x p} (-2 0. 4 \times 1 0^{3} / R T )
$$
I. $$
k_{\mathrm{b}}=3. 5 4 \times 1 0^{7} \ \mathrm{e x p} (-1 7. 3 \times 1 0^{3} / R T )
$$
J. $$
k_{\mathrm{b}}=4. 6 0 \times1 0^{8} \ \mathrm{e x p} (-1 8. 9 \times1 0^{3} / R T )
$$
Answer: J
The thorium decay series involves the following processes (in the reaction formula, the quantity below "→" indicates the half-life):
$$
{}^{228}Ac \xrightarrow[6.13 h]{\beta^-} {}^{228}Th \xrightarrow[1.913 a]{\alpha} {}^{224}Ra \xrightarrow[3.64 d]{-\alpha} {}^{220}Rn\xrightarrow[55 s]{\alpha} {}^{216}Po
$$
What is the ratio $N({}^{228}Ac)/N({}^{228}Th)$ at $t=2.0\times10^{5}$ s?

Options:
A. $$
1. 8 8 \times1 0^{-3}
$$
B. $$
2. 4 3 \times1 0^{-3}
$$
C. $$
1. 7 6 \times1 0^{-3}
$$
D. $$
1. 9 2 \times1 0^{-2}
$$
E. $$
9. 7 4 \times1 0^{-4}
$$
F. $$
1. 2 5 \times1 0^{-3}
$$
G. $$
5. 5 6 \times1 0^{-4}
$$
H. $$
8. 6 5 \times1 0^{-3}
$$
I. $$
2. 0 3 \times1 0^{-2}
$$
J. $$
2. 1 5 \times1 0^{-2}
$$
Answer: A
At 298K, the ionic mobility of $H^{+}$ is $3.61 \times 10^{-8} m^{2} \cdot V^{-1} \cdot s^{-1}$. If the viscosity of water at the same temperature is $8.94 \times 10^{-3} P$, what is the effective radius r of $H^{+}$?

Options:
A. $$
2. 6 3 \times1 0^{-9} \mathrm{~ c m}
$$
B. $$
5. 6 2 \times1 0^{-9} \mathrm{~ c m}
$$
C. $$
2. 1 7 \times1 0^{-9} \mathrm{~ c m}
$$
D. $$
4. 0 1 \times1 0^{-9} \mathrm{~ c m}
$$
E. $$
3. 0 9 \times1 0^{-9} \mathrm{~ c m}
$$
F. $$
1. 8 2 \times1 0^{-9} \mathrm{~ c m}
$$
G. $$
3. 7 5 \times1 0^{-9} \mathrm{~ c m}
$$
H. $$
4. 8 0 \times1 0^{-9} \mathrm{~ c m}
$$
I. $$
2. 9 6 \times1 0^{-9} \mathrm{~ c m}
$$
J. $$
1. 5 4 \times1 0^{-9} \mathrm{~ c m}
$$
Answer: A
At 298 K, using the same conductivity cell, the resistance of a $0.0200 \, \mathrm{mol \cdot dm^{-3}}$ KCl aqueous solution is measured to be 100 $\Omega$, the resistance of a saturated AgCl aqueous solution is $1.00 \times 10^{5} \, \Omega$, and the resistance of distilled water is $2.00 \times 10^{5} \, \Omega$. It is also known that at 298 K, the molar conductivity of the $0.0200 \, \mathrm{mol \cdot dm^{-3}}$ KCl aqueous solution is $\Lambda_\mathrm{m}($KCl$) = 0.01383 \, \Omega^{-1} \cdot \mathrm{m}^2 \cdot \mathrm{mol}^{-1}$, and the molar conductivity of the saturated AgCl aqueous solution is $\Lambda_\mathrm{m}($AgCl$) = 0.01268 \, \Omega^{-1} \cdot \mathrm{m}^2 \cdot \mathrm{mol}^{-1}$. What are the solubility and solubility product of AgCl in water at 298 K?

Options:
A. $$
8. 7 4 \times1 0^{-5} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
9. 8 6 \times1 0^{-9}
$$
B. $$
6. 4 8 \times1 0^{-6} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
8. 9 1 \times1 0^{-9}
$$
C. $$
1. 1 0 3 \times1 0^{-5} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
1. 0 8 \times1 0^{-1 0}
$$
D. $$
1. 5 0 2 \times1 0^{-7} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
4. 2 3 \times1 0^{-1 1}
$$
E. $$
9. 9 3 \times1 0^{-7} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
2. 0 5 \times1 0^{-1 1}
$$
F. $$
7. 2 6 \times1 0^{-8} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
6. 1 7 \times1 0^{-1 2}
$$
G. $$
1. 2 0 3 \times1 0^{-7} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
3. 7 9 \times1 0^{-1 1}
$$
H. $$
9. 3 5 \times1 0^{-6} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
5. 6 4 \times1 0^{-1 0}
$$
I. $$
2. 0 8 5 \times1 0^{-5} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
2. 3 2 \times1 0^{-1 0}
$$
J. $$
1. 0 9 1 \times1 0^{-5} \mathrm{~ m o l ~ \cdot~ d m^{-3} ~}
$$
$$
1. 1 9 \times1 0^{-1 0}
$$
Answer: J
Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir. The purpose of the venturi is to create a vacuum in the reservoir when the venturi is placed in an airstream. (The vacuum is defined as the pressure difference below the outside ambient pressure.) The venturi has a throat-to-inlet area ratio of 0.85. Calculate the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of 90 m/s at standard sea level conditions.

