...difference between the 2 types of standards is that the primary standard is a has powerful reactants and isn't sensitive to the the environment and the secondary standard is something that is determined will react with a highly pure primary standard that can be standardized. It was necessary to standardize the NaOH solution because NaOH absorbs moisture from the air, making the compound not 100% NaOH, so to obtain precise concentrations, the NaOH needed to be standardized with KHP. The molarity of the NaOH solution was determined by dividing the average number of moles of NaOH (determined by taking the moles of KHP and using the 1:1 ratio) and diving it by the average volume (number of L) at the indicator point. The difference between the equivalence point and the indicator point is that the indicator point is where the indicator changes color, and the equivalent point is the place in a titration where the amount of indicator added neutralizes the solution. Yes the calculated molarities of NaOH for the indicator method and 2nd derivative plot method were within a reasonable range of each other. The molarities calculated for each method were the same (0.5138M) The actual molarity of the NaOH solution was 0.5138, which is in the the 15% range. To determine the molarity of NaOH, the 2nd derivative method was used because it is...
Words: 655 - Pages: 3
...Exercises 7.1: Work: Let A = {a1,a2,…an} If R is a relation on A , R AxA and if R is antisymmetric also: If (ai,aj)R and (aj,ai)R then a i = a j Then if (ai,aj)R with i j then (aj,ai) R So we can construct R where Ris maximum taking: R = {(ai,ai) , 1 i n} U {(ai,aj) , 1 i < j n} R= n + n2 = n+n(n-1)/2 = n[1+(n-1)/2] = n(n+1)/2 Answer: Maximum value for R= n(n+1)/2 and it´s only 1 antisymmetric relation with this property (the one constructed above) Exercises 7.2: Work: We know that (a,a)R for all aA Since (a,a) R and (a,a) R if we apply definition of R2 (a,a) R2 for all aA then R2 is a reflexive relation on the set A Work: Each entry is 0 or 1 and we must count how many matrices 6x6 satisfy A = A tr The number of matrices that A = Atr is equal to the number of matrices that i construct selecting the entries in the diagonal and above. 1+2+3+4+5+6 = 21 Answer: 21 Exercise 7.3: a) Relation matrix is M=1 1 1 1 10 1 0 1 10 0 1 1 10 0 0 1 10 0 0 0 1 b) I will do it by hand c) Work: Using the Fig 7.23 For k = 1 e is the only vertex that has no edges starting at it Then we remove e Result: e For k = 2 d is the only vertex that has no edges starting at it Then we remove d Result: d < e For k = 3 b and c are the only vertices that has no edges starting at it we select c (it can be b too) Then we remove c Result: c<d<e For k = 4 b is the only vertex that has no edges starting at it Then...
Words: 2227 - Pages: 9
...Week 3 homework Exercise 7.1, problem 10a The total number of pairs we can either include or not is 4^2-4=16. Any reflexive relation is a subset of this set of 12 elements; we know there are 2^12 such subsets. Problem 10b The number of decisions we can make for any symmetric relation is 4+ (16-4)/2=4+6=10. The number of possible symmetric relations is 210. Exercise 7.2, Problem 15a a) Draw the digraph G1 (V1, E1) where V1 {a, b, c, d, e, f } and E1 {(a, b), (a, d), (b, c), (b, e), (d, b), (d, e), (e, c), (e, f), (f, d)}. Exercise 7.3, Problem 1 Draw the Hasse diagram for the poset ⊆, where{1, 2, 3, 4}. (1,1)<(1,2)<(1,3)<(1,4)<(2,1)<(2,2)<...<(4,3)<(4,4 ). o------o------o------o------o------o--- ... ---o------o (1,1) (1,2) (1,3) (1,4) (2,1) (2,2) ... (4,3) (4,4) 3 1 2 4 3 1 2 4 Problem 6a . For A {a, b, c, d, e}, the Hasse diagram for the poset (A, R) is shown in Fig. 7.23. (a) Determine the relation matrix for R. a b c d e a 1 1 1 0 0 b 1 1 0 1 0 c 1 0 1 1 0 d 0 1 1 1 0 e 1 0 0 0 1 Exercise 7.4, Problem 1a Determine whether each of the following collections of sets is a partition for the given set A. If the collection is not a partition, explain why it fails to be. a) A {1, 2, 3, 4, 5, 6, 7, 8}; A1 {4, 5, 6}, A2 {1, 8}, A3 {2, 3, 7}. The Question A is a partition. Problem 2a There are only three ways, corresponding to the partition A_1 = {1,2,8}, A_2 = {3,4}, A_3 = {5...
