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124
Fall Chem 2009
Name
Computer #
Types of Reactions
Perform all the reactions listed in the procedure and write your observations for each: write either the evidence for a reaction (such as bubbles, heat, precipitate, changed from blue to yellow etc.) or NR for No Reaction. For each reaction that was observed to occur, write out the formulas for reactants and resulting products, and balance each chemical equation. Use appropriate state labels (s for solid, aq for aqueous solution, etc). (NOTE: you may need to consult solubility rules in order to determine what the solid is that forms in some exchange reactions). For reactants that you observe to NOT REACT, write out just the reactant formulas and then write NR after the arrow to signify no reaction. For each reaction that occurred, classify the reaction type using the following initials: C for Composition, Dec for Decomposition, M for Metathesis, and SD for Single Displacement.
Reactants
Nickel(II) nitrate + sodium phosphate Magnesium + water Hydrochloric acid + sodium hydroxide Copper + silver nitrate Nickel(II) nitrate + sodium chloride
Evidence for a Reaction
Write your observations
Complete Balanced Chemical Equation
INCLUDE PHASE LABELS
Classify Reaction
Chem 124
Fall 2009
Nitric acid + potassium hydroxide
Reactants
Magnesium + oxygen gas Copper(II) sulfate + copper Hydrochloric acid + aluminum Calcium + water Zinc + copper(II) sulfate Copper(II) nitrate + sodium carbonate Iron(III) nitrate + sodium hydroxide
Evidence a for Reaction
Write your observations
Complete Balanced Chemical Equation
INCLUDE PHASE LABELS
Classify Reaction
Chem 124
Fall 2009
Chem 124 Experiment 1 Post-Lab Exercise:
Using the results of your completed experiment, solubility rules and the reactivity series found in your text, predict the products of the following reactions if there would be visible evidence of a reaction. If no reaction occurs, write NR after the arrow. 1. 2. 3. 4. 5. 6. 7.
Na3PO4 (aq) + KNO3 (aq) Fe(NO3)3 (aq) + CrCl3 (aq) + Al (s) + Cu (s) + Ag (s) + K3PO4 (aq)
Li2CO3 (aq)
H2O (l) HNO3 (aq) CuSO4 (aq) KOH (aq)
.
H2SO4 (aq) +
Chem 124
Fall 2009
8. Complete the following stoichiometry calculations. Note that the second one involves the Ideal Gas Law. See your textbook for review on both stoichiometry and gas laws. a. Hydrogen gas can be produced by reaction of many metals, like magnesium, with acids, like hydrochloric acid. How many grams of hydrogen gas could be produced from the complete reaction of 1.35 g magnesium with excess hydrochloric acid, according to the reaction below? Show all your work!
Mg (s) +
HCl(aq)
MgCl2(aq) + H2(g)
NOT BALANCED
b. Oxygen can be generated by the thermal decomposition of potassium chlorate, as described by the following equation: heat KClO3 (s) KCl (s) + O2 (g) NOT BALANCED Calculate the volume of oxygen gas (in L), measured at 26.5C and 768 torr, produced by the decomposition of 2.41 g of potassium chlorate. Show all your work!

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Cal Poly - CHEM - 124

Chem 124 Fall 2009Dr. RetsekName: Partner:Gurpal Bhoot Suyi OonnunnyComputer Number05The Heat of Sublimation of Dry IceInstructions: Each person is to submit an individual report after performing the experiment with your partner. This report is due

Cal Poly - CHEM - 124

Chem 124 Fall 2009Dr. RetsekName: Partner:Paul Bhoot Jay OonnunnyComputer # 05 Assigned Metal: MagnesiumThe Heat of Combustion of MetalsInstructions: Each person is to submit an individual report after performing the experiment with your partner. Th

Cal Poly - CHEM - 124

Tube Number 1 2 3 4 5Molarity of Nickel (II) Nitrate1.000 0.800Absorbance0.600 0.400 0.200Absorbance 0 0.922 0 0.810 f(x) = 9887.9x + 0.07 0 0.559 R = 0.99 0 0.295 0 0.170AbsorbanceLinear Regression for Absorbance0.000 0.0E+002.0E-054.0E-056.0E

