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Alright, last one: What type of hybridization results in an angle of 180° between the bonds? a) $sp^2$ b) $sp$ c) $sp^3$ d) $sp^4$

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\[ \begin{array}{l} q=\text { Normind Gol } \\ =\frac{\text { GDP D } 0 \times \text { Real Gpr }}{100} \\ =\frac{84 \times 123000}{100}=103320 \end{array} \] \[ \text { - Real GDP }=\frac{\text { norminal }}{\text { GDP } D} \times 100 \] Assume base year is. 2011 EMO e base year is. 2011 $ \qquad $ GDP DEFLATOR C) Shows the

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n that $f(x) = x^2 - 7$ and $g(x) = 4x + 20$, find $(g - f)(-4)$, if it exists.

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Question 35 --- is a factor that contributes to the agentic state. 1 pts Tuning. Redefining the meaning of the situation Super-ego functions shift from an evaluation of the goodness or badness of the act to how well or poorly one is functioning in the authority system. All the above.

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This is a full-answer question. You must show all work. The correct answer alone will receive NO marks. Suppose that $f(x)$ and $g(x)$ are two differentiable functions such that $f(1) = 1$, $f(2) = 2$, $f'(1) = -2$, $f'(2) = -1$, $g(1) = -3$, $g(2) = -4$, $g'(1) = 4$, $g'(2) = 3$. Let $P(x) = f(x) \cdot e^{g(x)}$. Find $P'(1)$.

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A B C Br OH Br H-Br H2 OH OOR Pd/C D E LI Br R1 R2 OH LI acid workup RR2 estion 1 Tool E is the most challenging of the tools in the toolbox above. Before you can master it going backwards, try one going In the forward direction. Draw the major product of the following reaction. uestion 2 Question 3 Li acid workup Create OscerSketch Answer 1 Draw the correct reactant that you must use in the next step going backwards in the synthesis below. OH ? LI acid workup Create OscerSketch Answer 2 Propose a synthesis of the following compound using compounds containing four carbon atoms or less. Input the order of the reactions from the synthetic toolbox that you should use (input them going forward from starting material to product). For example, if you feel that they synthesis should start with reaction A, followed by B, and then C, your answer should be: ABC. HO Enter Your Answer:

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The Schrödinger equation for a particle on a ring is \frac{-\hbar^2}{2I} \frac{d^2 \psi}{d\phi^2} = E\psi where I is the moment of inertia (a) Show that the wavefunction $\Psi_m = Ne^{im\phi}$ is a solution (6 marks) (b) State the boundary condition.... (4 marks) (c) ....and use it to prove that the quantum number m can only take the values $m = 0, \pm 1, \pm 2, \pm 3.......$ (8 marks) (d) Thence derive an expression for the energies $E_m$ (6 marks) (e) .....and prove that the normalization constant is given by $N = \left(\frac{1}{2\pi}\right)^{1/2}$ (6 marks)

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Using Maclaurin series, determine to exactly what value the series converges. $\sum_{n=0}^{\infty} \frac{(\ln(3))^n}{n!}$ (Use symbolic notation and fractions where needed.) $\sum_{n=0}^{\infty} \frac{(\ln(3))^n}{n!} = $

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2 40 gmole/s of solution containing 20 mole % monoethanolamine, MEA, (HO-CH$_2$-CH$_2$-NH$_2$) and 80 mole % water is circulated from the bottom of the Regenerator to the top of the Absorber. The Regenerator is a pressure of 150 kPa and 115°C. The Absorber is at a pressure of 500 kPa and 40°C. The Heat Exchanger decreases the temperature of the circulating MEA solution from 115°C to 40°C. The pump increases the pressure from 150 kPa to 500 kPa. The MEA solution enters the Absorber at a position that is 35 m vertically above the bottom of the Regenerator. The velocity of the MEA solution in the bottom of the Regenerator is virtually negligible. The MEA solution enters the top of the Absorber column through a nominal 6 inch schedule 40 pipe (cross-sectional area = 0.2006 ft²). Determine the following: top of the Absorber (a) The change in potential energy [J/s] of the MEA solution between the bottom of the Regenerator and the (b) The change in kinetic energy [J/s] of the MEA solution between the bottom of the Regenerator and the top of the Absorber (c)The change in enthalpy [kJ/s] of the MEA solution between the bottom of the Regenerator and the top of the Absorber (d) The shaft work [kJ/kg] performed by the pump - Assume there is no friction. Component Molecular Weight Density [g/cm³] Specific heat Cp [J/(g °C)] MEA 61.09 1.016 2.549 Water 18.016 1.000 4.185 Solution density $\rho = \sum_k x_k \rho_k$ where $x_k$ = mass fraction of component K $\rho_k$ = density of component K

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Problem 1. Use the formal definition of limit to prove that $\lim_{x \to 1} 2x + 4 = 6$ (5 marks in K/U, 1 mark in COM)

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