I believe it would be A.Gram
~~~hope this helps~~~
~~davatar~~
Answer:
First Method: Vacuum Distillation and Chromatographic separation of the remains that were precipitated out from the peel.
Second Method: Extraction of components from orange peels by help of precipitation procedures that are mostly done <em>In Situ. </em>Those components can be recovered using the saponification process. Then these are examined under UV light spectroscopy. Now, the existence and extent of carotenoids can be determined by checking the levels of anti-oxidants.
Answer:
0.5 × 10²³ atoms of iodine
Explanation:
Given data:
Mass of calcium iodide = 12.75 g
Number of atoms of iodine = ?
Solution:
First of all we will calculate the number of moles of calcium iodide.
Number of moles = mass/ molar mass
Number of moles = 12.75 g/ 293.9 g/mol
Number of moles = 0.04 mol
In one mole of calcium iodide there are two moles of iodine.
Thus in 0.04 moles:
0.04 mol × 2 = 0.08 moles of iodine
Now we will use the Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
0.08 moles of iodine × 6.022 × 10²³ atoms / 1 mol
0.5 × 10²³ atoms of iodine.
Answer:
elements in same group have same valence electron.
period of element is equal to valence shell
Explanation:
if the elements are in a same group then they will be having same number of valence electron.
the period of an element in periodic table is equal to the valence shell of the element. that is if the valence electron are in 3 rd shell then the element will be in third period.
This reaction is known as
Wittig Reaction. A powerful reaction for the synthesis of
Alkene. In question the starting materials are
aldehyde and a Phosphorous
Ylide. Ylide when reacted with aldehyde produces a four membered ring which on
rearrangement gives Alkene and triphenylphosphine oxide. Phosphorous having great
affinity toward the oxygen is the driving force of this rearrangement. The reaction along with product (highlighted
blue) is as follow,