Cal Poly - CPE - 229

C PE 229 FINAL STUDY GUIDELECTURE NOTES 1A computer is a device that sequentially executes a stored program. The two basic components of a CPU: the Control Unit and the D ata Path.LECTURE NOTES 3P ICOBLAZEThe line begins with ADDRESS as an assembler

Cal Poly - PHYS - 132

INTRODUCTIONThis course extends the ideas of Newton's laws and energy concepts to oscillations and waves, including sound; to heat and temperature, the thermal properties of materials and the principles of thermodynamics; and to light, geometrical optics

Cal Poly - PHYS - 132

HARMONIC O SCILLATIONSOscillatory motion is everywhere in nature. Any object which has both inertia and a restoring force will oscillate around an equilibrium position if displaced from that equilibrium. As we will see, the descriptions of essentially al

Cal Poly - PHYS - 132

List as of September 22, 2008. Suggested problems for Chapter 14 workbook problems are: 5, 6, 7, 8, 11, 12, 13, 17, 19, 21, 22, 25, and 26. Suggested problems for Chapter 20 workbook are: 4, 6, 7, 8, 9, 11, 12, 21, 22, 24 a-g, 25, 27, 29, 31, 32. Suggeste

Cal Poly - PHYS - 132

WAVE MOTION AND SOUNDThe general discussion of wave motion is important because the ideas of wave propagation are ubiquitous. In nearly all areas of science (and therefore real life) energy is transferred via the vibrations that make up waves. Examples o

Cal Poly - EE - 255

Chapter 1: Introduction to Machinery Principles1-1. A motors shaft is spinning at a speed of 3000 r/min. What is the shaft speed in radians per second? SOLUTION The speed in radians per second is = ( 3000 r/min ) 1-2. 1 min 2 rad = 314.2 rad/s 60 s 1

Cal Poly - EE - 255

H=F 425 A t = = 163 A t/m lc 2.60 mFrom the magnetization curve,B = 0.15 T and the total flux in the core isTOT = BA = (0.15 T ) (0.15 m ) ( 0.15 m ) = 0.0033 WbThe relative permeability of the core can be found from the reluctance as follows:R=FTO

Cal Poly - EE - 255

Chapter 2: Transformers2-1.The secondary winding of a transformer has a terminal voltage of vs ( t ) = 282.8 sin 377t V . The turns ratio of the transformer is 100:200 (a = 0.50). If the secondary current of the transformer is is ( t ) = 7.07 sin (377t

Cal Poly - EE - 255

(c) T he magnetization current is a higher percentage of the full-load current for the 50 Hz case than for the 60 Hz case. This is true because the peak flux is higher for the 50 Hz waveform, driving the core further into saturation.2-6.A 15-kVA 8000/23

Cal Poly - EE - 255

SOLUTION (a) T he transformer is connected Y-Y, so the primary and secondary phase voltages are the line voltages divided by 3 . The turns ratio of each autotransformer is given by VH N C + N SE 13.8 kV/ 3 = = VL NC 13.2 kV/ 3 13.2 N C + 13.2 N SE = 13.8

Cal Poly - EE - 255

Chapter 3: Introduction to Power Electronics3-1.Calculate the ripple factor of a three-phase half-wave rectifier circuit, both analytically and using MATLAB. SOLUTION A three-phase half-wave rectifier and its output voltage are shown below/65/62/3v

Cal Poly - EE - 255

(Note: T he above discussion assumes that transformer T3 is never in either state long enough for it to saturate.)3-8.Figure P3-3 shows a relaxation oscillator with the following parameters:R1 = variable C = 1.0 FVBO = 30 VR2 = 1500 VDC = 100 VI H =

Cal Poly - EE - 255

B1 =VM and 1 + 2 R 2C 2B2 = RC VM 1 + 2 R 2C 2Therefore, the forced solution to the equation is vC , f ( t ) = VM RC VM sin t cos t 1 + 2 R 2C 2 1 + 2 R 2C 2and the total solution is vC (t ) = vC ,n ( t ) + vC , f ( t )vC (t ) = Aet RC+ RC VM VM