Options:
A. $$
{2 0 5 0 \, \mathrm{N / m}^{2}}
$$
B. $$
{1 9 0 0 \, \mathrm{N / m}^{2}}
$$
C. $$
{1 9 8 0 \, \mathrm{N / m}^{2}}
$$
D. $$
{1 6 0 0 \, \mathrm{N / m}^{2}}
$$
E. $$
{1 8 6 0 \, \mathrm{N / m}^{2}}
$$
F. $$
{1 9 1 3 \, \mathrm{N / m}^{2}}
$$
G. $$
{2 1 0 0 \, \mathrm{N / m}^{2}}
$$
H. $$
{2 2 5 0 \, \mathrm{N / m}^{2}}
$$
I. $$
{1 7 5 0 \, \mathrm{N / m}^{2}}
$$
J. $$
{2 0 0 0 \, \mathrm{N / m}^{2}}
$$
Answer: F
If a hydrometer weighs 0.2 $\mathbf{N}$ and has a stem diameter of $5$ mm, find the distance between markings corresponding to relative densities of (i) 1.0 and 0.95 and (ii) 1.0 and 1.05.

Options:
A. $$
5 4. 9 0 \ \mathrm{m m}
$$
$$
- 4 9. 3 8 \mathrm{~ m m}
$$
B. $$
5 5. 0 2 \ \mathrm{m m}
$$
$$
- 4 9. 4 0 \mathrm{~ m m}
$$
C. $$
5 3. 8 1 \ \mathrm{m m}
$$
$$
- 5 0. 2 7 \mathrm{~ m m}
$$
D. $$
5 4. 5 7 \ \mathrm{m m}
$$
$$
- 5 0. 0 3 \mathrm{~ m m}
$$
E. $$
5 4. 2 4 \ \mathrm{m m}
$$
$$
- 4 9. 6 5 \mathrm{~ m m}
$$
F. $$
5 4. 6 8 \ \mathrm{m m}
$$
$$
- 4 8. 9 9 \mathrm{~ m m}
$$
G. $$
5 4. 7 6 \ \mathrm{m m}
$$
$$
- 4 9. 5 4 \mathrm{~ m m}
$$
H. $$
5 5. 6 3 \ \mathrm{m m}
$$
$$
- 4 8. 7 1 \mathrm{~ m m}
$$
I. $$
5 3. 9 7 \ \mathrm{m m}
$$
$$
- 4 9. 8 6 \mathrm{~ m m}
$$
J. $$
5 3. 7 4 \ \mathrm{m m}
$$
$$
- 5 0. 1 6 \mathrm{~ m m}
$$
Answer: G
A person of mass 70kg and body temperature 37°C breathes in 0.5l of air at a temperature of 20°C. Assume that there are no other sources of heat (turn off metabolism for a moment), and the body as a whole is insulated so no heat is lost to the environment Find the equilibrium temperuture that the air and body will ultimately attain is________. Useful data: $\rho_{a i r} \,=\,1.3 \,kg^{-3},\rho_{water}=1000kg m^{-3},c_{air}=1000J K^{-1} kg^{-1},c_{water}=4200J K^{-1} kg^{-1}.$ Assume that the person$^{\prime}$s body tissue has the same heat capacity and density as water.

Options:
A. $$
3 6. 9 0 0 0 0^{\circ}C
$$
B. $$
3 6. 8 9 9 9 9^{\circ}C
$$
C. $$
3 7. 0 0 0 0 0^{\circ}C
$$
D. $$
3 5. 8 8 8 8 9^{\circ}C
$$
E. $$
3 6. 9 9 9 9 6^{\circ}C
$$
F. $$
3 6. 8 8 8 8 8^{\circ}C
$$
G. $$
3 7. 0 0 1 1 1^{\circ}C
$$
H. $$
3 6. 7 7 7 7 7^{\circ}C
$$
I. $$
3 5. 9 9 9 9 8^{\circ}C
$$
J. $$
3 7. 1 1 1 1 1^{\circ}C
$$
Answer: E
The density of atoms in the interstellar medium is most similar to().

