Extra Credit 49

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Q2.5

Given the molar mass of a gas, solve for the density variable using the ideal gas equation

$$PV=nRT $$

Q2.65:

Given the speeds of 8 molecules are $3.0\dfrac{cm}{s}$, $3.0\dfrac{cm}{s}$, $3.1\dfrac{cm}{s}$, $3.4\dfrac{cm}{s}$, $3.7\dfrac{cm}{s}$, $3.8\dfrac{cm}{s}$, $4.1\dfrac{cm}{s}$, and $4.4\dfrac{cm}{s}$^, find the following:

Average speed
Root mean square speed

Q2.95

Lisa is doing an experiment about the kinetics of atmospheric gases. She chooses $CO_2$, $O_2$, and $N_2$. For her experiment, it is crucial to find the gas with the lowest average speed. Calculate the average speeds of $CO_2$, $O_2$, and $N_2$ at $T = 298K$. Which gas has the lowest average speed?

Given information:

$CO_2$ = $44.01\dfrac{g}{mol}$

$O_2$ = $32.00\dfrac{g}{mol}$

$N_2$= $28.00\dfrac{g}{mol}$

Q9.4

Given a first order reaction

$\mathrm{A} \xrightarrow{} \mathrm{B + C}$

If half of the initial concentration of $A$ is consumed in 115 seconds, what percent of $A$ will be remaining after 300 seconds?

Q9.20

What is the difference between steady state equilibrium and pre-equilibrium? Given the following equation, apply the steady state approximation to the $H_2$ and $I_2$ atoms.

\[\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightarrow 2 HI_{\large{(g)}}}\]

$\ce{I_{2\large{(g)}}} \xrightarrow{\Large{k_1}} \ce{2 I_{\large{(g)}}}$
$\ce{2 I_{\large{(g)}}} \xrightarrow{\Large{k_2}} \ce{I_{2\large{(g)}}}$
$\ce{H_{2\large{(g)}} + 2 I_{\large{(g)}}} \xrightarrow{\Large{k_3}} \ce{2 HI_{\large{(g)}}}$

Q10.12

Given an enzyme with $K_m = 1.5\times{10}^{-5}$ and a $V_{max} = 66\dfrac{μM}{min}$, find a value of initial velocity $v_0$ if $[S] = 4.1\times{10}^{-4}M$ and $[I] = 5.38\times{10}^{-4} M$ for a

Competitive inhibitor
Noncompetitive inhibitor
Uncompetitive inhibitor

Hint: $K_I = 0.9\times{10}^{-5} M$ for all three inhibitors.

Q11.10

If an electron has an uncertainty in its velocity of $1\times{10}^6\dfrac{m}{s}$, what is the uncertainty in its position?

Q12.1

Order the following bonds from weakest to strongest. Which of the bonds would be the shortest? The longest?

$CO_2$
$CH_3CONH_2$
$CH_3OH$
$CH_2O$

Q13.5

Name all of the possible intramolecular and intermolecular forces that could occur between the following molecules:

$CO_2$
$CH_2O$
$FeCl_2$
$NaF$

Q14.8

Regarding the Doppler effect and collision broadening when analyzing visible and UV spectra, what is one possible way for us to make the spectra more precise? Why does it work?

S2.5:

Starting from the Ideal Gas law, we can say that $n =\dfrac{mass\:of\:gas\:in\:kg}{molecular\:weight\:in\:kg/mol} = \dfrac{m}{M}$. We plug that in the ideal gas law to get

\[ PV=\dfrac{m}{M}RT \]

Because density $d = \dfrac{m}{V}$, we can Isolate $m$ and $V$

\[ PV=\dfrac{m}{M}RT \]

\[ \dfrac{PM}{RT}=\dfrac{m}{V} \]

\[ \dfrac{PM}{RT}= d \]

And we have isolated density.

S2.65

a: Average speed:

to find the average speed, we simply take the averages of the numbers

\[ =\dfrac{3.0 + 3.0 + 3.1 + 3.4 + 3.7 + 3.8 + 4.1 + 4.4}{8} \]

\[= 3.5625\]

\[= 3.6\dfrac{m}{s}\]

b: Root mean squared speed:

\[= \sqrt{\dfrac{3.0^2 + 3.0^2 + 3.1^2 + 3.4^2 + 3.7^2 + 3.8^2 + 4.1^2 + 4.4^2}{8}} \]

\[= 3.5964\]

\[= 3.6\dfrac{m}{s}\]

S2.95

Average speed of $CO_2 = \sqrt{\dfrac{8RT}{44.01π}}$

\[= \sqrt{\dfrac{8(8.314)(298)}{44.01π}}\]

\[= 11.97\dfrac{m}{s}\]

