do they fill up in 1s,2s,2p2p2p,3s,3p3p3p,4s,4p4p4p,4d5) and if your asked to hybridize ASf5 How do u know how many you have to hybridize when it starts at 4s
200615
You proffered the example of XeF4.  Now the Lewis structure for XeF4 gives an expanded octet, viz: AX4E2, where A represents the central atom (Xe) in this example; X represents a shared pair of electrons; and E represents an unshared pair.  Thus for XeF4 we have 4 shared pairs and 2 unshared pairs, i.e. 6 pairs.  \n\nNow whatever the principal energy level(n), there is 1 s orbital; and for n=2 and higher, there are 3 p orbitals; and for n=3 and higher there are 5 d orbitals. For n=4 and higher there are 7 f orbitals.\n\nNow considering XeF4 hybridization, for the 6 pairs of electrons there will be 6 orbitals [each orbital holds a maximum of two electrons with opposite spin]. That means sp3d2 hybridization.  That is the one s orbital, the three p orbitals and two of the five d orbitals form six sp3d2 orbitals.  Of course, we are talking here about valence shell electrons.  How would these orbitals fill?  Well, electrons are added to orbitals such that electrons with the same spin are added until there are no more empty orbitals, then the orbitals with one electron fill (i.e. add a second electron) by acquiring an electron of opposite spin.\n\nThis gives XeF4 sp3d2 hybrid orbitals and an octahedron orientation. \n\nConsidering the Lewis structure for water, as another example, we obtain a AX2E2 form. Thus sp3 hydrid orbitals and a tetrahedron orientation. The 4 hybrid orbitals come from 1 s and 3 p atomic orbitals.
400310
hybridization.com\n\nor just type the keyword and the information you needed wud appear..
548739
At the four strand stage 50% of the product at meiosis would be recombinant. At the two strand stage, all the products would be recombinants. Providing that the crossover in question happens every time.\n\nWithout knowing the location of the gene and which chromosome it is on then it is impossible to tell how many map units are affected.
373868
1, 2, 4, 8, 16, or 32
544230
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
111201
Maybe this information will answer your question.......\n\nIs this still correct (25/10/16): 3N 46 47 48 43 46 47 40 3N 3F 43 40 45 4H 4H 4L 41 22 4F 40 42 2E 4L 3N 44 4B 4B 1G 3O 4B 49\n\nWTF ??
550386
If the P phenotype is\nPPSS x ppss\nThen all of F1 is PpSs\nAll would be Purple Smooth\n\nb F1 crossed with ppss: PpSs x ppss\nthe ratio would be 1:1 for PpSs and ppss\nor, Purple Smooth to Yellow Wrinkled.\n\nYou need to learn to use the Punnet square!
370096
For starters, S can only be a 1 or a 2 as any other number would have made the P bigger than 10, and that does not work.\nBut then, since 4 A is a number that ends with S, it cannot be 1, as S needs to be even, so S is 2.\n2oda+2oda+2oda+2oda=pop2\nWe can also look at the "p": it can only be 8 or 9, depending if the 4 "o" create a carry or not.  But we know that 4 "o" create a number that ends with a "o" unless the 4 "d" also create a carry.\nZero do that (without the carry) 3 would create a number ending with 2 but ending with 3 if there is a carry from the sum of "d", 4 would make un number ending by 6, so would need a carry of 8 to get to 24, but then we would have 2 to carry over the toal of "s" and that would make "p" 10 which is not acceptable, so "o" is either zero or 3 with a carry of one from the total of "d". But if there is no carry (if "o" is zero) then "p" has to be 8, and "d" would have to be 2, but 2 is already "s"; so "o" has to be 3, and "p" is 9.\n\n23da+23da+23da+23da=9392.\n\nand "d" has to be 4 and "a" is 8.\n\n2348+2348+2348+2348=9392
158694
Suppose the men are A, B, C, D, E, F, G, H, I.  Then one grouping would be ABC DEF GHI\n\nBAC FED HIG would not be a different grouping since the same 3 men are in the same group.  Neither would GHI DEF ABC.\n\nBut ABC DEG FHI would be different.\n\nI believe your formula should have multiplication where you have addition.  You'd have 9C3 combinations for the first group, 6C3 for the second, and 3C3 for the third.  Each selection of 3 groups could be ordered in 3! or 6 ways so you divide by 3! to discount the repeats.  So the formula is\n9C3 times 6C3 times 3C3 divided by 3!.\n\nI'll check this and edit it if wrong.