Options:
A. the surface of the Moon
B. the center of a dense star
C. the deep interior of a neutron star
D. the core of a gas giant planet
E. the edge of a quasar disk
F. the corona of the Sun
G. the surface of a white dwarf
H. the interior of a TV tube
I. the atmosphere of Earth
J. the interior of a black hole
Answer: H
Let
$$
R=\left\{\begin{matrix} {{{1}}} & {{{0.4}}} & {{{0.9}}} & {{{0.6}}} \\ {{{0.4}}} & {{{1}}} & {{{0.7}}} & {{{0.5}}} \\ {{{0.9}}} & {{{0.7}}} & {{{1}}} & {{{0.8}}} \\ {{{0.6}}} & {{{0.5}}} & {{{0.8}}} & {{{1}}} \\ \end{matrix} \right\}
$$
What is $\hat{R}$?

Options:
A. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.8 & 0.6 & 0.7 \\
0.8 & 1 & 0.8 & 0.6 \\
0.6 & 0.8 & 1 & 0.7 \\
0.7 & 0.6 & 0.7 & 1
\end{matrix}\right\}$$
B. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.7 & 0.8 & 0.9 \\
0.7 & 1 & 0.8 & 0.6 \\
0.8 & 0.8 & 1 & 0.7 \\
0.9 & 0.6 & 0.7 & 1
\end{matrix}\right\}$$
C. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.9 & 0.8 & 0.7 \\
0.9 & 1 & 0.6 & 0.9 \\
0.8 & 0.6 & 1 & 0.9 \\
0.7 & 0.9 & 0.9 & 1
\end{matrix}\right\}$$
D. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.6 & 0.8 & 0.7 \\
0.6 & 1 & 0.6 & 0.6 \\
0.8 & 0.6 & 1 & 0.8 \\
0.7 & 0.6 & 0.8 & 1
\end{matrix}\right\}$$
E. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.5 & 0.9 & 0.7 \\
0.5 & 1 & 0.6 & 0.8 \\
0.9 & 0.6 & 1 & 0.7 \\
0.7 & 0.8 & 0.7 & 1
\end{matrix}\right\}$$
F. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.7 & 0.9 & 0.8 \\
0.7 & 1 & 0.7 & 0.7 \\
0.9 & 0.7 & 1 & 0.8 \\
0.8 & 0.7 & 0.8 & 1
\end{matrix}\right\}$$
G. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.6 & 0.7 & 0.8 \\
0.6 & 1 & 0.9 & 0.7 \\
0.7 & 0.9 & 1 & 0.7 \\
0.8 & 0.7 & 0.7 & 1
\end{matrix}\right\}$$
H. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.8 & 0.9 & 0.7 \\
0.8 & 1 & 0.8 & 0.7 \\
0.9 & 0.8 & 1 & 0.6 \\
0.7 & 0.7 & 0.6 & 1
\end{matrix}\right\}$$
I. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.9 & 0.7 & 0.8 \\
0.9 & 1 & 0.6 & 0.8 \\
0.7 & 0.6 & 1 & 0.7 \\
0.8 & 0.8 & 0.7 & 1
\end{matrix}\right\}$$
J. $$\hat{R}=
\left\{\begin{matrix}
1 & 0.7 & 0.8 & 0.6 \\
0.7 & 1 & 0.8 & 0.9 \\
0.8 & 0.8 & 1 & 0.6 \\
0.6 & 0.9 & 0.6 & 1
\end{matrix}\right\}$$
Answer: F
Let $G$ be a group.\
Let $G \,=\, D_{8}$ be the dihedral group with 8 elements. What are the dimensions of its irreducible representations over C?

Options:
A. 1
B. 0
C. 9
D. 4
E. 7
F. 8
G. 6
H. 2
I. 3
J. 5
Answer: F
What is $U$ for relativistic quark matter with $n_{u}=n_{d}=n_{s}=n?$

Options:
A. $$
9 \times1 0^{4 7} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-1 / 3} T_{9}^{2}
$$
B. $$
6 \times1 0^{4 5} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-3 / 3} T_{5}^{5}
$$
C. $$
1 \times1 0^{4 9} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-1 / 4} T_{6}^{1}
$$
D. $$
2 \times1 0^{4 8} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-2 / 4} T_{2}^{2}
$$
E. $$
8 \times1 0^{4 7} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-1 / 6} T_{0}^{0}
$$
F. $$
8 \times1 0^{4 6} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-1 / 2} T_{8}^{3}
$$
G. $$
4 \times1 0^{4 5} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-1 / 5} T_{1}^{1}
$$
H. $$
5 \times1 0^{4 8} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-2 / 3} T_{7}^{2}
$$
I. $$
7 \times1 0^{4 6} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-4 / 3} T_{3}^{3}
$$
J. $$
3 \times1 0^{4 9} \, \mathrm{e r g} \, ( M / M_{\odot} ) ( n / n_{\mathrm{n u c}} )^{-1 / 1} T_{4}^{4}
$$
Answer: A
Evaluate $I$ in the low-temperature limit appropriate for the late stages of collapse when the neutrinos become trapped in the core is(). In this regime, $E_{e} \gg m_{e} c^{2}$ , $\mu_{e} \gg k T$ , and $\mu_{\nu} \gg k T.$

