This is called a dehydrationreaction. a) H2SO4 H20Solution.pdf

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This is called a dehydrationreaction. a) H2SO4 / H20 Solution This is called a dehydrationreaction. a) H2SO4 / H20.

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package com.test; public class Team { private String teamId; private String teamName; private String pFirstName; private String pLastName; private String pemail; private String phoneNumber; public String getTeamId() { return teamId; } public void setTeamId(String teamId) { this.teamId = teamId; } public String getTeamName() { return teamName; } public void setTeamName(String teamName) { this.teamName = teamName; } public String getpFirstName() { return pFirstName; } public void setpFirstName(String pFirstName) { this.pFirstName = pFirstName; } public String getpLastName() { return pLastName; } public void setpLastName(String pLastName) { this.pLastName = pLastName; } public String getPemail() { return pemail; } public void setPemail(String pemail) { this.pemail = pemail; } public String getPhoneNumber() { return phoneNumber; } public void setPhoneNumber(String phoneNumber) { this.phoneNumber = phoneNumber; } public String toString() { return \"Team [teamId=\" + teamId + \", teamName=\" + teamName + \", pFirstName=\" + pFirstName + \", pLastName=\" + pLastName + \", pemail=\" + pemail + \", phoneNumber=\" + phoneNumber + \"]\"; } public Team(String teamId, String teamName, String pFirstName, String pLastName, String pemail, String phoneNumber) { super(); this.teamId = teamId; this.teamName = teamName; this.pFirstName = pFirstName; this.pLastName = pLastName; this.pemail = pemail; this.phoneNumber = phoneNumber; } public int hashCode() { final int prime = 31; int result = 1; result = prime * result + ((pFirstName == null) ? 0 : pFirstName.hashCode()); result = prime * result + ((pLastName == null) ? 0 : pLastName.hashCode()); result = prime * result + ((pemail == null) ? 0 : pemail.hashCode()); result = prime * result + ((phoneNumber == null) ? 0 : phoneNumber.hashCode()); result = prime * result + ((teamId == null) ? 0 : teamId.hashCode()); result = prime * result + ((teamName == null) ? 0 : teamName.hashCode()); return result; } public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; Team other = (Team) obj; if (pFirstName == null) { if (other.pFirstName != null) return false; } else if (!pFirstName.equals(other.pFirstName)) return false; if (pLastName == null) { if (other.pLastName != null) return false; } else if (!pLastName.equals(other.pLastName)) return false; if (pemail == null) { if (other.pemail != null) return false; } else if (!pemail.equals(other.pemail)) return false; if (phoneNumber == null) { if (other.phoneNumber != null) return false; } else if (!phoneNumber.equals(other.phoneNumber)) return false; if (teamId == null) { if (other.teamId != null) return false; } else if (!teamId.equals(other.teamId)) return false; if (teamName == null) { if (other.teamName != null) return false; } else if (!teamName.equals(other.teamName)) return false; return true; } } package com.test; import java.util.Date; public class Game { private String gameId; priv.

