C) 0.3275SolutionC) 0.3275.pdf

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The periodic table of the chemical elements (also known as the periodic table or periodic table of the elements) is a tabular display of the 118 known chemical elements organized by selected properties of their atomic structures. Elements are presented by increasing atomic number, the number of protons in an atom\'s atomic nucleus. While rectangular in general outline, gaps are included in the horizontal rows (known as periods) as needed to keep elements with similar properties together in vertical columns (known as groups), e.g. alkali metals, alkali earths, halogens, noble gases. Periods A period is a horizontal row in the periodic table. Although groups are the most common way of classifying elements, there are some regions of the periodic table where the horizontal trends and similarities in properties are more significant than vertical group trends. This can be true in the d- block (or \"transition metals\"), and especially for the f-block, where the lanthanides and actinides form two substantial horizontal series of elements. Periodic trend for ionization energy. Each period begins at a minimum for the alkali metals, and ends at a maximum for the noble gases. Elements in the same period show trends in atomic radius, ionization energy, electron affinity, and electronegativity. Moving left to right across a period, atomic radius usually decreases. This occurs because each successive element has an added proton and electron which causes the electron to be drawn closer to the nucleus. This decrease in atomic radius also causes the ionization energy to increase when moving from left to right across a period. The more tightly bound an element is, the more energy is required to remove an electron. Electronegativity increases in the same manner as ionization energy because of the pull exerted on the electrons by the nucleus. Electron affinity also shows a slight trend across a period. Metals (left side of a period) generally have a lower electron affinity than nonmetals (right side of a period) with the exception of the noble gases. Hope this explains your daughter.... :) Solution The periodic table of the chemical elements (also known as the periodic table or periodic table of the elements) is a tabular display of the 118 known chemical elements organized by selected properties of their atomic structures. Elements are presented by increasing atomic number, the number of protons in an atom\'s atomic nucleus. While rectangular in general outline, gaps are included in the horizontal rows (known as periods) as needed to keep elements with similar properties together in vertical columns (known as groups), e.g. alkali metals, alkali earths, halogens, noble gases. Periods A period is a horizontal row in the periodic table. Although groups are the most common way of classifying elements, there are some regions of the periodic table where the horizontal trends and similarities in properties are more significant than vertical group trends. This can be t.

The periodic table of the chemical elements (also known as the perio.pdf

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The answer is c, Polonium. An alpha particle contains two protons hence the original atom must have an atomic number 2 greater than Pb (lead). Looking at the periodic table we can see that that element is Po. Solution The answer is c, Polonium. An alpha particle contains two protons hence the original atom must have an atomic number 2 greater than Pb (lead). Looking at the periodic table we can see that that element is Po..

The answer is c, Polonium. An alpha particle contains two protons .pdf

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Solution : class IntNode { int data; public IntNode next,head; public IntNode(int value) { data=value; next=null; } public void setNext(IntNode n) { next=n; } public int getData() { return data; } public IntNode getNext() { return(next); } } public class CircularLinkedList { public static void main(String[] args) { //test your code here } //private variables: private IntNode head; private IntNode tail; //constructor public CircularLinkedList(){ //empty linked list head=null; tail=null; } //add a node to the end of the list public void add(int value){ //create new Node IntNode newNode=new IntNode(value); //if list is empty if (isEmpty()){ //set new node as head head=newNode; //set next link to the head newNode.setNext(head); //set tail as the head tail=head; }else{ //set next link of the tail to the new node tail.setNext(newNode); //set next link of the new node to the head newNode.setNext(head); //update tail - new node is the new list tail tail=newNode; } } //check if linked list is empty public boolean isEmpty(){ if (head ==null) return true; else return false; } //print the values in the linked list public void print(){ //if list is empty if(isEmpty()){ System.out.println(\"empty\ \"); return; } //if list has one node if (head.getNext() == head) { System.out.println(head.getData()); return; } //start from the head node IntNode currentNode=head; while(currentNode.getNext() != head){ //print the value of the current node System.out.print(currentNode.getData() +\"\\t\"); //go to the next node currentNode=currentNode.getNext(); } System.out.print(currentNode.getData()); System.out.println(); } public IntNode delete() { if(head.next == null) tail = null; IntNode temp = head; head = head.next; return temp; } // end delete }.