Options:
A. $$
\frac{1} {5} \left( \mu_{\nu}^{5}+\left( Q+\mu_{e} \right)^{5} \right)-\frac{Q} {2} \left[ \mu_{\nu}^{4}+\left( Q+\mu_{e} \right)^{4} \right]+\frac{Q^{2}} {3} \left[ \mu_{\nu}^{3}+\left( Q+\mu_{e} \right)^{3} \right]
$$
B. $$
\frac{1} {6} \left( \mu_{e}^{5}+\left( Q-\mu_{\nu} \right)^{5} \right)-\frac{Q} {3} \left( \mu_{e}^{4}+\left( Q-\mu_{\nu} \right)^{4} \right)+\frac{Q^{2}} {4} \left( \mu_{e}^{3}+\left( Q-\mu_{\nu} \right)^{3} \right)
$$
C. $$
\frac{1} {3} \left( \mu_{\nu}^{5}-\left( Q+\mu_{e} \right)^{5} \right)-\frac{Q} {4} \left( \mu_{\nu}^{4}-\left( Q+\mu_{e} \right)^{4} \right)+\frac{Q^{2}} {5} \left( \mu_{\nu}^{3}-\left( Q+\mu_{e} \right)^{3} \right)
$$
D. $$
\frac{1} {5} \left[ \mu_{e}^{5}-\left( Q+\mu_{\nu} \right)^{5} \right]-\frac{Q} {2} \left[ \mu_{e}^{4}-\left( Q+\mu_{\nu} \right)^{4} \right]+\frac{Q^{2}} {3} \left[ \mu_{e}^{3}-\left( Q+\mu_{\nu} \right)^{3} \right]
$$
E. $$
\frac{1} {7} \left[ \mu_{\nu}^{5}-\left( Q-\mu_{e} \right)^{5} \right]-\frac{Q} {2} \left[ \mu_{\nu}^{4}-\left( Q-\mu_{e} \right)^{4} \right]+\frac{Q^{2}} {3} \left[ \mu_{\nu}^{3}-\left( Q-\mu_{e} \right)^{3} \right]
$$
F. $$
\frac{1} {8} \left[ \mu_{\nu}^{5}-\left( Q-\mu_{e} \right)^{5} \right]-\frac{Q} {2} \left( \mu_{\nu}^{4}+\left( Q-\mu_{e} \right)^{4} \right)+\frac{Q^{2}} {5} \left( \mu_{\nu}^{3}+\left( Q-\mu_{e} \right)^{3} \right)
$$
G. $$
\frac{1} {7} \left[ \mu_{e}^{5}+\left( Q-\mu_{\nu} \right)^{5} \right]-\frac{Q} {4} \left[ \mu_{e}^{4}-\left( Q-\mu_{\nu} \right)^{4} \right]+\frac{Q^{2}} {5} \left[ \mu_{e}^{3}+\left( Q-\mu_{\nu} \right)^{3} \right]
$$
H. $$
\frac{1} {4} \left[ \mu_{\nu}^{5}+\left( Q+\mu_{e} \right)^{5} \right]-\frac{Q} {3} \left[ \mu_{\nu}^{4}-\left( Q+\mu_{e} \right)^{4} \right]+\frac{Q^{2}} {2} \left[ \mu_{\nu}^{3}-\left( Q+\mu_{e} \right)^{3} \right]
$$
I. $$
\frac{1} {5} \left[ \mu_{e}^{5}+\left( Q+\mu_{\nu} \right)^{5} \right]-\frac{Q} {2} \left( \mu_{e}^{4}+\left( Q+\mu_{\nu} \right)^{4} \right)+\frac{Q^{2}} {3} \left( \mu_{e}^{3}+\left( Q+\mu_{\nu} \right)^{3} \right)
$$
J. $$
\frac{1} {9} \left( \mu_{\nu}^{5}+\left( Q-\mu_{e} \right)^{5} \right)-\frac{Q} {2} \left[ \mu_{\nu}^{4}+\left( Q-\mu_{e} \right)^{4} \right]+\frac{Q^{2}} {3} \left[ \mu_{\nu}^{3}+\left( Q-\mu_{e} \right)^{3} \right]
$$
Answer: D
A sphere of radius $R$ carries a charge density $\rho( r )=k r$ (where $k$ is a constant).

