COMP115 Lecture Notes - Lecture 5: Ellipse

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raspberrycaribou903

28 Aug 2018

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Break; exits the loop that was before it. More complex structures there is absolutely no restriction on how the two control structures can be used (n/2)*n. 1. i=1 -> k loop executed 1 time i=2 -> k loop executed 2 time i=3 -> k loop executed 3 time i=10 -> k loop executed 10 time. State(cid:373)e(cid:374)t executes a total of (cid:1005) + (cid:1006) + (cid:1007) (cid:1005)(cid:1004) = (cid:1005)(cid:1004)*(cid:1005)(cid:1005)/(cid:1006) = (cid:1009)(cid:1009) (cid:1005)+(cid:1006)+(cid:1007) +(cid:374) = (cid:374) *(cid:894)(cid:374)+(cid:1005)(cid:895)/(cid:1006) I = 1 -> k = (cid:1005)(cid:1004), (cid:1013), (cid:1012), (cid:1006) (cid:894)(cid:1013) ti(cid:373)es(cid:895) I = 2 -> k = (cid:1005)(cid:1004), (cid:1013), (cid:1012), (cid:1007) (cid:894)(cid:1012) ti(cid:373)es(cid:895) I = 5 -> k = (cid:1005)(cid:1004), (cid:1013), (cid:1012), (cid:1010) (cid:894)(cid:1009) ti(cid:373)es(cid:895) Statement executed 9 (i=1) +8 (i=2) +7 (i=3) + 6 (i=4) +5 (i=5) Making loop circles size (300, 300); background (255); for(float x= width/8; x < width; x+=width/4) { println ("x = "+x); for(float y = height/8; y < height; y+=height/4) { ellipse (x, y, width/4, height/4);

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Please show your work and steps. I have the answers. I just want to make sure I work it out correctly. Thanks!

- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P.	Step 1.
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P.	Step 2.
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- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P.	Step 8.
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P.	Step 9.
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P.	Step 10.
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- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P.	Step 12.
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A.	Don appropriate personal protective equipment (PPE).
B.	Aseptically transfer 1 mL from dilution tube 2 to dilution tube 3 and mix well. This is a 10^-5 cumulative dilution.
C.	Aseptically transfer 1.0 mL from dilution tube 4 to dilution tube 5 and mix well. This is a 10^-7 cumulative dilution.
D.	Aseptically transfer 0.1 mL from dilution tube 1 to dilution tube 2 and mix well. The cumulative dilution of the culture is 10^-4.
E.	Allow the inocula to soak into the plates before proceeding.
F.	Aseptically transfer 0.1 mL from dilution tube 2 to plate A₁. Using the spread plate technique, disperse the sample evenly over the entire surface of the agar.
G.	Aseptically transfer 1 mL from dilution tube 3 to dilution tube 4 and mix well. This is a 10^-6 cumulative dilution.
H.	Obtain eight plates, organize them int four pairs and label them A₁, A₂, B₁, B₂, etc.
I.	Keep accurate records in your lab notebook. Clean up the lab benches, doff PPE, and wash hands.
J.	Repeate the procedure with plate A₂ and label both plates "10^-5mL of original sample," or simply "10^-5" Notice that transferring 0.1 mL counts as a dilution!! The tube was a 10^-4 dilution, but the plate is a 10^-5 dilution!!
K.	Invert the plates so they are "agar up" to prevent condensation from ruining the plates.
L.	Repeat the aseptic transfer, spread plate technique, and label schematic for dilutions tubes 3, 4, and 5, and plates B, C, and D.
M.	Incubate the plates at 35â for 24 to 48 hours.
N.	Obtain five dilution tubes and label them 1â5, respectively. Make sure they remain covered until needed.
O.	Aseptically add 9.9 mL sterile water to dilution tubes 1 & 2. Cover when finished. Aseptically add 9.0 mL sterile water to dilution tubes 3, 4, and 5. Cover when finished.
P.	Mix the broth culture an aseptically transfer 0.1 mL to dilution tube 1. Mix dilution tube 1 well. This is a 10^-2 dilution

1) What are the possible orbitals for n=3?

s, p, d
s, p, d, f
s
s, p

2) What value of ℓ is represented by a d orbital?

3) How many orbitals are contained in the third principal level (n=3) of a given atom?

4) Which of the following statements is TRUE?

The principal quantum number (n) describes the shape of an orbital.
The angular momentum quantum number (ℓ) describes the size and energy associated with an orbital
The magnetic quantum number (mℓ) describes the orientation of the orbital.
An orbital is a path that an electron follows during its movement in an atom
All of the above are true

5) Which of the following transitions represents the emission of a photon with the largest energy?

n=2 to n=1
n=3 to n=1
n=6 to n=4
n=1 to n=4
n=2 to n=4

6) Choose the transition (in a hydrogen atom) below that represents the absorption of the shortest wavelength photon.

n=1 to n=2
n=2 to n=3
n=4 to n=5
n=6 to n=3
n=3 to n=1

7) Each of the following sets of quantum numbers is supposed to specify an orbital. Choose the one set of quantum numbers that does not contain an error.

n=2, ℓ =2, mℓ= +1
n=2, ℓ=2, mℓ= 0
n=3, ℓ=2, mℓ = -3
n=4, ℓ=3, mℓ= -2
n=4, ℓ=2, mℓ= +4

8) The element that corresponds to the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ is:

titanium
vanadium
chromium
manganese
iron

9) Give the ground-state electron configuration for I.

[Kr] 5s² 4d¹⁰ 5p⁶
[Kr] 5s² 4d¹⁰ 5p⁵
[Kr] 4d¹⁰ 5p⁶
[Kr] 5s² 5p⁶
[Kr] 5s² 5d¹⁰ 5p⁶

10) Place the following elements in order of increasing atomic radius

P Ba Cl

Ba < P < Cl
P < Cl < Ba
Cl < P < Ba
Cl < Ba < P
Ba < Cl < P

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COMP115 Lecture Notes - Lecture 5: Ellipse

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