5

Determining the Heat of ReactionExperiment 3 Determining Heat of Reaction:Ice Calorimeter Pre-Lab Readings:. Introduction In this experiment,you will determine the ...

Question

Determining the Heat of ReactionExperiment 3 Determining Heat of Reaction:Ice Calorimeter Pre-Lab Readings:. Introduction In this experiment,you will determine the amount of heat evolved per mole ofmagnesium metal when it reacts with an acid according to equationE1 below: E1: Mg(s) + 2 H+ (aq) → Mg2+(aq) + H2(g) Apply! Why mighta chemist want to know the amount of heat given off by a reaction?______________________________ _ ______________________________ _One method to accomplish this is with

Determining the Heat of Reaction Experiment 3 Determining Heat of Reaction: Ice Calorimeter Pre-Lab Readings: . Introduction In this experiment, you will determine the amount of heat evolved per mole of magnesium metal when it reacts with an acid according to equation E1 below: E1: Mg(s) + 2 H+ (aq) → Mg2+(aq) + H2(g) Apply! Why might a chemist want to know the amount of heat given off by a reaction? ______________________________ _ ______________________________ _ One method to accomplish this is with the use of an ice calorimeter. The simple setup you will use is shown in Figure 1. The reaction takes place in a small test tube suspended in a sealed beaker filled with ice and water. As heat given off by an exothermic reaction is absorbed by the surroundings, the ice melts. Since ice and liquid water have very different densities, melting results in a change in volume of the ice-water mixture. The heat required to melt a given mass of ice is shown in equation E2: E2: where q is heat absorbed by ice , w is mass of ice melted (in grams), and Hf is the latent heat of fusion of ice (333.5 J/g) This means that for every gram of ice melted, 333.5 J of heat is absorbed by the ice and, therefore, this same amount of heat must be given off by the reaction. Since it would be very difficult to directly measure the weight of ice that gets melted, another approach is in order. You can measure the change in volume (caused by the melting ice): E3: where Vi is the volume of ice, and Vw is the volume of liquid water (after melting w grams of ice) Remember that you can relate mass and volume by density and derive equation E4: E4: where Di is the density of ice at 0°C (0.917 g/mL), and Dw is the density of water at 0°C (1.000 g/mL) Then rearrange E4 to solve for w (which is the unknown term in E2 that you need): E5: q  w  H f V  Vi Vw i Dw w D w V   ( ) w i i w D D D D w V Substitute E5 into E2 to get: E6: This equation means that you can calculate the heat change of a reaction easily by measuring the change in volume of the icewater mixture in your calorimeter. Conceptualize! Since a sample of ice occupies (more / less) space than the same weight of liquid water, the final volume measured will be (more / less) than the initial volume and ΔV will be (positive / negative). This will also result in a (positive / negative) value for q, which is expected since heat is (released / absorbed) by an (exothermic / endothermic) reaction. The heat per mole (ΔH) of Mg reacted is: E7: where y is the number of moles of Mg reacted in the experiment The accuracy of any calorimeter depends on being able to minimize heat exchange between all other sources except for the reaction in question. To do this, you will insulate your ice calorimeter as much as possible from the heat of the room with paper towels, water, and ice. After your setup has been assembled (without any air bubbles inside the beaker), and acid has been added to the test tube, the entire apparatus must come to equilibrium and sit for 10 minutes before you start to take readings. Method (to be done in pairs) The ice calorimeter setup is shown in Figure 1. A: 1.000 mL pipette B: Tygon tubing with screwpinch clamp C: loose stopper on test tube D: stopper assembly E: 200 mL beaker with ice/water F: 1000 mL beaker with ice/water and paper towel Figure 1: Ice calorimeter setup Crumple some paper towels and pack them in the bottom of a 1000 mL beaker. Thoroughly dry the inside of the test tube that is a part of the stopper assembly using a Kim Wipe wrapped around a pen. Add 5mL of 2.0 M HCl and stopper the test tube tightly. Fill the 200 mL beaker 3/4 full of crushed ice, then add cold water until it overflows. Gently stir the 200 mL beaker with a glass rod to allow as many air bubbles to escape as possible. Slowly push the stopper assembly into the 200 mL beaker so that the ice is pushed down and water spills out onto the counter top. f w i i w H D D D D q V      ( ) J mol y q H  / A B C D E F 27 Do not trap any air bubbles under the stopper! With the clamp open, push the stopper assembly firmly into the 200 mL beaker until water rises into the tubing and remains at a fairly constant level. If there are large air bubbles trapped under the stopper, or the test tube is not completely surrounded by crushed ice, remove the stopper assembly, add more crushed ice, refill the beaker with cold water, and repeat the above procedure