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Yes. The electronic configuration of Si is 1s2 2s2 2p6 3s2 3p2. All orbitals from 1s2 to 3s2 are being completely filled. There are two more electrons in the 3p orbitals, one which occupies 3px orbital and the other one 3py. They are both unpaired electrons.
depending on the number of electrons an element has determines the number of arrows in the orbitals. there is always one going up and one going down if the energy level is filled. Remember hund's rule in that each level must contain one arrow before another gets two. and the bottom figures are 1s^1 2s^2 3px 3py 3pz 4d.... and so forth
You need to state how many atoms are in the 4s. I.e. is it 4s1 i.e. Potassium, or 4s2 i.e. calcium
The s orbital is lower in energy than the p orbital.
From some point to the left and to the right. For example: 3px Radius mean range of 3px to the right and 3px to the left from some point like selection border.
CSS stands for Cascading Style SheetsStyles define how to display HTML elements on a web pageCSS3 is the third generation (level 3) of CSS.CSS3 offers a bunch of new ways you can write CSS rules with new CSS selectors, as well as a new combinator, and some new pseudo-elements.The box-shadow property of CSS3 allows designers to easily implement multiple drop shadows (outer or inner) on box elements, specifying values for color, size, blur and offset.For example,#myshadow1 {box-shadow: 10px 10px 5px #8a8a8a;}The first two attributes of each of the properties are the horizontal and vertical offsets for the shadow, the third is the blur radius, and the final one sets the color of the drop shadow.For browser compatibilities, you will need to write something like this:-moz-box-shadow: 3px 3px 3px #8a8a8a;-webkit-box-shadow: 3px 3px 3px #8a8a8a;box-shadow: 3px 3px 3px #8a8a8a;
{| ! style="padding-left: 3px" | 2 Dr Turbo Hatchback ! style="padding-left: 3px" | 2 Dr 2+2 Hatchback ! style="padding-left: 3px" | 2 Dr STD Hatchback | Original MSRP $34,570 $31,270 $28,175 |}
Atoms used to be thought to look like little solar systems. Now they are thought to be nesting shells of probability density functions known as orbitals. The S1 and P1 orbitals are the first two filled with other S and P orbitals following along with the various D and F orbitals. The S orbitals are essentially spheres with the P orbitals looking like infinity signs.Aside: A probability density function maps out where the electron is most likely to be in relation to the nucleus of the atom - not that it is there, just likely to be.
There are 2, 6 and 10 electrons in the 3s (1 suborbital), 3p (with 3 suborbitals: 3px, 3py, 3pz) and 5 sub orbitals in the 3d orbital: this makes a total of 18 electrons in 9 suborbitals
Yes. The electronic configuration of Si is 1s2 2s2 2p6 3s2 3p2. All orbitals from 1s2 to 3s2 are being completely filled. There are two more electrons in the 3p orbitals, one which occupies 3px orbital and the other one 3py. They are both unpaired electrons.
In its general form it works out as: 3px+qy-12p2-2q2 = 0Improved Answer:-Points:(p, q) and (7p, 3q)Slope: q/3pPerpendicular slope: -3p/qMidpoint: (4p, 2q)Equation: y-2q = -3p/q(x-4p) => yq = -3px+12p2+2q2Perpendicular bisector equation in its general form: 3px+qy-12x2-2q2 = 0
Points: (p, q) and (7p, 3q) Midpoint: (4p, 2q) Slope: q/3p Perpendicular slope: -3p/q Perpendicular bisector equation:- => y-2q = -3p/q(x-4p) => qy-2q^2 = -3p(x-4p) => qy-2q^2 = -3px+12p^2 => qy = -3px+12p^2+2q^2 In its general form: 3px+qy-12p^2-2q^2 = 0
x2 + 3px + p = 0 x2 + p(3x + 1) = 0 p(3x+1) = -x2 p = -x2/(3x+1) So p can have any value at all. In fact, around x = -1/3, p goes asymptotically to + and - infinity.
depending on the number of electrons an element has determines the number of arrows in the orbitals. there is always one going up and one going down if the energy level is filled. Remember hund's rule in that each level must contain one arrow before another gets two. and the bottom figures are 1s^1 2s^2 3px 3py 3pz 4d.... and so forth