Cal Poly - EE - 255

Chapter 4: AC Machinery Fundamentals4-1.The simple loop is rotating in a uniform magnetic field shown in Figure 4-1 has the following characteristics:B = 0.5 T to the rightl = 0.5 mr = 01 m . = 103 rad/s(a) Calculate the voltage etot ( t ) induced

Cal Poly - EE - 255

Chapter 5: Synchronous Generators5-1.At a location in Europe, it is necessary to supply 300 kW of 60-Hz power. The only power sources available operate at 50 Hz. It is decided to generate the power by means of a motor-generator set consisting of a synch

Cal Poly - EE - 255

The magnitude of E A is 12,040 V.(b) (c)T he torque angle of the generator at rated conditions is = 17.6. Ignoring R A , the maximum output power of the generator is given byPMAX =3 V E AXS=3 ( 7967 V )(12,040 V ) = 24.0 MW 12 The power at maximum

Cal Poly - EE - 255

5-21.Assume that the generator is connected to a 480-V infinite bus, and that its field current has been adjusted so that it is supplying rated power and power factor to the bus. You may ignore the armature resistance R A when answering the following que

Cal Poly - EE - 255

Chapter 6: Synchronous Motors6-1.A 480-V, 60 Hz, four-pole synchronous motor draws 50 A from the line at unity power factor and full load. Assuming that the motor is lossless, answer the following questions: (a) What is the output torque of this motor?

Cal Poly - EE - 255

I A2 = 5036.87 AThe internal generated voltage required to produce this current would beE A2 = V R AI A2 jX S I A2E A2 = 1200 V j (0.8 )(5036.87 A )E A2 = 147.5 12.5 VThe internal generated voltage E A is directly proportional to the field flux, and

Cal Poly - EE - 255

Chapter 7: Induction Motors7-1.A dc test is performed on a 460-V -connected 100-hp induction motor. If VDC = 24 V and I DC = 80 A, what is the stator resistance R1 ? Why is this so? SOLUTION If this motors armature is connected in delta, then there will

Cal Poly - EE - 255

IA+R10.075 jX1j0.17 jX2j0.17 R20.065 V-j7.2 jXM1 s R2 s1.56 (a)T he easiest way to find the line current (or armature current) is to get the equivalent impedance Z Fof the rotor circuit in parallel with jX M , and then calculate the curr

Cal Poly - EE - 255

(d) T o determine the starting code letter, we must find the locked-rotor kVA per horsepower, which is equivalent to finding the starting kVA per horsepower. The easiest way to find the line current (or armature current) at starting is to get the equivale

Cal Poly - EE - 255

Chapter 8: DC Machinery Fundamentals8-1.The following information is given about the simple rotating loop shown in Figure 8-6:B = 0.8 TV B = 24 V R = 0.4 l = 0.5 mr = 0.125 m = 250 rad/s (a) Is this machine operating as a motor or a generator? Expl

Cal Poly - EE - 255

Chapter 9: DC Motors and GeneratorsProblems 9-1 to 9-12 refer to the following dc motor: Prated = 15 hp I L ,rated = 55 AVT = 240 V nrated = 1200 r/min RA = 0.40 RS = 0.04 N F = 2700 turns per pole N SE = 27 turns per pole RF = 100 Radj = 100 to 400 R

Cal Poly - EE - 255

I F* = I F N SE 27 turns I A = 0.873 A (40 A ) = 0.473 A NF 2700 turnsFrom Figure P9-1, this field current would produce an internal generated voltage E Ao of 197 V at a speed no of 1200 r/min. Therefore, n= EA E Ao no = 227.4 V (1200 r/min ) = 1385 r/m

Cal Poly - EE - 255

RF = 200 nm = 1200 r/min Radj = 0 to 300 , currently set to 120 This motor has compensating windings and interpoles. The magnetization curve for this motor at 1200 r/min is shown in Figure P9-6.Note:An electronic version of this magnetization curve can

Cal Poly - EE - 255

n=EA 230 V no = (1800 r/min ) = 1711 r/min E Ao 242 V9-22.The magnetization curve for a separately excited dc generator is shown in Figure P9-7. The generator is rated at 6 kW, 120 V, 50 A, and 1800 r/min and is shown in Figure P9-8. Its field circuit