until the water level in the tube remains relatively constant. If the water level remains relatively constant, place the 200 mL beaker inside the 1000 mL beaker so that the top of the stopper assembly is below the lip of the large beaker and pack as many ice cubes around the sides as you can fit. Add water until the top of the stopper assembly is submerged and covered with ice, but the mouth of the test tube remains above water. Allow the system to come to equilibrium for about 5 minutes. Use a distilled water bottle to force water (and no air bubbles!) into the system until the pipette is filled. Have your partner clamp the tubing closed. Wait 5 more minutes for the water level to settle. Accurately weigh a 0.06-0.08 g sample of magnesium metal turnings. Avoid touching the metal with your fingers! Loosen the small test tube stopper so that you can open and close it easily. Adjust the meniscus so that it is near the top of the pipette but not past the calibration marks. Take all pipette readings to three significant figures! The first pipette reading should be the largest. Notice! What is the total volume of the pipette? _________________________ Take a pipette reading at , and every 30 seconds thereafter for seven minutes, to measure the base rate of melting due to the room temperature environment. Despite your best efforts to insulate the ice calorimeter, you will observe this slight decrease in volume before ever starting an exothermic reaction, but the base rate should remain constant throughout the experiment and not interfere with your results. At the seven minute mark, take the pipette reading and then drop in the magnesium metal. Loosely replace the small stopper on the test tube. Hypothesize! Why should you not seal the test tube tightly with the small stopper? ______________________________ _ ______________________________ _ Continue to take pipette readings every 30 seconds. Four minutes after the magnesium is added, check to make sure that all of the metal made it into the acid. If there are any magnesium pieces stuck on the sides of the test tube, use a pen to push them down. Continue to take pipette readings until the rate of volume change returns to the initial base rate you observed (or for about 18 minutes of total timing). t  0 28 Rinse the test tube with lots of water and repeat the experiment using another sample of magnesium and 5 mL of 1.0 M H2SO4 . Details of the Experiment Your graph should end up looking roughly like the one shown in Figure 2 below: Figure 2: Sample graph of pipette volume (mL) vs. time (mins) The first 14 volume measurements you took were to determine the background melting rate of the ice, caused by a hopefully small but constant absorption of heat from the room. Since the temperature of the room is not fluctuating wildly, the changes in volume are expected to be constant and a line of best fit should be straight. This background melting rate should continue at the same rate even while the reaction is taking place. When the magnesium metal turnings are dropped into the acid, the heat produced is greatest at first when there is the most available metal to react. The rate of heat production then continues to decrease as the magnesium is used up. As the reaction proceeds and the reaction mixture is heated up, the total rate of melting is not constant and the line of best fit will be a smooth curve through the graphed points rather than a straight or zig-zag line. After the reaction is finished and the test tube has cooled back to the initial (nearly 0°C) conditions, the absorption of room heat again becomes the only reason the ice in the calorimeter is melting. If the rate of room heat melting has not changed during the experiment, a straight line of best fit through the last few points should have exactly the same slope as the first 14 points. Thus, any vertical distance between these two extrapolated parallel lines is the change in volume (ΔV) due to the reaction, as shown in Figure 2. This graphing method takes any effects from the room temperature environment out of the calculation for ΔV. If the initial and final melt rates are not the same (i.e. the straight lines are not parallel with each other) calculate the ΔV at the t = 12 minute mark. Thermodynamic Supplement Why do we use two symbols (ΔH and ΔE) to designate the heat of a reaction? Many chemical reactions produce a different amount of heat depending on whether they are done under constant pressure or constant volume conditions! ΔE is the heat change per mole of product produced under constant volume conditions, and is referred to as the internal energy change. ΔH is the heat change per mole of product produced under constant pressure conditions, and is referred to as the enthalpy change. The difference in magnitude between ΔE and ΔH depends on whether work (w = PΔV) is done during the reaction. This kind of work can only be done if the reaction volume changes, which is determined by examining only the gases in a balanced equation. 1) Question is Name 2 sources of errors (not personal errors) in this experiment.



