PART A The element Si belongs to IVA group. Therefore, four electro.pdf

•

0 likes•2 views

PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital. Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s- electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal. PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl. Therefore, the number of unpaired electrons is 1 Solution PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Th.

Education

electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal.
PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration
of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl.
Therefore, the number of unpaired electrons is 1

Similar to PART A The element Si belongs to IVA group. Therefore, four electro.pdf

4structureoftheatom

Rajan Pv

4structureoftheatom.ppt

seema chaturvedi

Electron configuration

gabriela de la garza

Class XII d and f- block elements (part 2)

Arunesh Gupta

structure of the atom

shiva prasad

D and F Block.pdf

Renuha0130

4structureoftheatom-150805145649-lva1-app6892.pptx

Nikhilyadav56801

8. the d and f block elements anil-hsslive

Tariq Beigh

Similar to PART A The element Si belongs to IVA group. Therefore, four electro.pdf (8)

4structureoftheatom

4structureoftheatom.ppt

Electron configuration

Class XII d and f- block elements (part 2)

structure of the atom

D and F Block.pdf

4structureoftheatom-150805145649-lva1-app6892.pptx

8. the d and f block elements anil-hsslive

More from aquazac

Answer: Every investor expects dividend from his investments.Dividend income helps investors in dealing with their routine expenses. Investors put their money to earn some positive returns and dividend is one of the way of getting returns for investors. So, It is a duty of management to satisfy dividend expectation of investors. If management do not pay dividend then it is known as retention of earnings. Retention earnings has it\'s own pro and cons, which are described as follows: Pros: Retained earnings reduces organization\'s dependence on external borrowings. It can be considered as an advantage when external borrowings are not available easily. Retained earnings indicates good business opportunities, It means organization is growing financially. It may result in better stock valuation of organization. In some situation dividend is subject to taxes (either distribution tax or income tax) in the hands of investors. So, in this case investors are not interested in dividend but they are more interested in capital gain(which may be tax exempt). So, In this situation retained earnings fulfill investor\'s objective. Cons: As explained earlier all the investors have some expectation of dividend, in case of higher retention of earnings it may become difficult to fulfill investor\'s expectations. Retention of earnings may affect investor\'s earning capacity when investors have available opportunities, which gives higher return than organization earns by retaining amount. In this case investor looses earning opportunities. In case of dividend management\'s decision is significantly influenced by media and institutional investors. Media is responsible for organization\'s image in society, such image significantly affects organization\'s financial performance. This is why management always considers media\'s point of view before making any dividend decision. Institutional investors are usually invests their money and takes part in decision making process of organization. Institutional investor\'s weight in decision making process is generally very high, so it is obvious that they influence management\'s dividend decision also. Solution Answer: Every investor expects dividend from his investments.Dividend income helps investors in dealing with their routine expenses. Investors put their money to earn some positive returns and dividend is one of the way of getting returns for investors. So, It is a duty of management to satisfy dividend expectation of investors. If management do not pay dividend then it is known as retention of earnings. Retention earnings has it\'s own pro and cons, which are described as follows: Pros: Retained earnings reduces organization\'s dependence on external borrowings. It can be considered as an advantage when external borrowings are not available easily. Retained earnings indicates good business opportunities, It means organization is growing financially. It may result in better stock valuation of organization. In some situ.

Answer Every investor expects dividend from his investments.Dividen.pdf

aquazac

Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A functional product can be either protein (or polypeptide) or RNA. mRNA transcribes from a gene is further translated to produce polypeptides. However, all other RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene regulations. SO, polypeptides are just one class of functional product from DNA. The genomic DNA also produces various classes of RNAs with diverse functions that are crucial for cellular activates and gene regulation. Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all genes (including those encoding polypeptides and RNAs). Solution Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A functional product can be either protein (or polypeptide) or RNA. mRNA transcribes from a gene is further translated to produce polypeptides. However, all other RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene regulations. SO, polypeptides are just one class of functional product from DNA. The genomic DNA also produces various classes of RNAs with diverse functions that are crucial for cellular activates and gene regulation. Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all genes (including those encoding polypeptides and RNAs)..

