sulfur trioxide is SO3 atomic mass S = 32.1 gmo.pdf

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sulfur trioxide is SO3 atomic mass: S = 32.1 g/mol O =16.0 g/mol molar mass of SO3 = 32.1 + 16.0x3 = 80.1 g/mol moles of SO3 = 2.15 mol mass of SO3 = moles of SO3 x molar mass of SO3 = 2.15 x 80.1 = 172 g Solution sulfur trioxide is SO3 atomic mass: S = 32.1 g/mol O =16.0 g/mol molar mass of SO3 = 32.1 + 16.0x3 = 80.1 g/mol moles of SO3 = 2.15 mol mass of SO3 = moles of SO3 x molar mass of SO3 = 2.15 x 80.1 = 172 g.

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//tested on eclipse /*************DoublyLinkedList class*******************/ public class DoublyLinkedList { /*Data member variable declaration*/ private Node head; private int count; /**default constructor*/ public DoublyLinkedList() { /*Initializing variable*/ head=null; count =0; } /*Insert at beginning in doubly linked list*/ public void insertBeginning(String name){ /*checking head is null then we will add node at head*/ if(head==null){ head=new Node(name, null, null); count++; }else{ /*else create temp node and setting next to head * and then making temp to head*/ Node temp=new Node(name,null,null); temp.setNext(head); head.setPrev(temp); head=temp; count++; } } public void printList(){ Node temp=head; /*Printing element of doubly list*/ while(temp!=null){ System.out.println(temp.getData()); temp=temp.getNext(); } } /*Returning count from doubly linked list*/ public int count(){ return count; } /*deleting specific node from doubly linked list*/ public void delete(String name){ Node current=head,previous=head; /*if it is head node to be deleted*/ if(head.getData().equals(name)){ current=head.getNext(); current.setPrev(null); head=current; previous=head; }else{ /*If node in middle of list*/ while(current.getNext()!=null){ if(current.getData().equals(name)){ current.getNext().setPrev(previous); previous.setNext(current.getNext()); count--; break; } previous=current; current=current.getNext(); } /*If node is at last*/ if(current.getNext()==null){ if(current.getData().equals(name)){ previous.setNext(null); count--; } } } } /*deleting all nodes from list by assigning head to null * and memory releasing take care by garbage collector*/ public void clear(){ head=null; count=0; } /*Inserting element in alphabetic order*/ public void insertAlpha(String name){ Node current=head,previous=head; /*creating new node*/ Node newNode=new Node(name,null,null); /*if head node is null*/ if(head==null){ head=newNode; count++; }else{ /*please read about compareTo() method about returning value*/ if(head.getData().compareTo(name)>0){ newNode.setNext(head); head=newNode; count++; }else{ /*condition for getting position of node to be inserted*/ while(current!=null&&current.getData().compareTo(name)<=0){ previous=current; current=current.getNext(); } /*if node need to be inserted at last*/ if(current==null){ previous=newNode; }else{ /*If node need to inserted in middle*/ newNode.setNext(current); previous.setNext(newNode); current.setPrev(newNode); newNode.setPrev(previous); } count++; } } } /*Main method start*/ public static void main(String args[]){ /*creating object of doubly linked list*/ DoublyLinkedList doublyLinkedList=new DoublyLinkedList(); doublyLinkedList.insertBeginning(\"sunita\"); doublyLinkedList.insertBeginning(\"lalchand\"); /*calling alphabetic order method*/ doublyLinkedList.insertAlpha(\"ramsingh\"); doublyLinkedList.insertBeginning(\"Lokesh\"); doublyLinkedList.insertBeginning(\"Pritam\"); /*****printing list********/ doublyLinkedList.printL.