Solutionclass IntNode { int data; public IntNode next,head;.pdf

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import turtle import time import random def draw_rectangle(length, height): turtle.up() x = -150 y = 150 C = height*(7/13) D = length*(2/5) L = stripe_width = float(round(height/13,1)) ## Draw rectangle first. turtle.color(0,0,0) turtle.begin_fill() turtle.setpos(x,y) turtle.down() turtle.forward(length) turtle.right(90) turtle.forward(height) turtle.right(90) turtle.forward(length) turtle.right(90) turtle.forward(height) turtle.end_fill() ## Then draw the stripes. x1 = -150 y1 = 150-L for z in range(13): if z%2 == 0: r = s = t = 0 else: r = s = t = 1 turtle.up() turtle.speed(100) turtle.setpos(x1,y1) turtle.setheading(90) turtle.down() turtle.color(r,s,t) turtle.begin_fill() turtle.forward(L) turtle.right(90) turtle.forward(length) turtle.right(90) turtle.forward(L) turtle.right(90) turtle.forward(length) turtle.end_fill() y1 -= L ## Finally draw the stars rectangle overlapping the stripes, next is stars. x2 = -150+D y2 = 150.5-C turtle.up() turtle.setpos(x2,y2) turtle.down() turtle.color(0,0,0) turtle.begin_fill() turtle.forward(D) turtle.right(90) turtle.forward(C) turtle.right(90) turtle.forward(D) turtle.right(90) turtle.forward(C) turtle.end_fill() turtle.up() turtle.bye draw_star(-length, height) def draw_star(l, h): for z in range(50): if z < 7: row = 140 draw_starrows(row) if z < 14: row = row - 20 draw_starrows(row) if z < 21: row = row - 20 draw_starrows(row) if z < 28: row = row - 20 draw_starrows(row) if z < 35: row = row - 20 draw_starrows(row) ## This gets the turtle pen out of the way at the very end. turtle.up() turtle.setpos(-180,100) break def draw_starrows(row): x = -160 y = 150 for z in range(10): x += 15 turtle.up() turtle.color(1,1,1) turtle.speed(100) turtle.setpos(x,row) turtle.begin_fill() turtle.down() turtle.forward(6.154) turtle.left(144) turtle.forward(6.154) turtle.left(144) turtle.forward(6.154) turtle.left(144) turtle.forward(6.154) turtle.left(144) turtle.forward(6.154) turtle.left(144) turtle.end_fill() turtle.bye ##def get_color(): ## r = g = b = 0 ## color = r = g = b ## return color def draw_flag(): A = 200 height = int(A) ## length = height*1.9 ## C = height*(7/13) ## D = length*(2/5) ## E = F = union_height/10 ## G = H = union_length/12 ## stripe_width = height/13 ## diameter_star = stripe_width*(4/5) draw_rectangle(height*1.9, height) draw_flag() Here is the get_color solution: def get_color(color2): ## If color2 equals 1, then make the color white. if color2 == 1: r = g = b = 1 return (r, g, b) ## If color2 equals 0, then make the color red. if color2 == 0: r = 1 g = 0 b = 0 return (r, g, b) ## If color2 equals 2, then make the color black. if color2 == 2: r = 0 g = 0 b = 1 return (r, g, b) Solution import turtle import time import random def draw_rectangle(length, height): turtle.up() x = -150 y = 150 C = height*(7/13) D = length*(2/5) L = stripe_width = float(round(height/13,1)) ## Draw rectangle first. turtle.color(0,0,0) turtle.begin_fill() turtle.setpos(x,y) turtle.down() turtle.forward(length) t.