package com.test;public class Team { private String teamId;.pdf

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Given below is the completed code along with comments. Output of the program is shown at the end. You will need to create an input file containing multiple lines of input as specified in the question. I have given a sample input file with just 1 line of input. Please do rate the answer if it helped. Thank you very much. #include /* extractBits extracts the specified number of bits starting specified start location from left. bit numbering starts with 0 from left. len specifies the number of bits to extract. So extract leftmost 4 bits of a number n, the call should be extractBits(n, 0, 4). Here start = 0 means the 0th bit from left. Similarly to extract bit 16 through 20 (i.e 5 bits), we use extractBits(n, 16, 5) The way this function works is - First calculate the remaining number of bits on the rightside . Since int takes 32 bits, we subtract (start+len) from 32 to get remaining bits on right. it first clears all the bits upto starting bit by left shift << by start bits. Now we shift back the same number of times to right + */ int extractBits(unsigned n, int start, int len) { int rightRemaining = 32 - (start + len); //shift left and then right by start no. of bits clears leftside bits n = (n << start) >> start; //now shifting by remaining number of bits on right will extract only needed bits n = n >> rightRemaining; return n; } int main (int argc, char *argv[]) { // Get the filename to open, then open the file. If we can\'t // open the file, complain and quit. char *filename; // *** Your Code *** if argv[1] exists, set filename to it, // otherwise set filename to \"Lab3_data.txt\"; if(argv[1] != NULL) filename = argv[1]; else filename = \"Lab3_data.txt\"; // Open the file; if opening fails, say so and return 1. // otherwise say what file we\'re reading from. FILE *in_file; in_file = fopen(filename, \"r\"); // NULL if the open failed // *** Your Code *** if in_file is NULL, complain and quit // otherwise say that we\'ve opened the filename if (in_file == NULL){ printf(\"Couldn\'t open file %s\ \", filename); return 1; } printf(\"Opened file %s for reading...\ \", filename); // Repeatedly read and process each line of the file. The // line should have a hex integer and two integer lengths. int val, len1, len2, len3; int nbr_vals_on_line = fscanf(in_file, \"%x %d %d\", &val, &len1, &len2); // Read until we hit end-of-file or a line without the 3 values. while (nbr_vals_on_line == 3) { // We\'re going to break up the value into bitstrings of // length len1, len2, and len3 (going left-to-right). // The user gives us len1 and len2, and we calculate // len3 = the remainder of the 32 bits of value. // All three lengths should be > 0, else we complain // and go onto the next line. // len3 = 32 - (len1 + len2); // *** Your Code*** // if any of the lengths aren\'t > 0, // print out the value and the lengths and complain // about the lengths not all being positive if(len1 <= 0 || len2 <= 0 || len3 <= 0) { printf(\"Invalid lengths : len1=%d, len2=%d, len.

Given below is the completed code along with comments. Output of the.pdf

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(a) The American Institute of Certified Accountants. (AICPA) (b) This Statement establishes accounting and financial reporting standards for public colleges and universities within the financial reporting guidelines of GASB Statement No. 34, Basic Financial Statements-and Management’s Discussion and Analysis-for State and Local Governments. The standards are designed to provide financial information that responds to the needs of three groups of primary users of general purpose external financial reports: the citizenry, legislative and oversight bodies, and investors and creditors. Generally, this amendment to Statement 34 permits public colleges and universities, in separately issued financial statements, to use the guidance for special-purpose governments engaged only in business-type activities, engaged only in governmental activities, or engaged in both governmental and business-type activities in their separately issued reports. Under this guidance, in its separately issued reports, a public institution is required to include management’s discussion and analysis (MD&A); basic financial statements, as appropriate for the category of special-purpose government reporting selected; notes to the financial statements; and required supplementary information other than MD&A. Unless otherwise specified, pronouncements of the GASB apply to financial reports of all state and local governmental entities, including general purpose governments, public benefit corporations and authorities, and public employee retirement systems, utilities, hospitals and other healthcare providers, and colleges and universities. Paragraph 2 discusses the applicability of this Statement. (c) A statement of financial position, including accompanying notes to financial statements, provides relevant information about liquidity, financial flexibility, and the interrelationship of an organization\'s assets and liabilities. That information generally is provided by aggregating assets and liabilities that possess similar characteristics into reasonably homogeneous groups. For example, entities generally report individual items of assets in homogeneous groups, such as cash and cash equivalents; accounts and notes receivable from patients, students, members, and other recipients of services; inventories of materials and supplies; deposits and prepayments for rent, insurance, and other services; marketable securities and other investment assets held for long-term purposes; and land, buildings, equipment, and other long-lived assets used to provide goods and services. Cash or other assets received with a donor-imposed restriction that limits their use to long-term purposes should not be classified with cash or other assets that are unrestricted and available for current use. (d) A statement of activities shall report revenues as increases in unrestricted net assets unless the use of the assets received is limited by donor-imposed restrictions. For example, fees from rendering servic.

(a) The American Institute of Certified Accountants. (AICPA)(b) Th.pdf

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Yes. The compound is actively active. It is mainly caused by the carbon to the right of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices over 3 possible carbons) which give you 8. And as I previously explained: To be optically active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3 such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally active since they have 4 different groups. However, since they are mirror images of each other, they cancel each other out. It is the third and final carbon to the right of both oxygens, that makes this compound chirally active. Since it has also 4 different groups attached to the carbon and is not canceled out by anything. One simple way to test if something is optically active is to see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center, then it is optically active Solution Yes. The compound is actively active. It is mainly caused by the carbon to the right of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices over 3 possible carbons) which give you 8. And as I previously explained: To be optically active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3 such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally active since they have 4 different groups. However, since they are mirror images of each other, they cancel each other out. It is the third and final carbon to the right of both oxygens, that makes this compound chirally active. Since it has also 4 different groups attached to the carbon and is not canceled out by anything. One simple way to test if something is optically active is to see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center, then it is optically active.