Ans. Gene is defined as the segment of DNA that gives a functional p.pdf

aquazac

additional optmization techniques for underlying IP network must: 1.Have bandwidth and performance to handle converged services. 2.Seprate VLAN for Voice traffic. 3.Meet the demand of high availability of voice services by providing resiliency to mitigate the effect of network outages. 4.Should be modular, hierarchical & consistent to promote consistency and manageability. 5.configure QoS on core switch. Solution additional optmization techniques for underlying IP network must: 1.Have bandwidth and performance to handle converged services. 2.Seprate VLAN for Voice traffic. 3.Meet the demand of high availability of voice services by providing resiliency to mitigate the effect of network outages. 4.Should be modular, hierarchical & consistent to promote consistency and manageability. 5.configure QoS on core switch..

additional optmization techniques for underlying IP network must1.pdf

aquazac

According to the given equation, aqueous carbon dioxide reacts with water to form carbonic acid, Carbonic acid is not stable in normal body temperatures, therefore, it ionizes into bicarbonate ion (HCO3-) and a hydrogen ion (H+) is liberated. The liberation of hydrogen ions decreases the pH. During breathing, oxygen is inhaled which is carried by the blood to the different tissues of the body. Oxygen facilitates the various metabolic reactions occurring inside our cells and helps in respiration, or the generation of energy. Carbon dioxide, formed as a by-product of these energy- generating metabolic reactions, diffuses out of the tissues and into the blood stream, which carries CO2 to the lungs, where it is exhaled out. a) In case of hypoventilation : The exchange of gases that occurs in the lungs while breathing is impaired. Therefore, the requisite amount of carbon dioxide is not being exhaled out. This leads to a build-up of carbon- dioxide in the blood plasma. As a result, the forward reaction is facilitated and more H+ ions are generated. The higher concentration of H+ ions decreases the pH (acidic pH). This can lead to acidosis. The build-up of carbon dioxide in the blood facilitates the forward reaction and therefore the equilibrium shifts to the right. The hydronium ion or H3O+ is the product of a hydrogen ion (H+) and water. Since a hydrogen ion is too small to exist freely in water, it therefore combines with a water molecule to form a hydronium ion which is more stable. In case of hypoventilation, the concentration of H+ ions increase. Therefore, the concentration of hydronium ions increase during hypoventilation. The stress concentration or partial pressure of CO2 during hypoventilation is significantly higher than 40mmHg (around 60-65 mmHg). b) In case of hyperventilation : The breathing rate is more than normal. Therefore, as a result of heavier and faster breathing, the CO2 concentration in the blood rapidly decreases, that is, CO2 easily leaves the blood plasma. Decrease in the CO2 concentration slows down the rate of the forward reaction, reducing the levels at which bicarbonate ion and H+ ion concentration were being generated. The bicarbonate and hydrogen ions recombine to yield CO2 and water. This lowering in hydrogen ion concentration consequently lowers the pH (alkaline pH). This can lead to alkalosis. As a result of the lowering of CO2 , the reverse reaction takes place. The carbonic acid changes back to CO2 and H2O, shifting the equilibrium to the left. The concentration of H3O+ decreases during hyperventilation, since the H+ ion concentration is lowered. The stress levels or partial pressure of CO2 during hyperventilation is significantly lower than 40 mmHg (around 25-30 mmHg). c) If sodium carbonate or Na2CO3 is added instead of sodium bicarbonate, then sodium carbonate would react with water and it would hydrolyse into Na+ and CO32- ions. The CO32- combine with the hydrogen (H) present in water to form H2CO3 (carbon.