import turtleimport timeimport randomdef draw_rectangle(length.pdf

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Effects of NaCl Concentration on Red Blood Cell Volume A red blood cell (rbc) will attempt to regulate its volume when placed in a solution of impermeable substances dissolved in water (Randall, 1997). A solution is considered to be isosmotic when it has an osmolality equal to that of normal plasma (Freedman, 1998). The osmolarity is the osmotic pressure that a cell effectively has exerted upon it (Randall, 1997). The osmotic pressure is the pressure that is created by osmosis between different solutions that are separated by a semipermeable membrane (Randall, 1997). If a rbc is placed in an isotonic solution, the cell will maintain its normal volume because no osmotic pressure is developed. A hypotonic solution has a lower concentration of solutes than the interior of the cell. Therefore, a hypotonic solution will cause water to flow into the cell, as the concentration of solutes inside of the cell is higher. If this water influx continues for a long period of time (with ineffective (or absence of) volume regulation by the cell), the cell my lyse (burst) as the pressure builds up inside to a magnitude greater than can be handled by the cellular membrane. here are a number of ways in which red blood cells (and animal cells in general) cope with changes in solute concentration on the outside of their membrane surfaces. There may be a volume regulatory decrease (RVD) or increase (RVI) depending upon the concentration of the solutes on the outside of the cellular membrane (Freedman, 1998). The response that is activated by the swelling of the cells usually involves an increase in the K+/Cl- transporters (see figure 2) in ducks, sheep, rabbits, and some other mammals (Freedman, 1998; Motais et al, 1997). The swelling response may also trigger the formation of a channel (in which band 3 is involved) that allows osmolytes such as taurine (and other small, organic, osmotically active solutes) to be transported out of the cell (Freedman, 1998; Motais et al, 1997). The shrink activated response in red blood cells involves the activation of Na+/K+/2 Cl- cotransporters (in general, and in the case of the duck (Freedman, 1998)). The molarity in plasma of NaCl is approximately 0.15M, therefore the concentration of 0.15M NaCl treatment would be considered isosmotic to red blood cell plasma. The change in concentration to 0.12 M NaCl would be a hypotonic condition, and the conditions of 0.21, 0.33, 0.157, and 1.05M NaCl would be hypertonic conditions. Solution Effects of NaCl Concentration on Red Blood Cell Volume A red blood cell (rbc) will attempt to regulate its volume when placed in a solution of impermeable substances dissolved in water (Randall, 1997). A solution is considered to be isosmotic when it has an osmolality equal to that of normal plasma (Freedman, 1998). The osmolarity is the osmotic pressure that a cell effectively has exerted upon it (Randall, 1997). The osmotic pressure is the pressure that is created by osmosis between different soluti.

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Answer:- Possible types of audit for this project: 1. Financial audit - It shows acccuracy, fairness and where the financial portion used and how much. As in this it show all the financial records which shows relaibility of data. 2. Operational audit - it is future oriented and independent evalution of organisational activites, in this internal activites and effeciency and control is evaluated. 3. Investigavtive audit - In this audit evaluted on the suspicious matters which may be include with individual or departement, all the members confidentially repord to director about this suspcious matter. Steps in Audit process:- 1. scheduling the conference to audit objective, timing, purpose, format and work distribution. 2. Assesing the relability of documents and internal effeciency and system. 3. Testing internal control for ensuring proper operations. 4. Disscusing all the thing with management all observation 5. draft a report according to responses of the queries. 6. Following critical issue which raised in audit report, that this is resolved or not. Solution Answer:- Possible types of audit for this project: 1. Financial audit - It shows acccuracy, fairness and where the financial portion used and how much. As in this it show all the financial records which shows relaibility of data. 2. Operational audit - it is future oriented and independent evalution of organisational activites, in this internal activites and effeciency and control is evaluated. 3. Investigavtive audit - In this audit evaluted on the suspicious matters which may be include with individual or departement, all the members confidentially repord to director about this suspcious matter. Steps in Audit process:- 1. scheduling the conference to audit objective, timing, purpose, format and work distribution. 2. Assesing the relability of documents and internal effeciency and system. 3. Testing internal control for ensuring proper operations. 4. Disscusing all the thing with management all observation 5. draft a report according to responses of the queries. 6. Following critical issue which raised in audit report, that this is resolved or not..