Yes. The compound is actively active. It is mainl.pdf

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The probability of A and its complement will sum to oneSolution.pdf

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H+,CN-Solution H+,CN-.pdf

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// Adding elements public void add(String element) { // For first element Head is null. So, create a new node and mark it as head. // and increase numElements if(head == null){ head = new StringNode(element, null); numElements++; } else { // if not, find its place first. // then create a node and mark position link to new node // mark new node as link to positional node StringNode node = head; while(node.getLink() != null){ if(node.getData().compareTo(element) < 0){ break; } } StringNode newNode = new StringNode(element, null); newNode.setLink(node.getLink()); node.setLink(newNode); numElements++; } } // Removing elements public boolean remove(String target) { StringNode targetNode = head; boolean found = false; while (targetNode!= null && !found) { if(targetNode.getData().equalsIgnoreCase(target)) found = true; else targetNode = targetNode.getLink(); } if(found) { // copy the head to targetNode // and then advance head to the next node. targetNode.setData(targetNode.getLink().getData()); targetNode.setLink(targetNode.getLink()); numElements --; } return found; } Solution // Adding elements public void add(String element) { // For first element Head is null. So, create a new node and mark it as head. // and increase numElements if(head == null){ head = new StringNode(element, null); numElements++; } else { // if not, find its place first. // then create a node and mark position link to new node // mark new node as link to positional node StringNode node = head; while(node.getLink() != null){ if(node.getData().compareTo(element) < 0){ break; } } StringNode newNode = new StringNode(element, null); newNode.setLink(node.getLink()); node.setLink(newNode); numElements++; } } // Removing elements public boolean remove(String target) { StringNode targetNode = head; boolean found = false; while (targetNode!= null && !found) { if(targetNode.getData().equalsIgnoreCase(target)) found = true; else targetNode = targetNode.getLink(); } if(found) { // copy the head to targetNode // and then advance head to the next node. targetNode.setData(targetNode.getLink().getData()); targetNode.setLink(targetNode.getLink()); numElements --; } return found; }.

Adding elements public void add(String element) { For fi.pdf

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1) Investment Grade Domestic Bonds As bond bears a fixed rate of interest hence Mutual Fund or ETFS can\'t Represent Bonds as an asset class. 2) High Yield Bonds 3) Domestic stock Both are consist of shares and debentures so Mutual Fund or ETFS can\'t Represent them as an asset class. 4) International Stock 6) Energy Similarly ETF and Mutual fund never meet the defination of these 3 asset class so can\'t represent them 7) Commodities 8) Real Estate 5)E merging Market Stock An emerging market fund is a mutual fund or ETF that invests the bulk of its assets in stocks of developing countries. E merging countries are those which are not developed yet in the stage of development i.e. china, india etc. so invetment in these fund brings highest reurn for prospective investor. So ETF and Mutual fund Represents E merging market stock. Solution 1) Investment Grade Domestic Bonds As bond bears a fixed rate of interest hence Mutual Fund or ETFS can\'t Represent Bonds as an asset class. 2) High Yield Bonds 3) Domestic stock Both are consist of shares and debentures so Mutual Fund or ETFS can\'t Represent them as an asset class. 4) International Stock 6) Energy Similarly ETF and Mutual fund never meet the defination of these 3 asset class so can\'t represent them 7) Commodities 8) Real Estate 5)E merging Market Stock An emerging market fund is a mutual fund or ETF that invests the bulk of its assets in stocks of developing countries. E merging countries are those which are not developed yet in the stage of development i.e. china, india etc. so invetment in these fund brings highest reurn for prospective investor. So ETF and Mutual fund Represents E merging market stock..

1) Investment Grade Domestic Bonds As bond bears a fixed rate of i.pdf

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When atoms share one or more pairs of electrons t.pdf

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This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the solution. Now the two are separate. To get the NaCl out of the water just boil it off. Solution This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the solution. Now the two are separate. To get the NaCl out of the water just boil it off..