Average speed of $O_2 = \sqrt{\dfrac{8RT}{32.00π}}$

\[= \sqrt{\dfrac{8(8.314)(298)}{32.00π}}\]

\[= 14.04\dfrac{m}{s}\]

Average speed of $N_2 = \sqrt{\dfrac{8RT}{28.00π}}$

\[= \sqrt{\dfrac{8(8.314)(298)}{28.00π}}\]

\[= 15.01\dfrac{m}{s}\]

The gas with the slowest average speed is $CO_2$. The gas with the fastest average speed is $N_2$

S9.4

We find $k$ from the half-life formula for first-order reactions

\[k = \dfrac{ln 2}{115\:s}\]

\[k = 6.03\times{10}^{-3}\:s^{-1}\]

After 300s, the percent of $A$ that will remain is

\[\dfrac{[A]}{[A]_o} = e^{-kt}\]

\[\dfrac{[A]}{[A]_o} = e^{-(6.03\times{10}^{-3})(300s)}\]

\[= 0.1638\]

\[= 16.4 \% \:of\:A\:will\:be\:remaining\]

S9.20

Steady state approximation describes a system that uses up the enzyme-complex system as soon as it is made. It has a slow first step and a fast second step. Pre-equilibrium describes the opposite system that has a fast first step and a slow second step.

$\dfrac{d[I_2]}{dt} = -k_1[I_2] + k_2[I]^2$

$\dfrac{d[H_2]}{dt} = -k_3[H_2][I]^2$

S10.12

a. Competitive inhibition:

\[ v_o=\dfrac{V_{max} [S]}{K_M\times{(1 + \dfrac{[I]}{K_I})} +[S]} \]

\[ v_o=\dfrac{(66) (4.1\times{10}^{-4})}{1.5\times{10}^{-5}\times{(1 + \dfrac{5.38\times{10}^{-4}}{0.9\times{10}^{-5}})} +4.1\times{10}^{-4}} \]

\[ v_o= 20.47\:\dfrac{\mu M}{min}\]

b. Noncompetitive inhibition:

\[ v_o=\dfrac{\dfrac{V_{max}}{(1 + \dfrac{[I]}{K_I})}[S]}{K_M + [S]} \]

\[ v_o=\dfrac{\dfrac{66}{(1 + \dfrac{5.38\times{10}^{-4}}{0.9\times{10}^{-5}})}4.1\times{10}^{-4}}{1.5\times{10}^{-5} + 4.1\times{10}^{-4}} \]

\[ v_o= 1.048\:\dfrac{\mu M}{min}\]

c. Uncompetitive inhibitor

\[ v_o=\dfrac{\dfrac{V_{max}}{(1 + \dfrac{[I]}{K_I})}[S]}{\dfrac{K_M}{(1 + \dfrac{[I]}{K_I})}+[S]} \]

\[ v_o=\dfrac{\dfrac{66}{(1 + \dfrac{5.38\times{10}^{-4}}{0.9\times{10}^{-5}})}4.1\times{10}^{-4}}{\dfrac{1.5\times{10}^{-5}}{(1 + \dfrac{5.38\times{10}^{-4}}{0.9\times{10}^{-5}})}+4.1\times{10}^{-4}} \]

\[ v_o= 1.085\:\dfrac{\mu M}{min}\]

S11.10

Starting with the Heisenberg uncertainty principle:

\[\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi}\]

we know that $p = mv$. Substituting that into the equation will give us the uncertainty in the position

\[m\Delta{v}\Delta{x} \ge \dfrac{h}{4\pi}\]

\[(9.019\times{10}^{-31})(1\times{10}^{6})\Delta{x} \ge \dfrac{h}{4\pi}\]

\[\Delta{x} \ge \dfrac{6.626\times 10^{-34}}{4\pi \times (9.019\times{10}^{-31})\times(1\times{10}^{6})}\]

\[\Delta{x} \ge 5.85\times{10}^{-11}m\]

S12.1

The order from weakest to strongest bonds is

\[CH_3OH < CH_2O < CH_3CONH_2 < CO_2\]

A shorter bond indicates that the bond is stronger and his greater in energy. A double bond is shorter than a single bond and hence has greater bond energy.

The molecule with the shortest bond is $CO_2$. The one with the longest bond is $CH_3OH$

S13.5

First, all molecules have london dispersion forces (an intermolecular force).

$CO_2$ - covalent bonds
$CH_2O$ - covalent bonds, dipole-dipole forces, hydrogen bonding
$FeCl_2$ - ionic bond, coulombic forces
$NaF$ - ionic bond, coulombic forces

S14.8

Lowering the temperature would improve the spectra resolution. By lowering the temperature, the speeds of the molecular are reduced, which in turn reduces the effects of the Doppler and collisional broadening.