According to the given equation, aqueous carbon dioxide reacts with .pdf

aquazac

a) mean = 1.4*3 Thus distribution is Poisson(4.2) P(X = 4) = 4.2^4 * e^{-4.2} / 4! = 0.1944 b) average number of ghosts in 90min (1.5hrs) interval = 1.4*1.5 = 2.1 c) Probability that 4th ghost will appear before 10 am = Probability that number of ghosts appearing from 8 to 10 am >= 4 As it is a 2 hr period, the distribution of number of ghosts is Poisson(1.4*2) i.e Poisson(2.8) P(X >= 4) = 1 - P(X < 4) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 - [(2.8^0 * e^{-2.8} / 0!) + (2.8^1 * e^{-2.8} / 1!) + (2.8^2 * e^{-2.8} / 2!) + (2.8^3 * e^{- 2.8} / 3!)] = 1 - [0.0608 + 0.1703 + 0.2384 + 0.2225] = 0.692 d) When arrival distribution is Poisson(lambda), inter-arrival time distribution is Exponential(lambda) Thus, interarrival time distribution is Exponential(1.4) Expected value of time between 5th and 6th ghost = 1/1.4 = 0.7143hr e) On average time taken for 10th ghost to arrive = 10*(1/1.4) = 7.143hr Thus on an average, 10th ghost will arrive at 3:09pm f) Expected value of time between ghost appearances = 1/1.4 = 0.7143hr (see part d) g) Probability that 3rd ghost appears after 9:45 am = Probablity that number of ghosts appearing till 9:45 am < 3 Duration from 8am to 9:45 am = 1.75hr Mean number of ghosts = 1.4*1.75 = 2.45 Thus X is Poisson(2.45) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = (2.45^0 * e^{-2.45} / 0!) + (2.45^1 * e^{-2.45} / 1!) + (2.45^2 * e^{-2.45} / 2!) = 0.0863 + 0.2114 + 0.2590 = 0.5567 h) As inter-arrival time is Exponential(1.4), inter-arrival time is memoryless Thus, time for 7th ghost will be calculated from 1:00 pm and not from 12:35pm due to memorylessness property. Time duration from 1:00pm to 1:15pm = 0.25hrs Let Y be random variable denoting inter-arrival time probability that the 7th ghost will appear before 1:15 p.m. = P(Y < 0.25) = 1 - e^{-1.4 * 0.25) = 0.2953 i) Expected value of time for 7th ghost to appear after 1:00pm = 1/1.4 = 0.7143hr = 42.86 min Time from 12:35pm to 1:00pm = 25min Thus expected interarrival time = 25 + 42.86 = 67.86 min = 1hr 7.86min j) Expected duration after 1:00pm for 9th ghost to appear = 3 * (1/1.4) = 2.143hr = 2hr 8.58min Thus expected time at which 9th ghost appears = 3:09pm k) As the inter-arrival time follows memorylessness property, the future probabilities do not change with the observed value at past. Thus, 4 ghosts appearing between 7:00pm to 10:00pm does not affect ghost appearance from 11:00pm to 11:30pm Duration from 11:00pm to 11:30pm = 0.5hrs mean number of ghosts = 1.4*0.5 = 0.7 Thus it follows Poisson(0.7) distribution P(X > 2) = 1 - P(X <= 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] = 1 - [(0.7^0 * e^{-0.7} / 0!) + (0.7^1 * e^{-0.7} / 1!) + (0.7^2 * e^{-0.7} / 2!)] = 1 - [0.4966 + 0.3476 + 0.1217] = 0.0341 l) Probability that 8 ghosts appear between 6pm and midnight = Probability that number of ghosts appear from 6pm to 7pm and 10pm to midnight = 4 as it is known that 4 ghosts appeared from 7pm to 10pm Let A denote number of ghost from 6pm to 7pm B denote number of ghosts .

a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf

aquazac

2. a. Wired Media Type and Explination: Twisted-Pair Cable Twisted-pair cable is a type of cabling that is used for telephone communications and most modern Ethernet networks. A pair of wires forms a circuit that can transmit data. The pairs are twisted to provide protection against crosstalk, the noise generated by adjacent pairs. When electrical current flows through a wire, it creates a small, circular magnetic field around the wire. When two wires in an electrical circuit are placed close together, their magnetic fields are the exact opposite of each other. Thus, the two magnetic fields cancel each other out. They also cancel out any outside magnetic fields. Twisting the wires can enhance this cancellation effect. Using cancellation together with twisting the wires, cable designers can effectively provide self- shielding for wire pairs within the network media. Two basic types of twisted-pair cable exist: unshielded twisted pair (UTP) and shielded twisted pair (STP). The following sections discuss UTP and STP cable in more detail. UTP Cable UTP cable is a medium that is composed of pairs of wires . UTP cable is used in a variety of networks. Each of the eight individual copper wires in UTP cable \\is covered by an insulating material. In addition, the wires in each pair are twisted around each other. UTP cable relies solely on the cancellation effect produced by the twisted wire pairs to limit signal degradation caused by electromagnetic interference (EMI) and radio frequency interference (RFI). To further reduce crosstalk between the pairs in UTP cable, the number of twists in the wire pairs varies. UTP cable must follow precise specifications governing how many twists or braids are permitted per meter (3.28 feet) of cable. UTP cable often is installed using a Registered Jack 45 (RJ-45) connector The RJ-45 is an eight- wire connector used commonly to connect computers onto a local-area network (LAN), especially Ethernets. When used as a networking medium, UTP cable has four pairs of either 22- or 24-gauge copper wire. UTP used as a networking medium has an impedance of 100 ohms; this differentiates it from other types of twisted-pair wiring such as that used for telephone wiring, which has impedance of 600 ohms. UTP cable offers many advantages. Because UTP has an external diameter of approximately 0.43 cm (0.17 inches), its small size can be advantageous during installation. Because it has such a small external diameter, UTP does not fill up wiring ducts as rapidly as other types of cable. This can be an extremely important factor to consider, particularly when installing a network in an older building. UTP cable is easy to install and is less expensive than other types of networking media. In fact, UTP costs less per meter than any other type of LAN cabling. And because UTP can be used with most of the major networking architectures, it continues to grow in popularity. Disadvantages also are involved in using twisted-pair cabling, howe.