This is quite simple to do. Barium sulfate is ins.pdf

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Step1 Cocentration of NaCl =.250 moles Step2 Mol.pdf

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Nitric acid is a strong acid... we can assume com.pdf

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NaCN is a salt and it completely gets ionized, NaCN -------------> Na+ +CN- 0.05L * 0.10 M moles =0.005 0.005 0.005 HCl --------------> H+ + Cl- moles 0.020 L * 0.2 M =0.004 0.004 0.004 Chemical reaction: CN- (aq) + H+ (aq) ------------------> HCN Before rxn (moles) 0.005 0.004 0 After rxn (moles) 0.001 0 0.004 [CN-] = 0.001 moles / 0.07 L = 0.01428 M [H+] = 0 M (acid is completelyconsumed) [HCN] = 0.004 moles / 0.07 L = 0.057 M [Cl-] = 0.004 moles / 0.07L = 0.0571 M Solution NaCN is a salt and it completely gets ionized, NaCN -------------> Na+ +CN- 0.05L * 0.10 M moles =0.005 0.005 0.005 HCl --------------> H+ + Cl- moles 0.020 L * 0.2 M =0.004 0.004 0.004 Chemical reaction: CN- (aq) + H+ (aq) ------------------> HCN Before rxn (moles) 0.005 0.004 0 After rxn (moles) 0.001 0 0.004 [CN-] = 0.001 moles / 0.07 L = 0.01428 M [H+] = 0 M (acid is completelyconsumed) [HCN] = 0.004 moles / 0.07 L = 0.057 M [Cl-] = 0.004 moles / 0.07L = 0.0571 M.

NaCN is a salt and it completely gets ionized, .pdf

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molar ratio 6 carbon 12 hydirgen 6 oxygen .pdf

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H2CO3 = H2O + CO2 .pdf

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density = masswater displaced or, 10.5 = 5.25wa.pdf

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as Na+ ions replace H+ from H2S,as 2moles of Na+ .pdf

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Dependent SolutionDependent .pdf

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Bear with me, I\'m trying to type the Lewis structure: ::O=Se-O::: and there are four electrons surrounding the selenium. So: Oxygen with four unpaired electrons and a double bond to Selenium. Selenium with four unpaired electrons bonded by a single bond to Oxygen with six unpaired electrons. Hope I helped! Solution Bear with me, I\'m trying to type the Lewis structure: ::O=Se-O::: and there are four electrons surrounding the selenium. So: Oxygen with four unpaired electrons and a double bond to Selenium. Selenium with four unpaired electrons bonded by a single bond to Oxygen with six unpaired electrons. Hope I helped!.

Bear with me, Im trying to type the Lewis structure O=Se-O.pdf

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ANS. A & B . There is no direct or particular source for gases like methane and carbon dioxide, they are the product of various chemical and biological reactions that occur in a land fill. These gases compose approximately 90 % of gases in a landfill. As for there harmful effects there are many for e.g. there is concern of global warming , they causes nausea, fatigue at a high concentration to humans, methane is flammable. C. Solvents : These can be various chemicals from industrial disposal or domestic disposal also. These becomes a concern because they tend to seep through the soil and if reaches ground water, then they can directly pollute it and which may cause various health problems. D. Organic Nitrogen: Source is mostly sewage or agricultural waste. Organic nitrogen is of concern as it tends to stimulate and maintain microbial activities in landfills which may increase total biomass and creating a swampy area which will cause more emission of harmful gases. E. Rodents: Rodents are the species that survive in landfills, they survive on waste or potentially hazardous matter present there, which can cause a serious impact on food chain from environment point of view. Bio-magnification and bio-accumulation are also some serious problems that can arise from this. Solution ANS. A & B . There is no direct or particular source for gases like methane and carbon dioxide, they are the product of various chemical and biological reactions that occur in a land fill. These gases compose approximately 90 % of gases in a landfill. As for there harmful effects there are many for e.g. there is concern of global warming , they causes nausea, fatigue at a high concentration to humans, methane is flammable. C. Solvents : These can be various chemicals from industrial disposal or domestic disposal also. These becomes a concern because they tend to seep through the soil and if reaches ground water, then they can directly pollute it and which may cause various health problems. D. Organic Nitrogen: Source is mostly sewage or agricultural waste. Organic nitrogen is of concern as it tends to stimulate and maintain microbial activities in landfills which may increase total biomass and creating a swampy area which will cause more emission of harmful gases. E. Rodents: Rodents are the species that survive in landfills, they survive on waste or potentially hazardous matter present there, which can cause a serious impact on food chain from environment point of view. Bio-magnification and bio-accumulation are also some serious problems that can arise from this..

ANS.A & B . There is no direct or particular source for gases like.pdf

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package com.test;public class Team { private String teamId;.pdf

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