Words: 947 - Pages: 4
...Mark Martin 103 ENGL 1001 Dr. Chege 28 October 2014 “Is Affirmative Action Still Relevant In 21st Century America?” My topic is Affirmative Action, which is discrimination, and whether or not it is still happening in 21st century America. I am interested in it because I am interested in history and Affirmative Action is a big part of this country. Any other people who are either curious about Affirmative Action or anybody that likes history in general I think would be interested in this topic. I’m so invested in this topic that I want to research it is because like I said earlier, I like history and Affirmative Action is a big part of this country whether you agree with it or disagree with it. Now, my research question is, is affirmative action still relevant in 21st century America? My chosen topic relates to the course theme of Justice because Affirmative Action is not fair or equal. It’s being in favor of one person over another because of gender or skin color and that’s Justice because it’s not right to do that at all no matter what. I am connecting the idea of Justice to my topic because I’m going to talk about how it was in the past, how and it’s happening right now, and how we can end it for the future. I’m going to talk about how it shouldn’t happen in the first place and how people are trying to stop it to bring equality to everybody and bring Justice to the matter. My question is significant because this is something that is huge in the history of the United...
Words: 326 - Pages: 2
... The reaction is generally carried out in a flask containing the liquid or dissolved sample. Titrant solution is volumetrically delivered to the reaction flask using a burette. Delivery of the titrant is called a titration. The titration is complete when sufficient titrant has been added to react with all the analyte. This is called the equivalence point. An indicator is often added to the reaction flask to signal when all of the analyte has reacted. The titrant volume where the signal is generated is called the end point. The equivalence and end points are rarely the same. SUCCESSFUL TITRATIONS A few rules of thumb for designing a successful titration are: 1. The titrant should either be a standard or should be standardized. 2. The reaction should proceed to a stable and well defined equivalence point. 3. The equivalence point must be able to be detected. 4. The titrant’s and sample’s volume or mass must be accurately known. 5. The reaction must proceed by a definite chemistry. There should be complicating side reactions. 6. The reaction should be nearly complete at the equivalence point. In other words, chemical equilibrium favors products. 7. The reaction rate should be fast enough to be practical. TYPES OF TITRATIONS 1. ACID-BASE TITRATION a. These titrations are based on the neutralization reaction that occurs between an acid and a base, when mixed in solution. b. The acid (resp....
Words: 402 - Pages: 2
...1. a specific task or duty assigned to a person or group of people their mission was to irrigate the desert 2. a person's vocation (often in the phrase mission in life) 3. (Government, Politics & Diplomacy) a group of persons representing or working for a particular country, business, etc., in a foreign country 4. (Government, Politics & Diplomacy) a. a special embassy sent to a foreign country for a specific purpose b. US a permanent legation 5. (Christianity / Ecclesiastical Terms) a. a group of people sent by a religious body, esp a Christian church, to a foreign country to do religious and social work 1 : a fair return or equivalent in goods, services, or money for something exchanged 2 : the monetary worth of something : market price 3 : relative worth, utility, or importance 4 : a numerical quantity that is assigned or is determined by calculation or measurement 5 : the relative duration of a musical note 6 a: relative lightness or darkness of a color : luminosity b: the relation of one part in a picture to another with respect to lightness and darkness 7 : something (as a principle or quality) intrinsically valuable or desirable 8 : denomination 2 — val·ue·less \-(ˌ)yü-ləs, -yə-\adjective — val·ue·less·nessnoun See value defined for English-language learners » See value defined for kids » Examples of VALUE 1. The company's stock continues to decline in value. 2. Real estate prices have doubled in value over the last decade...
Words: 282 - Pages: 2
...Exercises 7.1: Work: Let A = {a1,a2,…an} If R is a relation on A , R AxA and if R is antisymmetric also: If (ai,aj)R and (aj,ai)R then a i = a j Then if (ai,aj)R with i j then (aj,ai) R So we can construct R where Ris maximum taking: R = {(ai,ai) , 1 i n} U {(ai,aj) , 1 i < j n} R= n + n2 = n+n(n-1)/2 = n[1+(n-1)/2] = n(n+1)/2 Answer: Maximum value for R= n(n+1)/2 and it´s only 1 antisymmetric relation with this property (the one constructed above) Exercises 7.2: Work: We know that (a,a)R for all aA Since (a,a) R and (a,a) R if we apply definition of R2 (a,a) R2 for all aA then R2 is a reflexive relation on the set A Work: Each entry is 0 or 1 and we must count how many matrices 6x6 satisfy A = A tr The number of matrices that A = Atr is equal to the number of matrices that i construct selecting the entries in the diagonal and above. 1+2+3+4+5+6 = 21 Answer: 21 Exercise 7.3: a) Relation matrix is M=1 1 1 1 10 1 0 1 10 0 1 1 10 0 0 1 10 0 0 0 1 b) I will do it by hand c) Work: Using the Fig 7.23 For k = 1 e is the only vertex that has no edges starting at it Then we remove e Result: e For k = 2 d is the only vertex that has no edges starting at it Then we remove d Result: d < e For k = 3 b and c are the only vertices that has no edges starting at it we select c (it can be b too) Then we remove c Result: c<d<e For k = 4 b is the only vertex that has no edges starting at it Then...