Options:
A. $$
\frac{\pi k^{2} R^{6}} {6 \epsilon_{0}}.
$$
B. $$
\frac{\pi k^{2} R^{8}} {8 \epsilon_{0}}.
$$
C. $$
\frac{2 \pi k^{2} R^{7}} {7 \epsilon_{0}}.
$$
D. $$
\frac{\pi k^{2} R^{9}} {9 \epsilon_{0}}.
$$
E. $$
\frac{\pi k R^{7}} {7 \epsilon_{0}}.
$$
F. $$
\frac{\pi k^{2} R^{7}} {5 \epsilon_{0}}.
$$
G. $$
\frac{\pi k^{2} R^{7}} {14 \epsilon_{0}}.
$$
H. $$
\frac{\pi k^{2} R^{6}} {8 \epsilon_{0}}.
$$
I. $$
\frac{\pi k^{2} R^{7}} {7 \epsilon_{0}}.
$$
J. $$
\frac{\pi R^{7}} {7 \epsilon_{0}^{2}}.
$$
Answer: I
The nuclear masses of $^7Be$, $^9Be$, and $^{10}Be$ are 7.0147 amu, 9.0100 amu, and 10.0113 amu, respectively. What are the binding energies per nucleon of these nuclei?

Options:
A. $$^7Be=7.32\times10^{-13}J/nucleon$$
$$^9Be=9.98\times10^{-13}J/nucleon$$
$$^{10}Be=1.10\times10^{-12}J/nucleon$$
B. $$^7Be=8.61\times10^{-13}J/nucleon$$
$$^9Be=1.04\times10^{-12}J/nucleon$$
$$^{10}Be=1.04\times10^{-12}J/nucleon$$
C. $$^7Be=8.65\times10^{-13}J/nucleon$$
$$^9Be=1.00\times10^{-12}J/nucleon$$
$$^{10}Be=1.06\times10^{-12}J/nucleon$$
D. $$^7Be=9.12\times10^{-13}J/nucleon$$
$$^9Be=9.96\times10^{-13}J/nucleon$$
$$^{10}Be=1.00\times10^{-12}J/nucleon$$
E. $$^7Be=8.11\times10^{-13}J/nucleon$$
$$^9Be=1.08\times10^{-12}J/nucleon$$
$$^{10}Be=1.02\times10^{-12}J/nucleon$$
F. $$^7Be=8.40\times10^{-13}J/nucleon$$
$$^9Be=1.07\times10^{-12}J/nucleon$$
$$^{10}Be=1.00\times10^{-12}J/nucleon$$
G. $$^7Be=9.00\times10^{-13}J/nucleon$$
$$^9Be=8.77\times10^{-13}J/nucleon$$
$$^{10}Be=1.05\times10^{-12}J/nucleon$$
H. $$^7Be=8.70\times10^{-13}J/nucleon$$
$$^9Be=9.98\times10^{-13}J/nucleon$$
$$^{10}Be=1.10\times10^{-12}J/nucleon$$
I. $$^7Be=8.95\times10^{-13}J/nucleon$$
$$^9Be=1.03\times10^{-12}J/nucleon$$
$$^{10}Be=9.99\times10^{-13}J/nucleon$$
J. $$^7Be=7.50\times10^{-13}J/nucleon$$
$$^9Be=1.02\times10^{-12}J/nucleon$$
$$^{10}Be=1.08\times10^{-12}J/nucleon$$
Answer: B
The average concentration of carbon monoxide in air in an Ohio city in 2006 was 3.5 ppm. The number of CO molecules in 1.0 L of this air at a pressure of 755 torr and a temperature of 18 °C is().

Options:
A. $$7.5 × 10^{16} molecules\ CO$$
B. $$8.5 × 10^{16} molecules\ CO$$
C. $$8.8 × 10^{16} molecules\ CO$$
D. $$8.9 × 10^{16} molecules\ CO$$
E. $$6.8 × 10^{16} molecules\ CO$$
F. $$7.2 × 10^{16} molecules\ CO$$
G. $$9.5 × 10^{16} molecules\ CO$$
H. $$8.1 × 10^{16} molecules\ CO$$
I. $$9.9 × 10^{16} molecules\ CO$$
J. $$8.2 × 10^{16} molecules\ CO$$
Answer: C
The $K_{s p}$ value for PbS(s) is $8. 0 \times1 0^{-2 8}$ . By using this
value, determine the value of the standard reduc-tion potential for the reaction
$$
\mathrm{P b S} ( s ) \;+\; 2 \; \mathrm{e}^{-} \;\; \longrightarrow\; \mathrm{P b} ( s ) \;+\; \mathrm{S}^{2-} ( a q )
$$
The result is ____ .