tested on eclipseDoublyLinkedList class.pdf

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Please read the comment ins code ExpressionTree.java ---------------------------------- /** * This is the class for Expression Tree. * Used to create Expression Tree and Evaluate it */ /** * Following logic is used to construct a Tree * Here we use stack for Preparing Tree * Loop through given Expression String * If Character is Operand , Create node and push to stack * If Character is Operator then * 1)Create Node for Operator * 2)Pop 2 nodes from Stack and Made * OpretorNode--> left == first node pop * OpretorNode--> right == second node pop * At the end of creation of Expression Tree, Stack have only one node , which is root of Expression tree */ /** Class ExpressionTree **/ class ExpressionTree { /** class TreeNode * Stored Character ==> Digit(0..9) or a Operator +,-,*,/ * Left Node and Right Node * * **/ class TreeNode { char data; TreeNode left, right; /** constructor **/ public TreeNode(char data) { this.data = data; this.left = null; this.right = null; } } /** class StackNode **/ class StackNode { TreeNode treeNode; StackNode next; /** constructor **/ public StackNode(TreeNode treeNode) { this.treeNode = treeNode; next = null; } } private static StackNode top; /** constructor * Constructor takes input string like \"+-+7*935*82*625\" * Input should be in Prefix notation * **/ public ExpressionTree(String expression) { top = null; //Call Method for prepare expression tree buildTree(expression); } /** function to clear tree **/ public void clear() { top = null; } /** function to push a node **/ private void push(TreeNode ptr) { if (top == null) top = new StackNode(ptr); else { StackNode nptr = new StackNode(ptr); nptr.next = top; top = nptr; } } /** function to pop a node * When it find operator pop 2 elements from Stack * * **/ private TreeNode pop() { if (top == null) throw new RuntimeException(\"Underflow\"); else { TreeNode ptr = top.treeNode; top = top.next; return ptr; } } /** function to get top node **/ private TreeNode peek() { return top.treeNode; } /** function to insert character **/ private void insert(char val) { try { //If Operand , Create node and push to Stack if (isDigit(val)) { TreeNode nptr = new TreeNode(val); push(nptr); } //If Operator , Create node and popup 2 elements and make them its right and left else if (isOperator(val)) { TreeNode nptr = new TreeNode(val); nptr.left = pop(); nptr.right = pop(); push(nptr); } } catch (Exception e) { System.out.println(\"Invalid Expression\"); } } /** function to check if digit **/ private boolean isDigit(char ch) { return ch >= \'0\' && ch <= \'9\'; } /** function to check if operator **/ private boolean isOperator(char ch) { return ch == \'+\' || ch == \'-\' || ch == \'*\' || ch == \'/\'; } /** function to convert character to digit **/ private int toDigit(char ch) { return ch - \'0\'; } /** function to build tree from input */ public void buildTree(String eqn) { for (int i = eqn.length() - 1; i >= 0; i--) insert(eqn.charAt(i)); } /** function to evaluate tree */ public dou.

Please read the comment ins codeExpressionTree.java-------------.pdf

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Please find the answers and explanations below: Part A: 1. The ingestion of botulism toxin would best be described as a food poisoning. (ingestion of botulism toxin or any other gut-infecting agent is non-technically termed as food poisoning. It is different from infection in the sense that food poisoning may not always lead to sustained gut infection but diagnosis of gut infection results evidently from food poisoning) 2. The ingestion of a significant number of Clostridium cells would be a infection. (Immunologically, a sufficeint amount of pathogen is required for starting a degrading immune response in the body. This is called the threshold number of the pathogen which can elicit an immune response in the body. When the number of pathogeni cells increases beyond this threshold, the person feels sick and an infection persists) 3.Spoilage is a change in appearance, smell, or taste of a food that makes it undesirable to a consumer. (Technically, spoilage is defined as physcial and chemical change in the appearance,smell, taste or nutritional content of the food due to infection or residence of a pathogen in it.) Part B: For any pathogen to enter and cause infection, some specific steps need to be fulfilled by it. Firstly, the concentration of the pathogen must be large enough to survive in the food item. This infected food item will be ingested by the host and the pathogen will reach the gut. The pathogen survives the harsh environment of the gut and starts replicating. This replication of the pathogen comes with severe side effects on the host which include physical and physiological deterioration of the gut lining along with eliciting immune response in the host. This ultimately causes infection/disease. So, the correct order of events would be: 1. The potatoes were contaminated with soil containing Clostridium botulinum endospores. 2. The toxin remained in the pruno after straining. 3. The conditions in the bag of pruno were favorable for germination. 4. Clostridium botulinum replicated and increased the number of organisms in the pruno. 5. Bacterial cells produced and secreted toxin. 6. The inmates drank the contaminated pruno and became ill. Solution Please find the answers and explanations below: Part A: 1. The ingestion of botulism toxin would best be described as a food poisoning. (ingestion of botulism toxin or any other gut-infecting agent is non-technically termed as food poisoning. It is different from infection in the sense that food poisoning may not always lead to sustained gut infection but diagnosis of gut infection results evidently from food poisoning) 2. The ingestion of a significant number of Clostridium cells would be a infection. (Immunologically, a sufficeint amount of pathogen is required for starting a degrading immune response in the body. This is called the threshold number of the pathogen which can elicit an immune response in the body. When the number of pathogeni cells increases beyond this threshold, the pers.

