Unit 10 6/23 | Quizalize

In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level

Here’Transformation,Ratio,Proportion, Fractions and Algebraic Expressions,Transformation 1. Translation 2. Reflection 3. Rotation 4. Enlargement 5. Transformation 6. Congruence 7. Similarity 8. Scale Factor 9. Image 10. Pre-image 11. Symmetry 12. Isometry 13. Ratio 14. Proportion 15. Equivalent Ratios 16. Simplify 17. Unit Ratio 18. Scale 19. Part-to-Part 20. Part-to-Whole 21. Rate 22. Comparison 23. Proportional Relationship 24. Cross Multiplication 25. Direct Proportion 26. Inverse Proportion 27. Constant of Proportionality 28. Golden Ratio 29. Linear Relationship 30. Equal Proportions 31. Proportional Constant 32. Scale Drawing 33. Word Problems 34. Unitary Method 35. Percentage 36. Double Number Line 37. Fraction 38. Numerator 39. Denominator 40. Improper Fraction 41. Proper Fraction 42. Mixed Number 43. Simplified Fraction 44. Reciprocal 45. Least Common Denominator (LCD) 46. Greatest Common Factor (GCF) 47. Equivalent Fractions 48. Decimal 49. Variable 50. Coefficient 51. Constant 52. Algebraic Term 53. Polynomial 54. Monomial 55. Binomial 56. Expression 57. Equation 58. Like Terms 59. Simplify 60. Substitution -