Answers

FRIEDEL-CRAFTS REACTIONS Friedel-Crafts reactions provide a method for the preparation of alkylbenzenes (ArR) and acylbenzenes (ArCOR). These reactions are called Fricdel-Crafts alkylation and Friedel-Crafts acylation. A Friedel-Crafts Alkylation The following is a general equation for a Friedel-Crafts alkylation reaction: (FIGURE CANNOT COPY) The mechanism for the reaction starts with the formation of a carbocation. The carbocation then acts as an electrophile and is attacked by the benzene ring to form an arenium ion. The arenium ion then loses a proton. This mechanism is illustrated below using 2 -chloropropane and benzenc. (FIGURE CANNOT COPY) This is a Lewis acid-base reaction (see Section $15.3)$ The complex dissociates to form a carbocation and $\mathrm{AlCl}_{4}^{-}.$ (FIGURE CANNOT COPY) The carbocation, acting as an electrophile, reacts with benzene to produce an arenium ion. A proton is removed from the arenium ion to form isopropylbenzene. This step also regenerates the AlCl_ and liberates HCI. When $R-X$ is a primary halide, a simple carbocation probably does not form. Instead, the aluminum chloride forms a complex with the alkyl halide, and this complex acts as the electrophile. The complex is one in which the carbon-halogen bond is nearly broken-and one in which the carbon atom has a considerable positive charge: (equation can't copy) Even though this complex is not a simple carbocation, it acts as if it were and it transfers a positive alkyl group to the aromatic ring. These complexes react so much like carbocations that they also undergo typical carbocation rearrangements (Sction $15.6 \mathrm{C}).$ Friedel-Crafts alkylations are not restricted to the use of alkyl halides and aluminum chloride. Other pairs of reagents that form carbocations (or species like carbocations) may be used in Friedel-Crafts alkylations as well. These possibilities include the use of a mixture of an alkene and an acid: (FIGURE CANNOT COPY) Propene Isopropylbenzene (cumene) $(84 \%)$ (FIGURE CANNOT COPY) Cyclohexene Cyclohexylbenzene $(62 \%)$ A mixture of an alcohol and an acid may also be used: (FIGURE CANNOT COPY) Cyclohexanol Cyclohexylbenzene $(56 \%)$ There are several important limitations of the Friedel-Crafts reaction. These are discussed in Section $15.6 \mathrm{C}.$ Outline all steps in a reasonable mechanism for the formation of isopropylbenzene from propene and benzene in liquid HF (just shown). Your mechanism must account for the product being isopropylbenzene, not propylbenzene. THE CHEMISTRY OF.... industrial Styrene Synthesis Styrene is one of the most important industrial chemicalsmore than 11 billion pounds is produced each year. The starting material for a major commercial synthesis of styrene is ethylbenzene, produced by Friedel-Crafts alkylation of benzene: (FIGURE CANNOT COPY) Styrene $(90-92 \% \text { yield })$ Most styrene is polymerized (Special Topic C) to the familiar plastic, polystyrene: (FIGURE CANNOT COPY) Polystyrene B Friedel-Crafts Acylation The $R$ group is called an acyl group, and a reaction whereby an acy group is introduced into a compound is called an acylation reaction. Two common acyl groups are the acetyl group and the benzoyl group. (The benzoyl group should not be confused with the benzyl group, $\left.-\mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5} ; \text { see Section } 14.2 .\right)$ (FIGURE CANNOT COPY) Acetyl group (ethanoyl group) Benzoyl group The Friedel-Crafts acylation reaction is often carried out by treating the aromatic compound with an acyl halide (often an acyl chloride). Unless the aromatic compound is one that is highly reactive, the reaction requires the addition of at least one equivalent of a Lewis acid (such as $\mathrm{AlCl}_{3}$ ) as well. The product of the reaction is an aryl ketone: (FIGURE CANNOT COPY) Acetyl chloride Acetophenone (methyl phenyl ketone) $(97 \%)$ Acyl chlorides, also called acid chlorides, are easily prepared (Section 18.5 ) by treating carboxylic acids with thionyl chloride (SOCI_) or phosphorus pentachloride (PCI_): (FIGURE CANNOT COPY) Acetic acid Thionyl chloride Acetyl chloride $(80-90 \%)$ (FIGURE CANNOT COPY) Benzoic acid Phosphorus pentachloride Benzoyl chloride $(90 \%)$ Friedel-Crafts acylations can also be carried out using carboxylic acid anhydrides. For example, (FIGURE CANNOT COPY) Acetic anhydride (a carboxylic acid anhydride) Acetophenone $(82-85 \%)$ In most Friedel-Crafts acylations the electrophile appears to be an acylium ion formed from an acyl halide in the following way: (FIGURE CANNOT COPY) (FIGURE CANNOT COPY) An acylium ion (a resonance hybrid) Show how an acylium ion could be formed from acetic anhydride in the presence of $\mathrm{AlCl}_{3}.