Options:
A. $$-0.635V$$
B. $$-0.758V$$
C. $$-1.120V$$
D. $$-0.702V$$
E. $$-0.989V$$
F. $$-0.972V$$
G. $$-1.050V$$
H. $$-0.928V$$
I. $$-1.045V$$
J. $$-0.817V$$
Answer: H
The standard state of a substance used to be defined as 1 atm at the specified temperature. By how much does the current definition of standard chemical potential (at 1 bar differ from its former value at 298.15 K?

Options:
A. $$
1 8. 5 7 4 \mathrm{J m o l}^{-1}
$$
B. $$
2 5. 9 8 4 \mathrm{J m o l}^{-1}
$$
C. $$
2 9. 4 5 9 \mathrm{J m o l}^{-1}
$$
D. $$
2 3. 8 6 5 \mathrm{J m o l}^{-1}
$$
E. $$
3 5. 2 1 0 \mathrm{J m o l}^{-1}
$$
F. $$
3 7. 8 0 3 \mathrm{J m o l}^{-1}
$$
G. $$
3 4. 1 3 7 \mathrm{J m o l}^{-1}
$$
H. $$
4 0. 2 8 9 \mathrm{J m o l}^{-1}
$$
I. $$
4 1. 6 7 2 \mathrm{J m o l}^{-1}
$$
J. $$
3 2. 6 3 1 \mathrm{J m o l}^{-1}
$$
Answer: J
One of the most extensively studied reactions of industrial chemistry is the synthesis of ammonia, as its successful operation helps to govern the efficiency of the entire economy. The standard Gibbs energy of formation of $NH_{\mathrm{3}}$(g) is -16.5 kJ mol$^{-1}$ at 298 K. What is the reaction Gibbs energy when the partial pressure of the $\mathrm{N}_{2}, \mathrm{H}_{2},$ and$NH_{\mathrm{3}}$(treated as perfect gases) are 3.0 bar, 1.0 bar, and 4.0 bar, respectively? What is the spontaneous direction of the reaction in this case?

Options:
A. $$
{-2 5. 0 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
B. $$
{-5. 5 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
C. $$
{+2 0. 9 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
D. $$
{-1 1. 0 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
E. $$
{-1 4. 4 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
F. $$
{-9. 2 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
G. $$
{+3 0. 7 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
H. $$
{-1 0. 3 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
I. $$
{-1 8. 7 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
J. $$
{+1 5. 0 \mathrm{k J} \mathrm{m o l}^{-1}}
$$
Answer: E
The standard enthalpy of combustion of solid phenol, CgH5OH, is -3054 kJ mol-1 at 298 K and its standard molar entropy is 144.0 J $\mathrm{K}^{-1}$ mol1. Calculate the standard Gibbs energy of formation of phenol at 298 K.

Options:
A. $$
{-5 1. 0 \mathrm{~ k J ~ m o l^{-1}}}
$$
B. $$
{-4 7. 1 \mathrm{~ k J ~ m o l^{-1}}}
$$
C. $$
{-2 8. 7 \mathrm{~ k J ~ m o l^{-1}}}
$$
D. $$
{-7 0. 2 \mathrm{~ k J ~ m o l^{-1}}}
$$
E. $$
{-5 4. 3 \mathrm{~ k J ~ m o l^{-1}}}
$$
F. $$
{-4 9. 8 \mathrm{~ k J ~ m o l^{-1}}}
$$
G. $$
{-6 0. 4 \mathrm{~ k J ~ m o l^{-1}}}
$$
H. $$
{-4 3. 6 \mathrm{~ k J ~ m o l^{-1}}}
$$
I. $$
{-3 0. 2 \mathrm{~ k J ~ m o l^{-1}}}
$$
J. $$
{-3 5. 5 \mathrm{~ k J ~ m o l^{-1}}}
$$
Answer: F
Estimate the pH of an aqueous solution of sodium hydrogenoxalate. Under what conditions is this estimate reasonably reliable?

Options:
A. $$
{6. 8 0}
$$
B. $$
{2. 7 1}
$$
C. $$
{4. 1 9}
$$
D. $$
{3. 9 4}
$$
E. $$
{1. 4 8}
$$
F. $$
{8. 5 7}
$$
G. $$
{9. 0 5}
$$
H. $$
{0. 2 6}
$$
I. $$
{5. 6 3}
$$
J. $$
{7. 3 2}
$$
Answer: B
Derive an expression for the ratio of solubilities of AgCl at two different temperatures; assume that the standard enthalpy of solution of AgCl is independent of temperature in the range of interest.