2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf

aquazac

A person may not choose to participate in the labour force due to Laziness, lack of ambition, emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities. However, race, location, education, and marital status are irrelevant at times. They do affect the type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay home and raise children is an example of different priorities, and a spouse\'s income, if they are wealthy enough, may negate having to even make a choice. (a) with low education, a job might not pay much more than unemployment benefit. (b) with children, there is strong incentive to stay at home and take care of them. (c) with one spouse earning a lot, another spouse can stay at home. (d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times women are viewed as better workers than men from same race. (e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they prefer to do some small work nearby or choose not to work at all. Solution A person may not choose to participate in the labour force due to Laziness, lack of ambition, emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities. However, race, location, education, and marital status are irrelevant at times. They do affect the type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay home and raise children is an example of different priorities, and a spouse\'s income, if they are wealthy enough, may negate having to even make a choice. (a) with low education, a job might not pay much more than unemployment benefit. (b) with children, there is strong incentive to stay at home and take care of them. (c) with one spouse earning a lot, another spouse can stay at home. (d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times women are viewed as better workers than men from same race. (e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they prefer to do some small work nearby or choose not to work at all..

A person may not choose to participate in the labour force due to La.pdf

aquazac

Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of the element, that is, the electrons in outer most shell. Remember each element wants to reach a more stable state. To achieve this it will share electrons with another element through covalent bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal electronegativity. Hope this helps! Solution Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of the element, that is, the electrons in outer most shell. Remember each element wants to reach a more stable state. To achieve this it will share electrons with another element through covalent bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal electronegativity. Hope this helps!.

Well.. 1) Ionic bonds are almost always metal to .pdf

aquazac

1. The answer is: d) Environmental Effects Environmental effects can cause a single genotype to produce a range of potential Phenotypes. 2. The answer is: c) Differ in both mean and variance The Figure represents two distributions that differ in both mean and variance. Solution 1. The answer is: d) Environmental Effects Environmental effects can cause a single genotype to produce a range of potential Phenotypes. 2. The answer is: c) Differ in both mean and variance The Figure represents two distributions that differ in both mean and variance..