$ STRATEGY AND ANSWER: We recognize that AICl_, is a Lewis acid and that an acid anhydride, because it has multiple unshared electron pairs, is a Lewis base. A reasonable mechanism starts with a Lewis acid-base reaction and proceeds to form an acylium ion in the following way. (FIGURE CANNOT COPY) Acylium ion The remaining steps in the Friedel-Crafts acylation of benzene are the following: (FIGURE CANNOT COPY) other resonance structures (draw them for practice) The acylium ion, acting as an electrophile, reacts with benzene to form the arenium ion. (FIGURE CANNOT COPY) A proton is removed from the arenium ion, forming the aryl ketone. (FIGURE CANNOT COPY) The ketone, acting as a Lewis base, reacts with aluminum chloride (a Lewis acid) to form a complex. (FIGURE CANNOT COPY) Treating the complex with water liberates the ketone and hydrolyzes the Lewis acid. Limitations of Friedel-Crafts Reactions Several restrictions limit the usefulness of Fricdel-Crafts reactions: 1. When the carbocation formed from an alkyl halide, alkene, or alcohol can rearrange to one or more carbocations that are more stable, it usually does so, and the major products obtained from the reaction are usually those from the more stable carbocations. When benzene is alkylated with butyl bromide, for example, some of the developing butyl cations rearrange by a hydride shift. Some of the developing $1^{\circ}$ carbocations (see following reactions) become more stable $2^{\circ}$ carbocations. Then benzene reacts with both kinds of carbocations to form both butylbenzene and sec-butylbenzene: (FIGURE CANNOT COPY) Butylbenzene $(32-36 \% \text { of mixture })$ sec-Butylbenzene $(64-68 \% \text { of mixture })$ 2. Friedel-Crafts alkylation and acylation reactions usually give poor yields when powerful electron-withdrawing groups (Section $15.8 \mathrm{B}$ and Table 15.1 ) are present on the aromatic ring. (FIGURE CANNOT COPY) These usually give poor yields in Friedel-Crafts reactions because the ring is electron deficient. Poor yiclds are also the case when the ring bears an $-\mathrm{NH}_{2},-\mathrm{NHR},$ or $-\mathrm{NR}_{2}$ group because they become electron-withdrawing when they react with the Lewis acid in the reaction mixture. (FIGURE CANNOT COPY) Does not undergo a Friedel-Crafts reaction These groups are changed into powerful electron-withdrawing groups by the Lewis acids used to catalyze Friedel-Crafts reactions. 3. Aryl and vinylic halides cannot be used as the halide component because they do not form carbocations readily (see Section $6.14 \mathrm{A}$ ): (FIGURE CANNOT COPY) No Friedel-Crafts reaction because the halide is at an $s p^{2}$ carbon. 4. Polyalkylations often occur. Alkyl groups are inductive electron-donating groups (Sections $15.8 \text { and } 15.9),$ and once one is introduced into the benzene ring, it activates the ring toward further substitution: (FIGURE CANNOT COPY) Isopropylbenzene $(24 \%)$ $p$ -Diisopropylbenzene $(14 \%)$ Polyacylations are not a problem in Friedel-Crafts acylations. The acyl group (-COR) by itself is an electron-withdrawing group, and when it forms a complex with AlCl in the last step of the reaction (Section $15.6 \mathrm{B}),$ it is made even more electron withdrawing. This strongly inhibits further substitution and makes monoacylation easy. When benzene reacts with 1-chloro-2,2-dimethylpropane (neopentyl chloride) in the presence of aluminum chloride, the major product is 2 -methyl-2-phenylbutanc, not 2,2 -dimethyl-1-phenylpropane (neopentylbenzenc). Explain this result. STRATEGY AND ANSWER: The carbocation formed by direct reaction of AICl_, with 1-chloro-2,2-dimethylpropane would be a primary carbocation; however, it rearranges to the more stable tertiary carbocation before it reacts with the benzene ring. (FIGURE CANNOT COPY) (FIGURE CANNOT COPY) Provide a mechanism that accounts for the following result. (FIGURE CANNOT COPY)