Options:
A. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{\Delta H^{\ominus }} {R} \left( \frac{T} {T^{'}} \right)}
$$
B. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{\Delta_s H^{\ominus }} {2 R} \left( \frac{1} {T} - \frac{1} {T^{'}} \right)}
$$
C. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{\Delta H^{\ominus }} {2} \left( \frac{1} {R} - \frac{T} {T^{'}} \right)}
$$
D. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{-\Delta H^{\ominus }} {R} \left( \frac{1} {2T} \frac{1} {T^{'}} \right)}
$$
E. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{-\frac{\Delta H^{\ominus }} {R} \left( \frac{1} {T^{'}} - \frac{1} {T} \right)}
$$
F. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{\Delta H^{\ominus }} {R} \left( \frac{1} {T^{'}} \frac{1} {2T} \right)}
$$
G. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{\Delta H^{\ominus }} {4 R} \left( \frac{T} {1} \frac{T^{'}} {1} \right)}
$$
H. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{\Delta H^{\ominus }} {R} \left( \frac{T^{'}} {T} + \frac{T} {T^{'}} \right)}
$$
I. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{\Delta H^{\ominus }} {R} \left( \frac{1} {T} - \frac{1} {T^{'}} \right)}
$$
J. $$
\frac{S^{\prime}} {S}=\mathrm{e}^{\frac{2\Delta H^{\ominus }} {R} \left( \frac{1} {T} + \frac{1} {T^{'}} \right)}
$$
Answer: B
Tabulated thermodynamic data can be used to predict the standard potential of a cell even if it cannot be measured directly. The standard Gibbs energy of the reaction $K_{2} \mathrm{C r O}_{4} ( \mathrm{a c} )+2 \, \mathrm{A g ( s )}+2 \, \mathrm{F e C l}_{3} ( \mathrm{a c} ) \to\mathrm{A g}_{2} \mathrm{C r O}_{4} ( \mathrm{s} )+2 \, \mathrm{F e C l}_{2} ( \mathrm{a c} )$ + 2 KCl(aq) is -62.5 kJ mol' at 298 K.the standard potential of the $\mathrm{A g}_{2} \mathrm{C r O}_{4} / \mathrm{A g}, \mathrm{C r O}_{4}^{2-}$ couple is().

Options:
A. $$
{-1. 0 2 \mathrm{V}}
$$
B. $$
{+1. 3 5 \mathrm{V}}
$$
C. $$
{+0. 9 9 \mathrm{V}}
$$
D. $$
{+1. 2 3 \mathrm{V}}
$$
E. $$
{-0. 3 8 \mathrm{V}}
$$
F. $$
{-0. 5 4 \mathrm{V}}
$$
G. $$
{+0. 7 1 \mathrm{V}}
$$
H. $$
{-0. 8 6 \mathrm{V}}
$$
I. $$
{+0. 4 5 \mathrm{V}}
$$
J. $$
{+0. 6 7 \mathrm{V}}
$$
Answer: I
If ice homogeneously nucleates at $40^{°}C$, calculate the critical radius given values of $-3.1 \times 10^{8} J/m^2$ and $2 5 \times1 0^{-3} J / m^{2}$ respectively, for the latent heat of fusion and the surface free energy.

Options:
A. $$
4. 5 0 {\mathrm{~ n m}}
$$
B. $$
2. 0 5 {\mathrm{~ n m}}
$$
C. $$
2. 1 0 {\mathrm{~ n m}}
$$
D. $$
0. 5 0 {\mathrm{~ n m}}
$$
E. $$
1. 1 0 {\mathrm{~ n m}}
$$
F. $$
3. 0 5 {\mathrm{~ n m}}
$$
G. $$
1. 0 2 {\mathrm{~ n m}}
$$
H. $$
1. 0 5 {\mathrm{~ n m}}
$$
I. $$
1. 5 0 {\mathrm{~ n m}}
$$
Answer: E
The polarization P of a dielectric material positioned within a parallel-plate capacitor is to be $4.0$ x $1 0^{-6} C / m^{2}$.
What must be the dielectric constant if an electric field of $1 0^{5} V/m$ is applied?

Options:
A. $$
3. 6 2 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
B. $$
8. 9 1 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
C. $$
3. 0 6 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
D. $$
4. 8 9 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
E. $$
2. 7 3 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
F. $$
7. 8 9 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
G. $$
1. 5 3 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
H. $$
9. 2 4 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
I. $$
5. 1 4 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
J. $$
6. 0 5 \times1 0^{-6} \ \mathrm{C / m}^{2}
$$
Answer: D
How many collisions does a single Ar atom make in 1.0 s when the temperature is 25°C and the pressure is 10bar?