Please find the answers and explanations belowPart A1. The ing.pdf

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nSolutionn.pdf

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Mass of liquid = 22.5 gHeat added = 2.34 kJ = 2340 JTemperature .pdf

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It is given that we should use 4 bit input, i.e we will have a set of 16 different inputs. The output of the circuit will be 1 if and only if the decimal equivalent formed by the 4 input bits is an integer that belongs to the fibonacci series. a) The truth table according to the above requirement is A B C D O 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 B) K - map for the above truth table is c) According to the above K-map, prime implicants would be A\'B\', B\'C\'D\' and C\'D d) All the prime implicants are essential prime implicants e) A\'B\'+B\'C\'D\'+C\'D A B C D O 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 Solution It is given that we should use 4 bit input, i.e we will have a set of 16 different inputs. The output of the circuit will be 1 if and only if the decimal equivalent formed by the 4 input bits is an integer that belongs to the fibonacci series. a) The truth table according to the above requirement is A B C D O 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 B) K - map for the above truth table is c) According to the above K-map, prime implicants would be A\'B\', B\'C\'D\' and C\'D d) All the prime implicants are essential prime implicants e) A\'B\'+B\'C\'D\'+C\'D A B C D O 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0.

It is given that we should use 4 bit input, i.e we will have a set o.pdf

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IndexingConcatinationSolutionIndexingConcatination.pdf

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import java.util.ArrayList; import java.util.List; public class H2ClassA { List list = new ArrayList (); public static void main(String[] args) { H2ClassA y = new H2ClassA (); int [] v = {4, 3, 7, 5, 99, 3}; // change the for loop to old type of for loop for(int i=0;i Solution import java.util.ArrayList; import java.util.List; public class H2ClassA { List list = new ArrayList (); public static void main(String[] args) { H2ClassA y = new H2ClassA (); int [] v = {4, 3, 7, 5, 99, 3}; // change the for loop to old type of for loop for(int i=0;i.

import java.util.ArrayList; import java.util.List;public class H.pdf

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i dont knowSolutioni dont know.pdf

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Heterochromatin is a tightly packed form of DNA, which comes in multiple varieties. constitutive heterochromatin the chromatin in regions of the chromosomes that are invariably heterochromatic; it contains highly repetitive sequences of DNA that are genetically inactive and serves as a structural element of the chromosome. facultative heterochromatin the chromatin in regions of the chromosomes that become heterochromatic in certain cells and tissues; for example, it makes up the inactive X chromosome in female somatic cells. Homologous chromosomes are chromosome pairs, one from each parent, that are similar in length, gene position and centromere location. Duplicated chromosome The occurrence of a repeated section of genes in a chromosome. Haploid is Having a single set of unpaired chromosomes, as in germ cell, or half the diploid number of a somatic cell. Aneuploidy is the presence of an abnormal number of chromosomes in a cell, example : a human cell having 45 or 47 chromosomes instead of the usual 46. Nondisjunction is the failure of one or more pairs of homologous chromosomes or sister chromatids to separate normally during nuclear division, usually resulting in an abnormal distribution of chromosomes in the daughter nuclei. Solution Heterochromatin is a tightly packed form of DNA, which comes in multiple varieties. constitutive heterochromatin the chromatin in regions of the chromosomes that are invariably heterochromatic; it contains highly repetitive sequences of DNA that are genetically inactive and serves as a structural element of the chromosome. facultative heterochromatin the chromatin in regions of the chromosomes that become heterochromatic in certain cells and tissues; for example, it makes up the inactive X chromosome in female somatic cells. Homologous chromosomes are chromosome pairs, one from each parent, that are similar in length, gene position and centromere location. Duplicated chromosome The occurrence of a repeated section of genes in a chromosome. Haploid is Having a single set of unpaired chromosomes, as in germ cell, or half the diploid number of a somatic cell. Aneuploidy is the presence of an abnormal number of chromosomes in a cell, example : a human cell having 45 or 47 chromosomes instead of the usual 46. Nondisjunction is the failure of one or more pairs of homologous chromosomes or sister chromatids to separate normally during nuclear division, usually resulting in an abnormal distribution of chromosomes in the daughter nuclei..