Cal Poly - EE - 255

% the line "Ea_ar - Vt - i_a*r_a" goes negative. i_a = 0:1:37; for jj = 1:length(i_a) % Calculate the equivalent field current due to armature % reaction. i_ar = (i_a(jj) / 50) * 200 / n_f; % Calculate the Ea values modified by armature reaction Ea_ar = i

Cal Poly - EE - 255

plot(i_l,v_t,'b-','LineWidth',2.0); xlabel('\bf\itI_cfw_L \rm\bf(A)'); ylabel('\bf\itV_cfw_T \rm\bf(V)'); string = ['\bfTerminal Characteristic of a Cumulatively ' . 'Compounded DC Generator']; title (string); hold off; axis([ 0 50 0 120]); grid on;The r

Cal Poly - EE - 255

The resulting torque-speed characteristic is shown below:10-5.A 220-V, 1.5-hp 50-Hz, two-pole, capacitor-start induction motor has the following main-winding impedances: R1 = 1.40 X 1 = 2.01 X M = 105 R2 = 1.50 X 2 = 2.01 At a slip of 0.05, the motors r

Cal Poly - EE - 255

The resulting torque-speed characteristic is shown below:10-5.A 220-V, 1.5-hp 50-Hz, two-pole, capacitor-start induction motor has the following main-winding impedances: R1 = 1.40 X 1 = 2.01 X M = 105 R2 = 1.50 X 2 = 2.01 At a slip of 0.05, the motors r

Cal Poly - EE - 255

Appendix C: Salient Pole Theory of Synchronous MachinesC-1.A 480-V 200-kVA 0.8-PF-lagging 60-Hz four-pole Y-connected synchronous generator has a direct-axis reactance of 0.25 , a quadrature-axis reactance of 0.18 , and an armature resistance of 0.03 .

Cal Poly - EE - 255

Thus the line voltage lags the corresponding phase voltage by 30. The phasor diagram for this connection is shown below.Vbc VbnVanVcnVabA-5.Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure P23.SO

Cal Poly - EE - 255

cha65239_ch01.qxd10/23/20039:22 AMPage 6262ELECTRIC MACHINERY FUNDAMENTALS0.010 0.005 (Wb)012345678t (ms) 0.005 0.010FIGURE P112 Plot of flux as a function of time for Problem 116.4 cm i N=? N turns 4 cm Depth = 4 cmlr = 4 cm lg = 0.0

Cal Poly - EE - 255

Cal Poly - EE - 255

Electrical Engineering Department Cal Poly State University EE 255: Electric Energy Conversion D. Dolan- Spring 2009 Office Hours Room 20-205, Tel: 756-2495 mailto: dsdolan@calpoly.edu Electric Machinery Fundamentals, by Stephen J. Chapman, 4th Edition, M

Cal Poly - EE - 255

Cal Poly - EE - 255

UC Davis - PLS - 144

Trees and Forests TreesLecture 6: Aboveground Interactions: Lecture light, photosynthesis and carbon allocation allocationOutline OutlineI. Photosynthesis introduction II. Light -3 types of light; PAR; sunflecks; LAI and the understory light environmen

Academy of Design Chicago - COMMUNICAR - IEE 126

IEEE International Conference on Application-specific System, Architectures and Processors (ASAP'08). July 2008Configurable and Scalable High Throughput Turbo Decoder Architecture for Multiple 4G Wireless StandardsYang Sun , Yuming Zhu , Manish Goel , a

Cy-Fair College - PH - 1511

Chapter 6: Friction and Projectile MotionEquilibrium We will get into Newtons laws in great detail in chapter 7, but will apply his first law at this time. To paraphrase the first law: If an object is not accelerating, the vector sum of the forces acting

Cy-Fair College - PH - 1511

Chapter 2: Introduction to Kinematics The Constant Velocity Particle Model (CVP)Kinematics is the study of motion. In this chapter we will study constant velocity in one dimension. That means the object is moving with the same speed in a straight line. T

Cy-Fair College - SC - 1515

Chapter 11: WAVES AND SOUNDPart 1: IntroductionWave motion is the transfer of energy from a source to a distant receiver without the transfer of matter between the two points. Waves are an oscillation that moves through a medium. In our work we will be