Predict results for the two foreign reaction. One finger prep. In with its shell, it's you will return it while a figure prepping and bread use a more stable Ben Zizek crocodile ritual. Capture CR minus on Give the world the alternative. Kavika Tile at this carbon would be less stable because of the leg off education with it benzene for hundreds inoculation. The reaction proceeds through Make uranium on the press. The charge will be distributed between America and roaster gerbils, and this carbon will have more positive charge because it will be better. Gillick allies back in education with the benzene. So this position, well being attacked with water with the bunt with maker comes off on America is a benched through the other carbon and then deport the nation and removal, or for America by reducing agent Junior it in Alcohol River Bridge Group in the bin Zigic position

Predict products or the fallen reactions. Dorian L. Sulfuric. That's Ah. So finish which goes toe the pair in the positions because 63 group is a hearted contributed a dollar dessert. Is it liberation? The car parks. A group is a resonance except her. Yeah, the try substitution goes to the net position Bremen nation off matter benzene. The letter group is an inductive, except er and business, except er so substitution goes to the net position insulation or for isopropyl Benzia I support group. That's ah Harper conjugated donor. So substitution goes toe the parrot in Ortho position. However you to the stereo hinderance imposed by the I support group. The author I, Samir is almost not existent. Just traces its most repaired. I, Samir.

So to offer some contacts. Methanol is produced in the following reaction. CEO add to H two that's in equilibrium with C H 30 H. So the methanol is condensed and separated out in his un reacted hydrogen and carbon monoxide that is recycled back through the reactor. So the first part we're calculating and 0.6 I 75.7 kill miles. Our we have end up for 24.3 killer moles per hour. We have end up five. That is 58.6 killer moles power. Yep. And one that is able to end sex. About 75.7 killer models. Iowa An end to is equal to two and six. So we multiply 75.5 by two. We got one 51.4 kg. Our and I mean calculated volumetric flow rate after recycle stream. That is one common 858 m cubed per hour. So the equation I used for this Waas and for the dark sorry. And for the ad and $5 multiplied by rt over fee that gives me overpay. Sorry, that gives me volume. Okay, so the next thing you need to do is create spreadsheet, um, for using the above equations for the different temperatures, pressures and percent excess hydrogen and then Arriva different flow rate of methanol that's produced so to solve for the method off flow rate, we need to use gold seek rather than the tedious trial in our. So we see that an increase of the access hydrogen we decrease temperature or increasing the pressure increases the yield of methadone in the next part. So the high pressure will result in higher power compression pumps and higher pressure rate uh, Reiter, containers and pipes. So high pressure rating is usually achieved by thick pipes as well as a container walls, which result in a higher up front cost. So in the next part of the reaction may be to lowest. Temperature is lowered too much, so the flow rate of materials may push. The gas is out of the reactor before they actually reach equilibrium. So in the final part here, if the calculations ideal gas behavior are assumed, it would assume that the reaction chamber is large enough for the reaction to reach equilibrium and the method or completely condenses out, and nothing is fed back into the recycle stream