Heterochromatin is a tightly packed form of DNA, which comes in mult.pdf

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H2A (ACID) Molarity = 0.15 M Normality N1 = Molarity x Basicity of acid = 0.15x 2 = 0.3 N Volume V1 = 100 ml N1V1 = 0.3 x 100 = 30 NaOH (BASE) Molarity = 0.16 M Normality N2 = Molarity xAcidity of Base = 0.16 x 1 = 0.16 N Volume V2 = 100 ml N2V2 = 0.16X100 = 16 Since concentration of ACID is more than concentration of BASE, resulatant concentration is equal to H+ ion concentration. Normality at eqivalance point N = (N1V1 - N2V2)/ (V1+V2) = 30 - 16/(100+100) = 14/200 = 0.07 N = H+ ion concentartion Solution H2A (ACID) Molarity = 0.15 M Normality N1 = Molarity x Basicity of acid = 0.15x 2 = 0.3 N Volume V1 = 100 ml N1V1 = 0.3 x 100 = 30 NaOH (BASE) Molarity = 0.16 M Normality N2 = Molarity xAcidity of Base = 0.16 x 1 = 0.16 N Volume V2 = 100 ml N2V2 = 0.16X100 = 16 Since concentration of ACID is more than concentration of BASE, resulatant concentration is equal to H+ ion concentration. Normality at eqivalance point N = (N1V1 - N2V2)/ (V1+V2) = 30 - 16/(100+100) = 14/200 = 0.07 N = H+ ion concentartion.

H2A (ACID) Molarity = 0.15 M Normality N1 = Mo.pdf

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False , we should not always rename the worksheet.SolutionFals.pdf

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Difference between computer architecture and computer organization: A computer architecture refers as a set of attributes of a system as seen by programmer.The term \"computerarchitecture\" was originated by IBM.It was essentially acontract with software stating unambiguously what thehardware does. Computer oraganization was usually high-level description of the logic, memorey ....etc used to implement that contract.Mostly formulating what a cache,RAM,different I/O busses ,history of computers etc. a.) An instruction set of architecture is called as ISA.It is a part of computer that applicable to programming, which is basically machine language.All CPUs have instruction sets that enable commands to the processor directiong the CPU to switch the relevant transistors.The instructions setswere:reduced(RISC) and complex (CISC). CISC:Complex instruction set computer is rooted in the history of computing. RISC:Reduced instruction set computer was realized in the 1970s by IBM. b.) Memory to memor add13 x,y and z #x,y and z are memory addresses mov1 eax , x add1 eax , y mov1 z , eax load store: lw $t0, x lw $t1, y add $t0, $t0, $t1 sw $t0, z Accumulator-based: load x # AC x add y # AC AC + y store z # z - AC c.)The size size of instuction part of an ISA are 32 bits long. opcode tells the machine which format so add r1, r2, r3 has: ---opcode =0,fanct=32,rs=2,rt=3,rd=1,sa=0 Solution Difference between computer architecture and computer organization: A computer architecture refers as a set of attributes of a system as seen by programmer.The term \"computerarchitecture\" was originated by IBM.It was essentially acontract with software stating unambiguously what thehardware does. Computer oraganization was usually high-level description of the logic, memorey ....etc used to implement that contract.Mostly formulating what a cache,RAM,different I/O busses ,history of computers etc. a.) An instruction set of architecture is called as ISA.It is a part of computer that applicable to programming, which is basically machine language.All CPUs have instruction sets that enable commands to the processor directiong the CPU to switch the relevant transistors.The instructions setswere:reduced(RISC) and complex (CISC). CISC:Complex instruction set computer is rooted in the history of computing. RISC:Reduced instruction set computer was realized in the 1970s by IBM. b.) Memory to memor add13 x,y and z #x,y and z are memory addresses mov1 eax , x add1 eax , y mov1 z , eax load store: lw $t0, x lw $t1, y add $t0, $t0, $t1 sw $t0, z Accumulator-based: load x # AC x add y # AC AC + y store z # z - AC c.)The size size of instuction part of an ISA are 32 bits long. opcode tells the machine which format so add r1, r2, r3 has: ---opcode =0,fanct=32,rs=2,rt=3,rd=1,sa=0.