185784
How many ways to pick 5 from 20 if order isn't important?\nAnd, how many ways to pick 2 from 7 if order isn't important?\n\nFor the second question, you have... 1,2 1,3...1,7 or 6 that include 1. Then 2,3 2,4... 2,7 which is 5 more. Then 3,4 3,5...3,7 which is 4 more. Keep doing this and you will see the answer is 6+5+4+3+2+1 = 21\n\nFor the first part 5 out of 20, note that the total number of arrangements of 20 things is 20! which is 20 x 19 x 18 x17 ..... x 2 x 1. But for every arrangement that you take the first 5 of, there will be 15! ways to arrange the rest.\nAnd for those first 5, you can get them in any order which means that 1 2 4 3 5 is the same for you as 2 3 1 4 5. So, you need to divide the 20! by both 15! and 5! which looks like this-  20!/(15! x 5!) and equals 15,504.\n\nNow for each pick from the cable houses (15,504 possible) you have 21 different ways to pick a no cable set. Therefore, multiply 15,504 by 21 to get your final answer. 325,584
416680
It depends on if you have a scsi narrow or scsi wide device.  If you have a scsi narrow device jumpers 1-3 will assign an id and jumper 4 will usually be used to enable parity or disable it.  If the device is scsi wide than each jumper will represent a binary number.  I'll assume it's a scsi wide device because narrow is usually on really old machines.  Basically you have to look at the jumpers in binary. Jumper 1=1 Jumper 2=2 jumper 3=4 jumper 4=8.  You just add together the numbers to reach the scsi ID.    Your scsi id range for a narrow device is 0 through 15 which gives you a total of 16 unique ID's for that scsi bus.  So in your example you have jumper 2 on which equals 2 and jumper 3 on which equals 4.  2+4=6 so the scsi id would be 6.  No jumpers at all would give you scsi id 0.  I usually reserve zero for boot up drives.
434018
http://en.wikipedia.org/wiki/Atomic_orbital\nhttp://winter.group.shef.ac.uk/orbitron/AOs/1s/index.html\n\n1s2, \n2 s2  2p6\n3s2  3p6  3d10\n4s2 4p6 4d10 4f14\n\nthey dont fill in that order though\n4s2 is filled before 3d10   (look on periodic table)\n5s2 before 4d10 and so on
408166
OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
49668
2p, 5d, 4d, 3p, 3d exist.
436163
As I recall, the map unit is a 1% chance that two genes will sort independently. So, 13% will be al/+ phenotype, 13% dp/+, 37% are al/dp, and the other 37% are wildtype. I think... definitely check this before you accept it as true.
420160
Let us call the first 2 "2 sub 1", the second "2 sub 2" ... same thing for the 6's.\n\nLet us say that "2 sub 1" is distinct from "2 sub 2", and so on.\n\nWe have 8 choices for the first number, seven choices for the second number, etc. This means that there are 8*7*6*5*4*3*2*1 choices, or 8!. \n\nNow if you look at any number, take all of the 2's out. You see that you can rearrange the 2's in 3! different ways (including the original arrangement) using the procedure in the paragraph above. If we take out the 2's rearrange them, and put them back in we have created the a new arrangement. So for each arrangement of numbers, we have counted it 3! times. So we divide by 3!. The same thing for the sixes, we divide by 2!.\n\nThe result it (8!)/((3!)(2!)) = 3360 choices.
417322
Simply:\n\nyes\nyes\nyes\nyes\nand yes\n\nThey are all single replacement.\n\nGood job---Get an "A".\n\nI use this: \nA synthesis is an element + an element = one compound.\nA decomposition is one compound = an element + and element (synthesis backwards)\nA singe replacement has an element + a compound = an element + a compound.\nA double replacement  has compound + compound = compound + compound.