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Ans to Q4. Colloid is a particular state of matter. A colloidal system is a two-phase heterogeneous system consisting of two separate phases with distinct boundaries between them, viz., a continuous phase or dispersion medium made of one substance and a disperse phase of particles of another substance dispersed in the continuous phase. In contrast, a true solution is a single-phase homogeneous system without any physically demonstrable boundary between the solvent and the solute particles. In fact, dispersed particles of a colloidal system, though submicroscopic, are much larger than the solute particles of a true solution. So, they can be separated from true solutes and the dispersion medium by ultracentrifugation and dialysis. Dispersed colloidal particles are much smaller than the particles of a suspension; so, they do not sediment under gravity alone, and are not held back by ordinary filter papers. Ans to Q5. Natural Organic Matter (NOM) is the organic material present in surface or ground water. It includes both humic and non-humic fractions. The humic fraction includes high molecular weight organic molecules such as humic and fulvic acids. These substances with tanic acid are the major fraction of Dissolved Organic Matter (DOM) in water. These substances are oxidized very slowly and their solubility in water may vary with pH. The dissolved fraction of NOM may not be fully removed using conventional water treatment practices and have been shown to produce by-products such as trihalomethane during disinfection. On-line analysis of dissolved NOM is important for proper monitoring and control of enhanced or special treatment processes designed to remove prior to disinfection. An increase in the amount of NOM has been observed over the past 10-20 years in raw water supplies in several areas, which has a significant effect on drinking water treatment. The presence of NOM causes many problems in drinking water and drinking water treatment processes such as: i) negetive effect on water quality by causing colour, taste and odor problems, ii) increased coagulant and disinfectant doses which in turn results in increased sludge volumes and production of harmful disinfectant by-products, iii) promoted biological growth in distribution system and iv) increased levels of complexed heavy metals and absorbed organic pollutants. Solution Ans to Q4. Colloid is a particular state of matter. A colloidal system is a two-phase heterogeneous system consisting of two separate phases with distinct boundaries between them, viz., a continuous phase or dispersion medium made of one substance and a disperse phase of particles of another substance dispersed in the continuous phase. In contrast, a true solution is a single-phase homogeneous system without any physically demonstrable boundary between the solvent and the solute particles. In fact, dispersed particles of a colloidal system, though submicroscopic, are much larger than the solute particles of a true solut.

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Amphipathic means that the molecule has a polar, hydrophillic end and a non-polar hydrophobic end. In water, amphipathic molecules tend to form micelles, with the hydrophillic ends outward and the hydrophobic ends inward. Solution Amphipathic means that the molecule has a polar, hydrophillic end and a non-polar hydrophobic end. In water, amphipathic molecules tend to form micelles, with the hydrophillic ends outward and the hydrophobic ends inward..

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Moran Enterprises Inc. (All values in $) Multi Step Income Statement for the year ended 31 December 2012 Sales Revenue 220000 Cost of Goods Sold 120000 Gross Profit 100000 Less : Operating Expenses 47000 Net Profit before Interest, Depreciation 53000 and Taxation Less : Depreciation Expense 0 Less : Interest Expense 5000 5000 Net Profit before Taxation 48000 Taxation Expense 10000 Profit after Taxation 38000 Dividends Paid 14000 Transferred to Retained Earnings 24000 Classified Balance Sheet as on 31 December 2012 Current Assets Accounts Receivable 24000 Inventory 33000 Cash 12000 Total Current Assets 69000 Tangible and Intangible Assets Equipment, Net of Depreciation 75000 Total Assets 144000 Current Liabilities Accounts Payable 15000 Unearned Revenue 21000 Total Current Liabilities 36000 Equity and Retained Earnings Capital Stock 65000 Retained Earnings Opening Balance 19000 Profit for the year 24000 43000 Total Liabilities and Equity 144000 Solution Moran Enterprises Inc. (All values in $) Multi Step Income Statement for the year ended 31 December 2012 Sales Revenue 220000 Cost of Goods Sold 120000 Gross Profit 100000 Less : Operating Expenses 47000 Net Profit before Interest, Depreciation 53000 and Taxation Less : Depreciation Expense 0 Less : Interest Expense 5000 5000 Net Profit before Taxation 48000 Taxation Expense 10000 Profit after Taxation 38000 Dividends Paid 14000 Transferred to Retained Earnings 24000 Classified Balance Sheet as on 31 December 2012 Current Assets Accounts Receivable 24000 Inventory 33000 Cash 12000 Total Current Assets 69000 Tangible and Intangible Assets Equipment, Net of Depreciation 75000 Total Assets 144000 Current Liabilities Accounts Payable 15000 Unearned Revenue 21000 Total Current Liabilities 36000 Equity and Retained Earnings Capital Stock 65000 Retained Earnings Opening Balance 19000 Profit for the year 24000 43000 Total Liabilities and Equity 144000.

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