Binary from AP exam

Conversion from binary, octal, hexadecimal to decimal

When television first started, the system of broadcasting was analogue. This meant that a TV aerial received radio waves from the TV companies. These transformed into sound and images on the TV screen. At the time, the system was revolutionary, but it wasn't perfect. It couldn't transmit many channels and the quality of the pictures was sometimes poor because of interference. For a long time there was only analogue I, but at the end of the 1990s, digital television arrived. Digital TV doesn't use radio waves. Instead, it converts sound and images into binary code. This is a computer language which only consists of zeros and ones and broadcasters can use it to send a lot of information very quickly. This digital information arrives at a TV via an aerial, a cable, a satellite dish, or broadband. Then a decoder converts the information back to sound and images. Decoders are either inside a TV or computer, or in a separate box which is connected to it. Digital broadcasting is much better than the analogue system. There are a lot more channels because the TV companies can transmit much more information. The quality of the images and sound is much higher because there is less interference. Picture quality is even better on a high definition TV (an HDTV) than on a standard TV, so viewers can enjoy cinema-quality pictures in their own home.

Covalent Molecules and Compounds Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the diatomic (“two atoms”) molecules H2, N2, O2, F2, Cl2, Br2, and I2 (part (a) in Figure 3.1.1). Similarly, a few pure elements exist as polyatomic (“many atoms”) molecules, such as elemental phosphorus and sulfur, which occur as P4 and S8 (part (b) in Figure 3.1.1). Each covalent compound is represented by a molecular formula, which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as H2O. Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as CO2. Covalent compounds that predominantly contain carbon and hydrogen are called organic compounds. The convention for representing the formulas of organic compounds is to write carbon first, followed by hydrogen and then any other elements in alphabetical order (e.g., CH4O is methyl alcohol, a fuel). Compounds that consist primarily of elements other than carbon and hydrogen are called inorganic compounds; they include both covalent and ionic compounds. In inorganic compounds, the component elements are listed beginning with the one farthest to the left in the periodic table, as in CO2 or SF6. Those in the same group are listed beginning with the lower element and working up, as in ClF. By convention, however, when an inorganic compound contains both hydrogen and an element from groups 13–15, hydrogen is usually listed last in the formula. Examples are ammonia (NH3) and silane (SiH4). Compounds such as water, whose compositions were established long before this convention was adopted, are always written with hydrogen first: Water is always written as H2O, not OH2. The conventions for inorganic acids, such as hydrochloric acid (HCl) and sulfuric acid (H2SO4), are described elswhere. Note! For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up. Write the molecular formula of each compound. a. The phosphorus-sulfur compound that is responsible for the ignition of so-called strike anywhere matches has 4 phosphorus atoms and 3 sulfur atoms per molecule. b. Ethyl alcohol, the alcohol of alcoholic beverages, has 1 oxygen atom, 2 carbon atoms, and 6 hydrogen atoms per molecule. c. Freon-11, once widely used in automobile air conditioners and implicated in damage to the ozone layer, has 1 carbon atom, 3 chlorine atoms, and 1 fluorine atom per molecule. Solution: a. • A The molecule has 4 phosphorus atoms and 3 sulfur atoms. Because the compound does not contain mostly carbon and hydrogen, it is inorganic. • B Phosphorus is in group 15, and sulfur is in group 16. Because phosphorus is to the left of sulfur, it is written first. • C Writing the number of each kind of atom as a right-hand subscript gives P4S3 as the molecular formula. b. • A Ethyl alcohol contains predominantly carbon and hydrogen, so it is an organic compound. • B The formula for an organic compound is written with the number of carbon atoms first, the number of hydrogen atoms next, and the other atoms in alphabetical order: CHO. • C Adding subscripts gives the molecular formula C2H6O. c. • A Freon-11 contains carbon, chlorine, and fluorine. It can be viewed as either an inorganic compound or an organic compound (in which fluorine has replaced hydrogen). The formula for Freon-11 can therefore be written using either of the two conventions. • B According to the convention for inorganic compounds, carbon is written first because it is farther left in the periodic table. Fluorine and chlorine are in the same group, so they are listed beginning with the lower element and working up: CClF. Adding subscripts gives the molecular formula CCl3F. • C We obtain the same formula for Freon-11 using the convention for organic compounds. The number of carbon atoms is written first, followed by the number of hydrogen atoms (zero) and then the other elements in alphabetical order, also giving CCl3F. Write the molecular formula for each compound. a. Nitrous oxide, also called “laughing gas,” has 2 nitrogen atoms and 1 oxygen atom per molecule. Nitrous oxide is used as a mild anesthetic for minor surgery and as the propellant in cans of whipped cream. b. Sucrose, also known as cane sugar, has 12 carbon atoms, 11 oxygen atoms, and 22 hydrogen atoms. c. Sulfur hexafluoride, a gas used to pressurize “unpressurized” tennis balls and as a coolant in nuclear reactors, has 6 fluorine atoms and 1 sulfur atom per molecule. Answer: a. N2O b. C12H22O11 c. SF6. Ionic Compounds The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively-charged protons in the nucleus is equal to the number of negatively-charged electrons. In contrast, ions are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cations. Conversely, ions that contain more electrons than protons have a net negative charge and are called anions. Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge. Note! Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge.An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows: Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges. When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Mg2O2:Polyatomic Ions Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. P. The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K+ and NO3− ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42− form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3− form NH4NO3 (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43− ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43− ions.Summary • There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties. • The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit.Contributors The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most covalent compounds consist of molecules, groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The molecular formula of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called organic compounds, whereas compounds that consist primarily of elements other than carbon and hydrogen are inorganic compounds. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two. A structural formula indicates the composition and approximate structure and shape of a molecule. Single bonds, double bonds, and triple bonds are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions). Ions can consist of one atom (monatomic ions) or several (polyatomic ions). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure. An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration.

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

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