Words: 2227 - Pages: 9
...Assignments for Math 221, Discrete Structures J. Stanley Warford April 5, 2013 Assignment 1 1. 2. 3. 4. 5. Study Section 12.2. Do Exercises 12.4(b, e). Do Exercise 12.8. Math 221, Discrete Structures Do Exercise 12.9. In your induction case, you should start with (n + 1)2 and use the result from Exercise 12.8. Do Exercise 12.10. i is divisible by 3 means that i = 3k for some integer k. You may use the fact that the sum of two expressions, each one divisible by 3, is also divisible by 3. 1 Assignment 2 1. 2. 3. 4. Study Section 12.5. Do Exercise 12.5. Do Exercise 12.14. Math 221, Discrete Structures Prove (12.16.1). There are two base cases, one for n = 1 and one for n = 2. For the induction case, there are two inductive hypotheses–one with n − 1 and one with n. You can assume both of them to prove the case for n + 1. Start with the RHS, use (12.14), then the inductive hypotheses. Prove (12.35a). The base case is n = 1. Prove (12.35b). The base case is n = 1. 5. 6. 1 Assignment 3 1. 2. Study Section 10.1. Math 221, Discrete Structures Do Exercise 10.1(a, b, c, d, e, g). For 10.1(d), you will need an implication in the body of a universal quantification. For 10.1(g), it is easiest to translate “It is not the case that” as ¬. 1 Assignment 4 1. Do Exercise 10.1(h, i, j, k, l, m). For 10.1(h) and (l), you will need an implication with the ∈ symbol. For 10.1(m), you will need to quantify with Σ with a body of 1. Do Exercise 10.3. Math...
Words: 3138 - Pages: 13
.... . . . . . . . . . . . . . . . . . . . Methods of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proof by Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . Basic Notion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Operations on Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . Power Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cartesian Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Injective, Surjective and Bijective Functions . . . . . . . . . . . . . . 14 Composite Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . ....
Words: 6227 - Pages: 25
...titrate a solution of unknown concentration with another solution of unknown concentration and still get a meaningful answer? Explain your answer in a few sentences. 5) Explain the difference between an endpoint and equivalence point in a titration. Solutions to the Titrations Practice Worksheet For questions 1 and 2, the units for your final answer should be “M”, or “molar”, because you’re trying to find the molarity of the acid or base solution. To solve these problems, use M1V1 = M2V2 . 1) 0.043 M HCl 2) 0.0036 M NaOH For problem 3, you need to divide your final answer by two, because H2SO4 is a diprotic acid, meaning that there are two acidic hydrogens that need to be neutralized during the titration. As a result, it takes twice as much base to neutralize it, making the concentration of the acid appear twice as large as it really is. 3) 0.1 M H2SO4 4) You cannot do a titration without knowing the molarity of at least one of the substances, because you’d then be solving one equation with two unknowns (the unknowns being M1 and M2). 5) Endpoint: When you actually stop doing the titration (usually, this is determined by a color change in an indicator or an indication of pH=7.0 on an electronic pH probe) Equivalence point: When the solution is exactly neutralized. It’s important to keep in mind that...
Words: 377 - Pages: 2
...Homework Assignment 1.1 1. List the number and name of each runlevel as used by Fedora/Red Hat Linux Provide a brief explanation of each runlevel. ID | Name | Description | 0 | Halt | Shuts down the system. | 1 | Single-user Mode | Mode for administrative tasks. | 2 | Multi-user Mode | Does not configure network interfaces and does not export networks services. | 3 | Multi-user Mode with Networking | Starts the system normally. | 4 | Not used/User-definable | For special purposes. | 5 | Start the system normally with appropriate display manager. ( with GUI ) | As runlevel 3 + display manager. | 6 | Reboot | Reboots the system. | 2. Which two runlevel should never be set as the default? 0 (HALT) and 6 (reboot) should never be set as default. 3. What script contains the default runlevel? /etc/rc <tab> 4. The init scripts for each runlevel are contained in which directory tree? Check /etc/inittab 5. What would you name the link to an init script that would start the fictitious bigd daemon early in the boot process? Open /etc/rc.local and write the script to run at boot time. 6. What would you name the link to kill the same daemon? Kill -9 <pid> 7. How would you use the service command to manually kill the bigd daemon? Kill -9 <pid /// You can get through ps for service> 8. How would use chkconfig to set the bigd daemon to start at runlevel 2 and 3? chkconfig bigd --level 23 on 9. How would you use...