Fennel visited, which can be produced from funeral, is less reactive than funeral in automatic. There are a few exceptions fearful. Explain what it's this, but the group is still Order Pereira director. Just like hydroxy group, you know, So I don't see a group in. Fino is a force of credit director because over too long in the pants, which supplies elections, most it through the orphan and better positions, this group still has to look natural in prayer on oxygen, which pretty X the same way as the age group. In funeral, however, the compound is less reactive because for conjugation off long and compare with adjusting Gruber. No, which most away park the church former our since one So the little there compares on oxygen become less active.


Similar Solved Questions

5 answers
1 . Describe three factors that affect the tertiary structure of an enzyme: 2_ How does a change in pH change the tertiary structure on an enzyme? How does an increase in temperature change the tertiary structure of an enzyme? 3_ What is "induced fit"? How might activity at an allosteric site affect zinduced fitz and the tertiary structure of an active site?
1 . Describe three factors that affect the tertiary structure of an enzyme: 2_ How does a change in pH change the tertiary structure on an enzyme? How does an increase in temperature change the tertiary structure of an enzyme? 3_ What is "induced fit"? How might activity at an allosteric s...
5 answers
QUESTIONWhich of the following could not have driven the Cambrian explosion? Homplasty B. Duplication and proliferation of genes C.Competition D. Predation E. Increasing levels of free atmospheric 02
QUESTION Which of the following could not have driven the Cambrian explosion? Homplasty B. Duplication and proliferation of genes C.Competition D. Predation E. Increasing levels of free atmospheric 02...
5 answers
Do Not WRITEANXTHING on these7^5~iimict smders [o Fta 2a Jtd it Tnaea denJ Ron] Tho Onti 4 Mren'UShor In Htre nua[Ccnsruc[ Sr; contidercc
Do Not WRITE ANXTHING on these 7^5~iimict smders [o Fta 2a Jtd it Tnaea denJ Ron] Tho Onti 4 Mren'U Shor In Htre nua[ Ccnsruc[ Sr; contidercc...
5 answers
Tculale the perce#3C - C-cHz 36%f yule Z chloro 3mneluylbutare Part € Give the relatlonship same, different; constitutional between isomers, following pairs of structures. The stereoisomers. (2 ea) possible relati
Tculale the perce #3C - C- cHz 36%f yule Z chloro 3mneluylbutare Part € Give the relatlonship same, different; constitutional between isomers, following pairs of structures. The stereoisomers. (2 ea) possible relati...
5 answers
Task: Investigation of Simple Harmonic Motion Prove Asin(wt + $) = C2 sin Wt + C1 cos wt b. Solve for amplitude: Solve for the phase shift: d. Rewrite the equation in the form y(t) = Asin(wt + $) Describe the initial condition of the experiment: Given position, solve for time_
Task: Investigation of Simple Harmonic Motion Prove Asin(wt + $) = C2 sin Wt + C1 cos wt b. Solve for amplitude: Solve for the phase shift: d. Rewrite the equation in the form y(t) = Asin(wt + $) Describe the initial condition of the experiment: Given position, solve for time_...
5 answers
Let €,a,8 be nonzero real numbers such that € # B. Find & given that 102n ac2Tiirkce: C, 0, 8 sifirdan farkli reel sayilar c # B olsun: Egerac? Ise 0 'yi bulunuz;6990 7100 6900 7000 7010 6000102n
Let €,a,8 be nonzero real numbers such that € # B. Find & given that 102n ac2 Tiirkce: C, 0, 8 sifirdan farkli reel sayilar c # B olsun: Eger ac? Ise 0 'yi bulunuz; 6990 7100 6900 7000 7010 6000 102n...
5 answers