Ions Ions are charged substances that have formed through the gain or loss of electrons. Cations form from the loss of electrons and have a positive charge while anions form through the gain of electrons and have a negative charge. Cation Formation Cations are the positive ions formed by the loss of one or more electrons. The most commonly formed cations of the representative elements are those that involve the loss of all of the valence electrons. Consider the alkali metal sodium (Na) . It has one valence electron in the n=3 energy level. Upon losing that electron, the sodiu ion now has an octet of electrons from the second energy level and a charge of 1+ . The electron arrangement of the sodium ion is now the same as that of the noble gas neon. Consider a similar process with magnesium and aluminum. In this case, the magnesium atom loses its two valence electrons in order to achieve the same arrangement as the noble gas neon and a charge of 2+ . The aluminum atom loses its three valence electrons to have the same electron arrangement as neon and a charge of 3+ . For representative elements under typical conditions, three electrons is usually the maximum number that will be los. Representative elements will not lose electrons beyond their valence because they would have to "break" the octet of the previous energy level which provides stability to the ion. Anions Anions are the negative ions formed from the gain of one or more electrons. When nonmetal atoms gain elections, they often do so until their outermost principal energy level achieves an octet. For fluorine, which has an electron arrangement of (2, 7), it only needs to gain one electron to have the same electron arrangement as neon. Forming an octet (eight electrons in the outer shell) provides stability to the atom. Fluorine will gain one electron and have a charge of 1− . The electron arrangement of the fluoride ion (2, 8) will also change to reflect the gain of an electron. Oxygen has an electron arrangement of (2, 6) and needs to gain two electrons to fill the n=2 energy level and achieve an octet of electrons in the outermost shell. The oxide ion will have a charge of 2− as a result of gaining two electrons. Under typical conditions, three electrons is the maximum that will be gained in the formation of anions. Subatomic Particles in an Ion Since ions form from the gain or loss of electrons, we can also look at the number of subatomic particles (protons, neutrons, and electrons) found in an ion. Remember that the number of protons determines the identity of the element and will not change in a chemical process. Example 2.5.1 How many protons, neutrons, and electrons in a single oxide (O2−) ion? Solution Oxygen has the atomic number 8 so both the atom and the ion will have 8 protons. The average atomic mass of oxygen is 16. Therefore, there will be 8 neutrons (atomic mass−atomic number=neutrons) . A neutral oxygen atom would have 8 electrons. However, the anion has gained two electrons so O2− has 10 electrons. We can also use information about the subatomic particles to determine the identity of an ion. Example 2.5.2 An ion with a 2+ charge has 18 electrons. Determine the identity of the ion. Solution If an ion has a 2+ charge then it must have lost electrons to form the cation. If the ion has 18 electrons and the atom lost 2 to form the ion, then the neutral atom contained 20 electrons. Since it was neutral, it must also have had 20 protons. Therefore the element is calcium. Polyatomic Ions A polyatomic ion is an ion composed of two or more atoms that have a charge as a group (poly = many). The ammonium ion (see figure below) consists of one nitrogen atom and four hydrogen atoms. Together, they comprise a single ion with a 1+ charge and a formula of NH+4 . The hydroxide ion (see figure below) contains one hydrogen atom and one oxygen atom with an overall charge of 1− . The carbonate ion (see figure below) consists of one carbon atom and three oxygen atoms and carries an overall charge of 2− . The formula of the carbonate ion is CO2−3 . The atoms of a polyatomic ion are tightly bonded together and so the entire ion behaves as a single unit. The figures below show several examples. Soult Screenshot 2-2-1.png Figure 2.5.1 : The ammonium ion (NH+4) is a nitrogen atom (blue) bonded to four hydrogen atoms (white). Soult Screenshot 2-2-2.png Figure 2.5.2 : The hydroxide ion (OH−) is an oxygen atom (red) bonded to a hydrogen atom. Soult Screenshot 2-2-3.png Figure 2.5.3 : The carbonate ion (CO2−3) is a carbon atom (black) bonded to three oxygen atoms. The table below lists a number of polyatomic ions by name and by structure. The heading for each column indicates the charge on the polyatomic ions in that group. Note that the vast majority of the ions listed are anions - there are very few polyatomic cations. 1− 2− 3− 1+ Table 2.5.1 : Common Polyatomic Ions acetate, CH3COO− carbonate, CO2−3 arsenate, AsO3−3 ammonium, NH+4 bromate, BrO−3 chromate, CrO2−4 phosphite, PO3−3 chlorate, ClO−3 dichromate, Cr2O2−7 phosphate, PO3−4 chlorite, ClO−2 hydrogen phosphate, HPO2−4 cyanide, CN− oxalate, C2O2−4 dihydrogen phosphate, H2PO−4 peroxide, O2−2 hydrogen carbonate, HCO−3 silicate, SiO2−3 hydrogen sulfate, HSO−4 sulfate, SO2−4 hydrogen sulfide, HS− sulfite, SO2−3 hydroxide, OH− hypochlorite, ClO− nitrate, NO−3 nitrite, NO−2 perchlorate, ClO−4 permanganate, MnO−4 The vast majority of polyatomic ions are anions, many of which end in -ate or -ite. Notice that in some cases such as nitrate (NO−3) and nitrite (NO−2) , there are multiple anions that consist of the same two elements. In these cases, the difference between the ions is the number of oxygen atoms present, while the overall charge is the same. As a class, these are called oxyanions. When there are two oxyanions for a particular element, the one with the greater number of oxygen atoms gets the -ate suffix, while the one with the fewer number of oxygen atoms gets the -ite suffix. The four oxyanions of chlorine are shown below, which also includes the use of the prefixes hypo- and per-. ClO− , hypochlorite ClO−2 , chlorite ClO−3 , chlorate ClO−4 , perchlorate Not your usual ion Soult Screenshot 2-2-4.png "Drink you milk. It's good for your bones." We're told this from early childhood, and with good reason. Milk contains a good supply of calcium, part of the structure of bone. However, there are two other ionic components of hydroxyapatite, the mineral component. Phosphate ion and hydroxide ion make up the remainder of the inorganic material in bone. News You Can Use Bone is a very complex structure. It is composed of protein (mainly collagen), hydroxyapatite (a calcium-phosphate-hydroxide mixture), some other minerals, and contains 10 - 20% water. The calcium/phosphate ratios are not stoichiometric, but vary somewhat from one portion of bone to the next. Bones are very strong but will break under enough stress. Regular exercise and proper nutrition help to increase bone strength. Watch a video about bone structure at http://www.youtube.com/watch?v=d9owEvYdouk Nitrate is an anion with a complex bonding structure. Major sources for this ion in drinking water are runoff from fertilizer, septic tank leakage, sewage, and natural deposits. High concentrations of nitrates represent a significant health hazard, especially to infants. The nitrate in the body is converted to nitrite, which then binds to hemoglobin. This binding decreases the ability of hemoglobin to transport oxygen, thus depriving the cells of the O2 needed for proper functioning. Cyanide production is widespread throughout nature. Forest fires will produce significant amounts of cyanide. Many plants contain cyanide, and it is produced by a number of bacteria, algae, and fungi. Cyanide is used industrially in metal finishing, iron and steel mills, and in organic synthesis processes. This material is also an important component for the refining of precious metals. Formation of a complex between cyanide and gold allows extraction of this metal from a mixture.

Unit 10.6-7 Bellringer

FAIRYLAND AB Junior Unit 10 10.2-10.6, 10.10-10.13 ,10.16, 10.17 B

FAIRYLAND AB Junior Unit 10 10.2-10.6, 10.10-10.13 ,10.16, 10.17 A

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