Cy-Fair College - SC - 1515

Unit 9: Momentum, Impulse and Conservation of MomentumPart 1: Impulse and momentumIn this chapter we will be looking at further applications of Newtons 2nd and 3rd Laws. First a physics definition of the term momentum. Early scientists thought that when

SUNY Stony Brook - PHY - 02

Phy125MidtermI10/12/07Showallofyourworktoreceivefullcredit 1.Acannonballisfiredhorizontallyfromthetopofahill500mabovealevel valley.Theinitialvelocityofthecannonballis200m/s.Find: hittingthegroundinthevalley. a.Thetimetheballisintheairbefore b.Thehorizo

Central State University - CS - 5530

CDA5530: Performance Models of Computers and Networks (Fall 2009) Homework 1 Solution1. Bill and George go target shooting together. Both shoot at a target at the same time. Suppose Bill hits the target with prob. 0.7, George, independently, hits the tar

UCSB - BL ST - blst1

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 1.m = 5.4 kg,a = 600 mm = 0.6 m,b = 300 mm = 0.3 mTotal length of rod.l = 2a + 2b = 1.8 mm 5.4 = = 3 kg/m l 1.8Mass per unit length. =Use principal axes x, z as show

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 2.Use principal axes y, z as shown. y = cos 45, z = sin 45 x = 0I x =1 2 ma , 3I y =1 ma 2 12I z = I x + I y =5 ma 2 12H A = I x xi + I y y j + I z zk 1 5 = 0 +

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 3.m=3.6 = 0.1118 lb s 2 /ft 32.2l = 24 in. = 2 ft =(1200 )( 2 )60= 125.664 rad/sUse principal axes x, y as shown. x = - sin 20 = -42.980 rad/s y = cos 20 = 118.085

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 4.Use principal axes x, y ,z as shown. i = cos i + sin j j = - sin i + cos j i = cos i - sin j j = sin i + cos j = i = cos i - sin j Pr incipal moments of inertia: I x = 1

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 5.Body diagonal:=d =a 2 + ( 2a ) + a 2 =26ad( -ai + 2aj - ak ) = -6i+2 j- k 6 6Ix =Iy = Iz = (a)1 5 2 2 m ( 2a ) + a 2 = 12 ma 121 2 1 m a + a 2 = ma 2 6 12 1

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 6.Body diagonal=d =a 2 + ( 2a ) + a 2 =26ad( -ai + 2aj - ak ) = -6i+2 j- k 6 6Total area = 2 a 2 + 2a 2 + 2a 2 = 10a 2()For each square plate:m =1 m 10Ix =

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 7. = 2 j + 1k = ( 4 rad/s ) j + (12 rad/s ) kFor axes x, y, z parallel to x, y, z with origin at A,I x =1 2 1 2 mr = ( 8 )( 0.100 ) = 0.02 kg m 2 4 4I y = I x = 0.02 kg

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 8. = 1j + 2i = (8 rad/s ) i + (16 rad/s ) jFor axes x, y, z parallel to x, y, z with origin at A,I x = 1 2 1 2 mr = ( 6 )( 0.160 ) = 0.0384 kg m 2 4 4 I y = I x + I z = 0

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 9.m=60 = 1.86335 lb s 2 /ft, 32.2k x = 2.4 in. = 0.2 ft,2k y = 10 in. = 0.83333 ft.2 I x = mk x = (1.86335 )( 0.2 ) = 0.074534 lb s 2 ft2 I y = I z = mk y = (1.86335

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 10.m=60 = 1.86335 lb s 2 /ft, 32.2 11 rG/ A = ft i 12 v = v cos i - v sin j H A = HG + rG/ A mv 11 = 0.640i - 0.018j + i (1950cos 5i - 1950sin 5 j) 12 11 = 0.640i - 0.0

Purdue - CE - CE297

COSMOS: Complete Online Solutions Manual Organization SystemChapter 18, Solution 11.Using Equation (18.11)H O = r mv + H G= ( Li ) ( mr1k ) + = -mrL1j + 1 2 r mr 1 i - j 2 2L 1 2 1 r3 mr 1i - m 1j 2 4 LHO =1 2 1 mr 1i - m L2 + r 2 ( r1/L ) j 2 4 w