Let A be the following matrix: AFind 5 x 2 matrix B with linearly independent columns s0 that AB
Let A be the following matrix: A Find 5 x 2 matrix B with linearly independent columns s0 that AB...
1 answers
A die is weighted so that the probability of rolling each of the numbers $2,3,4,$ and 5 is still $1 / 6,$ but the probability of rolling 1 is $9 / 60$ and the probability of rolling 6 is $11 / 60$ What are the mean and standard deviation of the number $X$ rolled using this die? What is the probability that $X \leq 3 ?$
A die is weighted so that the probability of rolling each of the numbers $2,3,4,$ and 5 is still $1 / 6,$ but the probability of rolling 1 is $9 / 60$ and the probability of rolling 6 is $11 / 60$ What are the mean and standard deviation of the number $X$ rolled using this die? What is the probabili...
5 answers
Given f (2,y,2)I2 arcsin €, Find 8Itvz) Oydz812zarcsinz+arcsin €
Given f (2,y,2) I2 arcsin €, Find 8Itvz) Oydz81 2zarcsinz +arcsin €...
1 answers
Evaluate each expression. See Example 2 and $3 .$ $3 s^{2}-2 s+8$ for a. $\quad s=1$ b. $s=0$
Evaluate each expression. See Example 2 and $3 .$ $3 s^{2}-2 s+8$ for a. $\quad s=1$ b. $s=0$...
5 answers
Unich Stru Ju) SuleuAA 1 Structure ! E1 orzeIS ine paclcn XcOf , struciuc 2 1 Anmcrun Structure 2
Unich Stru Ju) SuleuAA 1 Structure ! E1 orzeIS ine paclcn XcOf , struciuc 2 1 Anmcrun Structure 2...
5 answers
Explain the lagrange equation in case of velocity dependentpotiential ?NOTE: PLZZ EXPLAIN IT PROPERLY.SUBJECT NAME ( CLASSICAL MECHANICS)
Explain the lagrange equation in case of velocity dependent potiential ? NOTE: PLZZ EXPLAIN IT PROPERLY. SUBJECT NAME ( CLASSICAL MECHANICS)...
5 answers
Zererel Prsics (2) 41-20202}42 210 nthe The equivalent resistance for the combination of resistors shown in figure IS: (2 Points)6,7 03302405,6 0
Zererel Prsics (2) 41-20202} 42 2 10 n the The equivalent resistance for the combination of resistors shown in figure IS: (2 Points) 6,7 0 330 240 5,6 0...
5 answers
If the Cauchy-Riemann equations are satisfied at point, then the function necessarily analytic there_TrueFalse
If the Cauchy-Riemann equations are satisfied at point, then the function necessarily analytic there_ True False...
5 answers
What is the difference between sequence and series?Determine whether the geometric series E5-, ( %" ! is convergent or divergent Ifit is convergent, find the sum:
What is the difference between sequence and series? Determine whether the geometric series E5-, ( %" ! is convergent or divergent Ifit is convergent, find the sum:...
4 answers
Show all your works to get full credits Not by Calculator in the Exam.Problem (20 points) NOT BY CALCULATORSolve the system of linear equations using Gauss elimination ( 15 points) x + y +2 Zx + 3y +2 = 3x -y - 2z = -3b) Find the rank ($ points)
Show all your works to get full credits Not by Calculator in the Exam. Problem (20 points) NOT BY CALCULATOR Solve the system of linear equations using Gauss elimination ( 15 points) x + y +2 Zx + 3y +2 = 3 x -y - 2z = -3 b) Find the rank ($ points)...
5 answers
Q10. Which of the following defines an inner product (11,12) , (91.92) 2F[41 31242 in R? ii) 0[ + 611,02 +bzr 0142 + b1bg in Pt iii) 01 +br + 9112, 02 b" + 012 (1(2 + 0102 iu PzOlly only: and (iii) oulyNone of the given ASWCTS is (Tlle.
Q10. Which of the following defines an inner product (11,12) , (91.92) 2F[41 31242 in R? ii) 0[ + 611,02 +bzr 0142 + b1bg in Pt iii) 01 +br + 9112, 02 b" + 012 (1(2 + 0102 iu Pz Olly only: and (iii) ouly None of the given ASWCTS is (Tlle....

-- 0.020978--