Words: 1402 - Pages: 6
...Assignment Title: Paper #1 Forensic Analysis of UEFI by Pui Yee (Pearl) Pang Pearl.pang.py@gmail.com By submitting this assignment I acknowledge that I have read and agree to abide by the Champlain College Academic Honesty Policy. I declare that all work within this assignment is my own or appropriately attributed. I accept that failure to follow the academic honesty policy may result in a failure grade, or expulsion from Champlain College. Date Due: _11/4/2015___ Date Submitted:_11/4/2015__ Abstract The Unified Extensible Firmware Interface (UEFI) Specification is an interface between the operating system (OS) and the platform firmware and is managed through the UEFI forum, a collection of chipset, hardware, system, firmware, and operating system vendors. One of many benefits of using UEFI is that it provides a more secure environment during the boot process by adding several security mechanisms such as secure boot, and update. With the added secure boot feature, the correct bootloader certification is needed and database key authentication is also required before the booting process. As a result, rootkit or other malware program have a hard time hijacking the boot process and concealing itself from the operating system. This paper will focus on the analysis of UEFI's secure boot feature and its implications and challenges for digital investigators conducting computer forensic investigation. Keywords: UEFI secure boot, boot firmware, malware, rootkit. Introduction ...
Words: 1677 - Pages: 7
...Assessment Worksheet Installing a Core Linux Operating System on a Server IS3440 - Linux Security Course Name and Number: _____________________________________________________ Student Name: ________________________________________________________________ Instructor Name: ______________________________________________________________ Lab Due Date: ________________________________________________________________ Overview The foundation of host-based security starts with the installation of the operating system (OS). Contrary to popular opinion, there is no such thing as a secure operating system, but in this lab, you learned how to install the Linux CentOS operating system in a secure manner. You created a new virtual machine, partitioned the hard drive, and installed the Linux operating system. You also created a non-root user account and verified that key services were (or were not) running. Lab Assessment Questions & Answers 1. During the Minimal install, NTP (Network Time Protocol) server was not installed. From a security perspective, why is it important for a system to keep accurate time? To keep the system in sync and up to date for logging purposes such as any incidents that occur. Otherwise a time may not be correct when checking logs for an incident. 2. During the install, you set a password for the root user. What is the root user, and when is it appropriate to use this account? The root user is the system administrator. It is only appropriate...
Words: 503 - Pages: 3
...interface, console user interface, and character user interface (CUI), is a means of interacting with a computer program where the user issues commands to the program in the form of successive lines of text (command lines). Summary: The video taught me that everything that was developed after series 9x of operating systems is built on the NT platform. Also the NT did not have a device manager or a hardware wizard as well as not supporting plug and play. I also learned that Linux was developed by a Finnish student named Linus Torvalds. This information is not really useful for me its old information that for me has no value unless I want to be a technology historian. Unit 8 Assignment 1 Video Summary 6 VIDEO #3.02 Topics: * Partitioning * Creating Partitions * Create/Delete Partitions Terms: * Partition - In personal computers, a partition is a logical division of a hard disk created so that you can have different operating systems on the same hard disk or to create the appearance of having separate hard drives for file management, multiple users, or other purposes. * FAT - File Allocation Table is a computer file system architecture and a family of industry-standard file systems utilizing it. * NTFS - New Technology File System, The primary Windows file system, starting with Windows NT. NTFS is required to implement various security and administrative features...
Words: 819 - Pages: 4
...Equality is something that we try to depend we all strive for. In reality equality is, in most people’s opinions, an extremely unfair goal when it comes to financial equality. It may seem like a double standard to some but in fact it is not. This subject can be looked at from three different points of view the first being individuals who are hard workers and still can’t seem to thrive. It is this group who really suffer the most, fairness to them would them having the same opportunities to do as well as the upper classes. Sometimes it is due to their own decisions that they are in the position they are in but many times they are just recipients of bad luck. The next group of people are the people who are lazy and still think they deserve to have the same success as the next level of the monetary food chain. These individuals believe fairness is them being handed what others work very hard for. This is not that same idea of fairness that most of the rest of the country subscribes to . The third group is the individuals who thrives. These people are the upper-middle class and upper class who worked hard to get educations and/or build up their skills in a particular field which allows them to thrive (Lawhead 588) Does Rawls have a point? Yes. It becomes apparent that those who work hard and have success should not be held back by those who do not have the same success. If everyone receives the same “equal” share the more successful people would have no reason to work hard and...
Words: 328 - Pages: 2