1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf

aquazac

clear clc close all %Use polyfit to solve for the phase lines point11=[0,15]; point12=[0,300]; line1=polyfit(point11,point12,1); m1=line1(1); b1=line1(2); xpf1=linspace(min(point11),max(point11),100); ypf1=m1*xpf1+b1; point21=[0,15]; point22=[700,300]; line2=polyfit(point21,point22,1); m2=line2(1); b2=line2(2); xpf2=linspace(min(point21),max(point21),100); ypf2=m2*xpf2+b2; point31=[0,50]; point32=[700,300]; line3=polyfit(point31,point32,1); m3=line3(1); b3=line3(2); xpf3=linspace(min(point31),max(point31),100); ypf3=m3*xpf3+b3; point41=[50,100]; point42=[300,800]; line4=polyfit(point41,point42,1); m4=line4(1); b4=line4(2); xpf4=linspace(min(point41),max(point41),100); ypf4=m4*xpf4+b4; point51=[100,85]; point52=[800,300]; line5=polyfit(point51,point52,1); m5=line5(1); b5=line5(2); xpf5=linspace(min(point51),max(point51),100); ypf5=m5*xpf5+b5; point61=[85,100]; point62=[300,0]; line6=polyfit(point61,point62,1); m6=line6(1); b6=line6(2); xpf6=linspace(min(point61),max(point61),100); ypf6=m6*xpf6+b6; point71=[15,85]; point72=[300,300]; %Allow the user to set the percent of B b=input(\'Enter the mass percent of B: \'); %Return an error if the percent of B is not a valid number if b<0 || b>100 error(\'Please enter a value between 0 and 100.\') end %Allow the user to set the temperature t=input(\'Enter the temperature [deg C]: \'); %Solve for percent composition of A a=100-b; %Determine phase based on temperature and percent composition of B using %the phase lines if t<300 if b<15 && t>m1*b+b1 phase=\'alpha\'; elseif b<15 && t=15 && b<=85 phase=\'alpha + beta\'; elseif b>85 && t>m6*b+b6 phase=\'beta\'; else phase=\'alpha + beta\'; end elseif b<15 if t=15 && b<=50 if t50 && b<=85 if t=15 && b<=85 && t==300 fprintf(\'The provided conditions fall on the eutectic line.\ \') end if b==50 && t==300 fprintf(\'The provided conditions fall on the eutectic point.\ \') end %Output a formatted statement to display the phase fprintf(\'For the composition of %0.2f%% A, %0.2f%% B and a temperature of %0.0f degrees Celsius, the phase is %s.\ \',a,b,t,phase) %Create a plot of the phase lines and the user\'s set point figure(\'color\',\'white\') %Set axes axis([0 100 0 1000]) %Plot the phase lines, eutectic line, eutectic point, and the user\'s point plot(xpf1,ypf1,\'-b\') hold on plot(xpf2,ypf2,\'-b\') plot(xpf3,ypf3,\'-b\') plot(xpf4,ypf4,\'-b\') plot(xpf5,ypf5,\'-b\') plot(xpf6,ypf6,\'-b\') plot(point71,point72,\'-k\') plot(b,t,\'ro\',\'MarkerFaceColor\',\'r\') %Assign a title and axes labels title(\'Phase Diagram of Elements A and B\') xlabel(\'Percent Composition of B\') ylabel(\'Temperature (T) [deg C]\') %Set scaling for the axes set(gca,\'XTick\',0:10:100,\'YTick\',0:100:1000) %Place the phase names on the graph text(40,200,\'alpha+beta\') text(45,600,\'Liquid\') text(90,300,\'beta\') text(3,300,\'alpha\') text(15,375,\'alpha + Liquid\') text(62,375,\'beta + Liquid\') Solution clear clc close all %Use polyfit to solve for the phase lines point11=[0,15]; point.

clear clc close all Use polyfit to solve for the phase l.pdf

aquazac

The oxygen appears in both step reactions. But, it would not appear in the final reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction is 2O3 ====> 3 O2 (No O) Solution The oxygen appears in both step reactions. But, it would not appear in the final reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction is 2O3 ====> 3 O2 (No O).

The oxygen appears in both step reactions. But, i.pdf

aquazac

The two contributions to the cohesive energy of the ionic crystal are : Vander waals attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero- point energy is the lowest possible energy that a quantum mechanical physical system may have; it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in their ground state and have an associated zero-point energy, a consequence of their wave-like interaction. Solution The two contributions to the cohesive energy of the ionic crystal are : Vander waals attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero- point energy is the lowest possible energy that a quantum mechanical physical system may have; it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in their ground state and have an associated zero-point energy, a consequence of their wave-like interaction..

The two contributions to the cohesive energy of t.pdf

aquazac

If you are talking about an extraction design, the the water solubility has to be greater in the aqueous base because of the greater solubility of the salts of the acids (with few exceptions), again presuming a differential extraction aqueous to organic. Solution If you are talking about an extraction design, the the water solubility has to be greater in the aqueous base because of the greater solubility of the salts of the acids (with few exceptions), again presuming a differential extraction aqueous to organic..

If you are talking about an extraction design, th.pdf

aquazac

Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near common ancestor of homosapiens. Solution Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near common ancestor of homosapiens..

Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf

aquazac

When something boils, it changes states of matter. It would go from .pdf

aquazac

We need to discuss why there is an importance of adding residents to the budget planning, we can go through below explaintion and can derived various conclusions from the same. In explaining why they did or did not seek input from citizens, respondents in- dicated that the process was guided by the desires of the leadership. Of those who said that they did seek citizen input, the desire (as opposed to the obligation through policy) of the staff or the governing board or staff to seek public opinion seemed to provide some motivation. Almost three-quarters of responding cities cited staff desire, and almost two-thirds, governing board desire. The pattern for counties was similar. The most often cited reason for extra effort to involve the public was board interest. Almost half of the counties also cited staff interest. Although citizen involvement is a hot topic in management literature these days, local governments in North Caro- lina are generally not making extensive fforts to involve residents in budgeting. Barely half of the cities responding to the survey and less than a third of the counties go beyond the single, mandated hearing to get citizen input In general, cities tend to seek input more than counties do. This result might be explained by the different types of services offered by the two forms of gov- ernment. Counties provide funding for, among other things, social services, public health services, mental health services, and schools, and the boards over- seeing these services may seek citizen input when they are preparing their own budget request for the county. Although there is informal interest in hearing from citizens, boards clearly want to remain flexible in when and how they obtain input. Only 13 percent of the city respondents had a formal policy or requirement for citizen involvement beyond the single, state-mandated hearing. Medium to large cities were more likely to have such requirements than small cities. Only one county, a larger one, reported having such a policy or requirement. Although citizen involvement is a hot topic in management literature these days, local governments in North Caro- lina are generally not making extensive efforts to involve residents in budgeting. Barely half of the cities responding to the survey and less than a third of the counties go beyond the single, mandated hearing to get citizen input Of those that did not seek extra citizen involvement, the most common reason cited, on both the city and the county level, was lack of governing board interest. The second most common reason was poor response in the past, when at- tempts to get citizens involved did not seem very effective. At the county level, lack of board interest was a far more important factor than anything else, including poor past response: 71 percent of respondents cited lack of board interest, whereas only 39 percent cited poor past response Although staff interest seemed to be a big motivator for involving ci.

We need to discuss why there is an importance of adding residents to.pdf

aquazac

What is the largest decimal integer that can be represented with the following number of bits? 1. 3 bits 2. 8 bits 3. 32 bits 4. 64 bits How do you find out the largest decimal integer? Solution In general, with k bits we can represent 2k values. Their range will depend on the system: Unsigned: 0 to 2k-1 1. 3 bits => 2^3 - 1 => 7 => 111 2. 8 bits => 2^8 - 1 => 255 3. 32 bits => 2^32 - 1 4. 64 bits => 2^64 - 1.

What is the largest decimal integer that can be represented with the.pdf

aquazac

Throwing.javaimport java.util.InputMismatchException; import jav.pdf

aquazac

This electron transport is accompanied by the protons transfer into intermembrane space. Protons release potential which drives the phosphorylation of ADP to ATP. Protons return to the matrix through a proton specific channel Solution This electron transport is accompanied by the protons transfer into intermembrane space. Protons release potential which drives the phosphorylation of ADP to ATP. Protons return to the matrix through a proton specific channel.

This electron transport is accompanied by the protons transfer into .pdf

aquazac

In linear or non cyclic electron transport of photosynthesis NADPH a.pdf

aquazac

More from aquazac (20)

Answer Every investor expects dividend from his investments.Dividen.pdf

Ans. Gene is defined as the segment of DNA that gives a functional p.pdf

additional optmization techniques for underlying IP network must1.pdf

According to the given equation, aqueous carbon dioxide reacts with .pdf

a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf

2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf

A person may not choose to participate in the labour force due to La.pdf

Well.. 1) Ionic bonds are almost always metal to .pdf

1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf

clear clc close all Use polyfit to solve for the phase l.pdf

The oxygen appears in both step reactions. But, i.pdf

The two contributions to the cohesive energy of t.pdf

If you are talking about an extraction design, th.pdf

Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf

When something boils, it changes states of matter. It would go from .pdf

We need to discuss why there is an importance of adding residents to.pdf

What is the largest decimal integer that can be represented with the.pdf

Throwing.javaimport java.util.InputMismatchException; import jav.pdf

This electron transport is accompanied by the protons transfer into .pdf

In linear or non cyclic electron transport of photosynthesis NADPH a.pdf

Recently uploaded

Photos from our Spring Gala 2024. Thank you for making our 2024 Spring Gala unforgettable! Whether you attended or supported us with a donation, we extend our heartfelt thanks for celebrating School-Community Partnerships alongside ExpandED Schools. Special appreciation to our honorees, Rachel Skaistis and Cravath, Swaine, & Moore LLP, for their unwavering support in advancing educational equity. Your contributions have profoundly impacted countless young New Yorkers, and we’re proud to have you as part of our community. Our gratitude also extends to our Board of Directors, sponsors, city partners, community-based organizations, and the inspiring young people who joined us. Your support fuels our mission, and we’re endlessly thankful for your dedication. While we celebrate our gala’s success, our work continues. As you view the evening’s highlights, consider how you can further empower young New Yorkers and educators with your donation, no matter the size. Donate today: https://www.expandedschools.org/support-our-work/donate/

Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships

expandedwebsite

How to Send Pro Forma Invoice to Your Customers in Odoo 17

Celine George

AIM of Education-Teachers Training-2024.ppt

Nishitharanjan Rout

Mattingly "AI and Prompt Design: LLMs with NER"

National Information Standards Organization (NISO)

會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文

中央社

How to Manage Website in Odoo 17 Studio App.pptx

Celine George

Supporting Newcomer Multilingual Learners

David Douglas School District

Contoh Aksi Nyata Refleksi Diri ( NUR ).pdf

cupulin

ESSENTIAL of (CS/IT/IS) class 07 (Networks)

Dr. Mazin Mohamed alkathiri

An overview of the various scriptures in Hinduism

Dabee Kamal

Improved Approval Flow in Odoo 17 Studio App

Celine George

Stl Algorithms in C++ jjjjjjjjjjjjjjjjjj

Mohammed Sikander

TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...

Nguyen Thanh Tu Collection

Analyzing and resolving a communication crisis in Dhaka textiles LTD.pptx

Limon Prince

Major project report on Tata Motors and its marketing strategies

AmanpreetKaur157993

e-Sealing at EADTU by Kamakshi Rajagopal

EADTU

Transport (British English) or Transportation (American English) ransportation has developed along three basic Mode (Media):- 1. Land Transportation (way)– (a) Road Transportation (b) Rail Transportation 2. Water Transportation 3. Air Transportation Tramway Inland water transport Ocean transport These may be classified as under: (a). Liners (b). Tramps Liners Vs Tramps Figure- Layout airport runway design TRAFFIC SIGNS

Basic Civil Engineering notes on Transportation Engineering & Modes of Transport

Denish Jangid

DEMONSTRATION LESSON IN ENGLISH 4 MATATAG CURRICULUM

ELOISARIVERA8

OS-operating systems- ch05 (CPU Scheduling) ...

Dr. Mazin Mohamed alkathiri

UChicago CMSC 23320 - The Best Commit Messages of 2024

Borja Sotomayor

Recently uploaded (20)

Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships

How to Send Pro Forma Invoice to Your Customers in Odoo 17

AIM of Education-Teachers Training-2024.ppt

Mattingly "AI and Prompt Design: LLMs with NER"

會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文

How to Manage Website in Odoo 17 Studio App.pptx

Supporting Newcomer Multilingual Learners

Contoh Aksi Nyata Refleksi Diri ( NUR ).pdf

ESSENTIAL of (CS/IT/IS) class 07 (Networks)

An overview of the various scriptures in Hinduism

Improved Approval Flow in Odoo 17 Studio App

Stl Algorithms in C++ jjjjjjjjjjjjjjjjjj

TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...

Analyzing and resolving a communication crisis in Dhaka textiles LTD.pptx

Major project report on Tata Motors and its marketing strategies

e-Sealing at EADTU by Kamakshi Rajagopal

Basic Civil Engineering notes on Transportation Engineering & Modes of Transport

DEMONSTRATION LESSON IN ENGLISH 4 MATATAG CURRICULUM

OS-operating systems- ch05 (CPU Scheduling) ...

UChicago CMSC 23320 - The Best Commit Messages of 2024

PART A The element Si belongs to IVA group. Therefore, four electro.pdf

1. PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital. Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s- electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal. PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl. Therefore, the number of unpaired electrons is 1 Solution PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital. Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s-

2. electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal. PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl. Therefore, the number of unpaired electrons is 1

PART A The element Si belongs to IVA group. Therefore, four electro.pdf

Recommended

Recommended

More Related Content

Similar to PART A The element Si belongs to IVA group. Therefore, four electro.pdf

Similar to PART A The element Si belongs to IVA group. Therefore, four electro.pdf (8)

More from aquazac

More from aquazac (20)

Recently uploaded

Recently uploaded (20)

PART A The element Si belongs to IVA group. Therefore, four electro.pdf