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How many feet of cloth is a piece of cloth How big is a piece of cloth? How many square meters?
1Pi = 10zhang = 100ft = 33.3333m
A piece of cloth is 240 meters in total. Every four fifths of a meter can be used to make a tablecloth. Now one third of this cloth has been cut. How much has it been cut
Big brothers and sisters, help,
240 / 3 = 80m
80 / 0.8 (4 / 5m) = 100 (pieces)
The formula of logarithm and exponent in Senior High School
It is difficult to understand the logarithm operation formula and the downward point of logarithm
I don't know the logarithm operation formula of the downward point log (a, b) = C. I say that B ^ C = log (a, c) + log (B, c) = log (AB, c) log (a, c) - log (B, c) = log (a / B, c) log (a ^ C, b) = (c) log (a, B ^ C) = (1 / C) log (a, b) log (a, b) = log (a, c) / log (B, c) word limit is not proved
On the formula of high school logarithm
I remember that LG20 = LG (4 * 5) = LG4 + lg5, and I knew that there was another one
log20=log4+log5
Does the formula LG4 * lg5 = LG (4 + 5) = lg9 hold?
And how to simplify the difference between LG square 5 and LG square 5
LG4 * lg5 = LG (4 + 5) = lg9 does not hold, LG1 * LG1 = 0 instead of = LG2
LG (5 * 5) = lg5 + lg5 = 2lg5 is definitely not equal to the square of (lg5)
The formula of changing base in logarithm,
What is it? I forgot
loga b=logc b/logc a
On logarithmic formula
Please list some common formulas of logarithm. Thank you
It seems a little simpler, but thank you
loga+logb=logab
loga-logb=loga/b
a=e^lna
The formula log (a) (n) = log (b) (n) / log (b) (a)
log(a^n)(b^m)=m/n*[log(a)(b)] .
Arithmetic of logarithm
Basic properties: 1. A ^ (log (a) (b)) = B 2. Log (a) (a ^ b) = B3. Log (a) (MN) = log (a) (m) + log (a) (n); 4. Log (a) (M △ n) = log (a) (m) - log (a) (n); 5. Log (a) (m ^ n) = NLog (a) (m) 6. Log (a ^ n) M = 1 / NLog (a) (m) derivation 1. Because n = log (a) (b), substituting a ^
What is the valence layer electron logarithm
Sum of formed σ bond and unbound electron pair
Or = (the number of valence electrons of the central atom + the number of electrons provided by the ligand) △ 2
How to find the valence layer electron logarithm of ion?
I know the molecular basis
The number of valence electron pairs = the number of bond pairs + the lone pair electrons of the central atom
And the lone pair electron of the central atom = 1-2 (the valence electron number of the central atom - the maximum number of electrons that can be accepted by the atom combined with the central atom)
I want to know how to calculate the number of valence layer electron pairs for ions like co2-so3-or acid radical ions
First calculate the number of lone electrons, anion: a = the number of outermost electrons + the number of charges of the central atom, cation: a = the number of outermost electrons - the number of charges of the central atom, then use the number of lone electrons = 1 / 2 (a-xb), and finally add the number of & bonds of the central atom?
How to determine the valence layer electron logarithm
If I want to know the geometry of a molecule, I need to use the valence layer electron pair repulsion theory, so I want to determine the valence layer electron logarithm of the central atom. For different molecules, how can I determine the valence layer electron logarithm? What is the electron layer structure, s, P, D? It's better to explain several different molecular structures, using several examples,
No
1. Count the valence electrons of the central atom
There are several valence centers of the molecule σ 2
3. Calculate the total number of electron pairs around the central atom, including lone pair electrons and common pair electrons (a σ bond is a common pair electrons)
Six -- octahedron
Five double triangle cones
Four -- tetrahedron
Three -- plane triangle
Two straight lines
This is the original shape. If there is a lone pair electron in the central atom, it will be filled into the position of the original bond and the shape will be changed
Ex: H2O has four pairs of electrons, so its original shape is tetrahedron, but there are two lone pairs of electrons on O, so it takes up two positions of tetrahedron and becomes a broken line shape with an angle close to 109.5 degrees
If the central atom is a transition metal atom, we should use orbital domain mixing as you said
SP -- straight line
SP2 -- plane triangle
SP3 -- tetrahedron
Dsp2 -- plane quadrilateral
Dsp3 -- double triangle cone
D2sp3 -- octahedron
1. Write the electronic configuration of the central atom
2. Determine the number of empty orbital domains in the d orbital domain of the central atom
3. The number ligand provides several electron pairs for bonding
4. Start with D, complete with s, and then P
The most troublesome one is the 4-coordinated one, because it may be dsp2 or SP3
For example, for bivalent Ni (CN) 4 (2 -), the valence electron configuration of Ni is d8s2, and a pair of electrons can be inserted in the d-orbital domain, so it is a mixture of dsp2, and the shape is a planar quadrilateral,
However, Ni (CO) 4 (2 +) with + 2 valence and d8s2 valence electron configuration will become SP3 mixed and tetrahedral, which is due to the different ability of ligands. In short, the more electron withdrawing group is, the easier it is to be dsp2 mixed. Cn is a very strong electron withdrawing group
The number ligands of other coordination numbers can be almost guessed
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