Difference between computer architecture and computer organization.pdf

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Coupling: Coupling is the act of combination of two things jointly. In software development, coupling submits to the degree to which software components are dependent ahead each other. For example, in a tightly-coupled architecture, every component and its associated components should be present within order for code to be performed or compiled. In a loosely coupled architecture, components can stay autonomous and allow middleware software to manage communication among them. In a decoupled architecture, the components can work completely separately and autonomously. Portability: An application that can be changed from one computer environment to another. An application amassed on a USB flash drive that can run on a computer with no leaving any trace of it at what time the application is finished. It is designed to reinstate all changed settings when the program is closed. For example, in general, software written in Java has this capability. Solution Coupling: Coupling is the act of combination of two things jointly. In software development, coupling submits to the degree to which software components are dependent ahead each other. For example, in a tightly-coupled architecture, every component and its associated components should be present within order for code to be performed or compiled. In a loosely coupled architecture, components can stay autonomous and allow middleware software to manage communication among them. In a decoupled architecture, the components can work completely separately and autonomously. Portability: An application that can be changed from one computer environment to another. An application amassed on a USB flash drive that can run on a computer with no leaving any trace of it at what time the application is finished. It is designed to reinstate all changed settings when the program is closed. For example, in general, software written in Java has this capability..

CouplingCoupling is the act of combination of two things jointly..pdf

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D. A trademark is a symbol, word, name, sound, or thing that identif.pdf

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Benefits of using VHDL Executable specification It is often reported that a large number of ASIC designs meet their specifications first time, but fail to work when plugged into a system. VHDL allows this issue to be addressed in two ways: A VHDL specification can be executed in order to achieve a high level of confidence in its correctness before commencing design, and may simulate one to two orders of magnitude faster than a gate level description. A VHDL specification for a part can form the basis for a simulation model to verify the operation of the part in the wider system context (eg. printed circuit board simulation). This depends on how accurately the specification handles aspects such as timing and initialization. Solution Benefits of using VHDL Executable specification It is often reported that a large number of ASIC designs meet their specifications first time, but fail to work when plugged into a system. VHDL allows this issue to be addressed in two ways: A VHDL specification can be executed in order to achieve a high level of confidence in its correctness before commencing design, and may simulate one to two orders of magnitude faster than a gate level description. A VHDL specification for a part can form the basis for a simulation model to verify the operation of the part in the wider system context (eg. printed circuit board simulation). This depends on how accurately the specification handles aspects such as timing and initialization..

Benefits of using VHDLExecutable specificationIt is often report.pdf

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Answer C. Fingertip Two-point discrimination measures the individual’s ability to perceive two points of stimuli presented simultaneously. It is the ability to discern that two nearby objects touching the skin are truly two distinct points, not one. The fingertips have resolution, so, low errors Solution Answer C. Fingertip Two-point discrimination measures the individual’s ability to perceive two points of stimuli presented simultaneously. It is the ability to discern that two nearby objects touching the skin are truly two distinct points, not one. The fingertips have resolution, so, low errors.

AnswerC. FingertipTwo-point discrimination measures the individu.pdf

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Addition of H2O2 after ozonolysis oxidises the product so the unoxidised product is 7-formaldehyde 2 heptanone now removing the oxygens and joining the carbon we get 1-methylcyclohex-1-ene option 4 or d Solution Addition of H2O2 after ozonolysis oxidises the product so the unoxidised product is 7-formaldehyde 2 heptanone now removing the oxygens and joining the carbon we get 1-methylcyclohex-1-ene option 4 or d.

Addition of H2O2 after ozonolysis oxidises the productso the unoxi.pdf

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Add mCPBA to make the epoxide, then use aqueous N.pdf

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1log 0.01= -22[H+]=1.6 x 10 -2pH=-log[H+]=-log(1.610-.pdf

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