Options:
A. $$
8. 0 \times1 0^{1 0} \mathrm{s}^{-1}
$$
B. $$
9. 4 \times1 0^{1 0} \mathrm{s}^{-1}
$$
C. $$
5. 3 \times1 0^{1 0} \mathrm{s}^{-1}
$$
D. $$
3. 2 \times1 0^{1 0} \mathrm{s}^{-1}
$$
E. $$
6. 1 \times1 0^{1 0} \mathrm{s}^{-1}
$$
F. $$
2. 9 \times1 0^{1 0} \mathrm{s}^{-1}
$$
G. $$
1. 3 \times1 0^{1 0} \mathrm{s}^{-1}
$$
H. $$
7. 6 \times1 0^{1 0} \mathrm{s}^{-1}
$$
I. $$
4. 2 \times1 0^{1 0} \mathrm{s}^{-1}
$$
J. $$
3. 5 \times1 0^{1 0} \mathrm{s}^{-1}
$$
Answer: C
When 229 J of energy is supplied as heat to 3.00 mol Ar(g), the temperature of the sample increases by 2.55 K. Calculate the molar heat capacities at constant volume and constant pressure of the gas.

Options:
A. $$
3 2 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
3 7 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
B. $$
3 0 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
3 8 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
C. $$
2 8 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
4 0 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
D. $$
3 1 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
3 6 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
E. $$
2 9 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
3 9 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
F. $$
3 3 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
3 5 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
G. $$
2 6 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
4 1 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
H. $$
3 4 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
3 4 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
I. $$
3 5 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
3 3 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
J. $$
2 7 \mathrm{~ J ~ K^{-1} ~ m o l^{-1} ~}
$$
$$
4 2 \mathrm{~ J ~ K^{-1} ~ m o l^{-1}}
$$
Answer: B
How much energy (at constant temperature and pressure) must be supplied as heat to 10.0 g of chlorine gas (as $Cl_{\mathrm{2}}$) to produce a plasma (a gas of charged particles, in this case ions) composed of Cl$^{-}$ and Cl$^{+}$ions? The enthalpy of ionization of Cl(g) is +1257.5 kJ mol$^{-1}$ and its electron-gain enthalpy is -354.8 kJmol$^{-1}$.

Options:
A. $$
-2. 8 0 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 1 7 0 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 1 9 4 \ \mathrm{k J}
$$
B. $$
-2. 4 1 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 2 3 5 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 1 6 5 \ \mathrm{k J}
$$
C. $$
-3. 4 5 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 2 0 0 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 1 5 7 \ \mathrm{k J}
$$
D. $$
-1. 1 8 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 3 2 2 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 1 2 5 \ \mathrm{k J}
$$
E. $$
-2. 0 5 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 1 6 8 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 2 0 2 \ \mathrm{k J}
$$
F. $$
-2. 3 0 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 2 5 0 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 1 7 9 \ \mathrm{k J}
$$
G. $$
-3. 2 0 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 2 1 2 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 1 3 2 \ \mathrm{k J}
$$
H. $$
-2. 4 8 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 1 5 4 \mathrm{~ k J ~ m o l}^{-1}
$$
$$
+ 1 6 3 \ \mathrm{k J}
$$
I. $$
-2. 7 8 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 1 9 5 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 1 8 7 \ \mathrm{k J}
$$
J. $$
-2. 6 5 \mathrm{~ k J ~ m o l^{-1} ~}
$$
$$
+ 1 1 7 8 \mathrm{~ k J ~ m o l^{-1}}
$$
$$
+ 2 1 8 \ \mathrm{k J}
$$
Answer: H
Calculate the change in entropy when 100 g of water at 80°C is poured into 100 g of water at 10°C in an insulated vessel given that $C_{p, \mathrm{m}}=7 5. 5 \mathrm{J} \; \mathrm{K}^{-1} \; \mathrm{m o l}^{-1}$ .

Options:
A. $$
{1. 4 7 \ J K^{-1}}
$$
B. $$
{0. 8 2 \ J K^{-1}}
$$
C. $$
{9. 3 5 \ J K^{-1}}
$$
D. $$
{4. 2 9 \ J K^{-1}}
$$
E. $$
{6. 0 9 \ J K^{-1}}
$$
F. $$
{8. 4 7 \ J K^{-1}}
$$
G. $$
{5. 1 1 \ J K^{-1}}
$$
H. $$
{7. 1 5 \ J K^{- 1}}
$$
I. $$
{2. 6 3 \ J K^{-1}}
$$
J. $$
{3. 8 4 \ J K^{-1}}
$$
Answer: G
You have a sample of gas at -33 °C. You wish to in-crease the rms speed by 10.0%. To what temperature should the gas be heated?

Options:
A. 85°C
B. 70°C
C. 60°C
D. 45°C
E. 40°C
F. 50°C
G. 65°C
H. 55°C
I. 80°C
J. 75°C
Answer: F