solution:
Weight of caffeine is W = 0.170 gm.
Volume of water is V= 10 ml
Volume of methylene chloride which extracted caffeine is v= 5ml
No of portions n=3
Distribution co-efficient= 4.6
Total amount of caffeine that can be unextracted is given by
![w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms](https://tex.z-dn.net/?f=w_%7Bn%7D%3Dw%5Ctimes%5B%5Cfrac%7Bk_%7BDx%7Dv%7D%7Bk_%7BDx%7Dv%2Bv%7D%5D%5En%5C%5C%3C%2Fp%3E%3Cp%3Ew_%7B3%7D%3D0.170%5B%5Cfrac%7B4.6%5Ctimes10%7D%7B%284.6%5Ctimes10%2B5%29%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B46%2B5%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B51%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B97336%7D%7B132651%7D%5D%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5Ctimes0.734%3D0.125gms)
amount of caffeine un extracted is 0.125gms
amount of caffeine extracted=0.170-0.125
=0.045 gms
Ikr behehbenekebe sgwhebejebeb
A significant figure is every symbol that made the number itself.
In this case, the number 40.00 has four figures but only two of them are significant 40, this is because you haven't got any more decimals than the first zero.
If you have a case with zeros in front, you take to the first non zero digits.
For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.
Answer:
See the explanation
Explanation:
1) The Lewis structure for
has a central Carbon<em> </em>atom attached to Oxygen atoms.
In the
we will have a structure: O=C=O the <u>central atom</u> "carbon" we will have <u>2 sigma bonds and 2 pi bonds</u>, therefore, we have an <u>Sp hybridization</u>. For O we have <u>1 pi and 1 sigma bond</u>, therefore, we have an <u>Sp2 hybridization</u>.
2) These atoms are held together by <u>double bonds.</u>
<u></u>
Again in the structure of
: O=C=O we only have double bonds.
3. Carbon dioxide has a Carbon dioxide has a <u>Linear</u> electron geometry.
Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.
4. The carbon atom is <u>Sp</u> hybridized.
We will have for carbon 2 pi bonds, so we will have an <u>Sp</u> hybridization.
5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.
(See figures)
Figure 1: Carbon hybridization
Figure 2: Oxygen hybridization