Glucose is the simplest sugar and carbohydrate that provides energy. The simplified model of glucose (C₆H₁₂O₆) shows carbon, hydrogen, and oxygen atoms linked together.
<h3>What is glucose?</h3>
Glucose is an example of a carbohydrate macromolecule that is further classified as a monosaccharide. They are crystalline and fundamental units of carbohydrates.
The molecular formula of glucose is C₆H₁₂O₆ and the mass is 180.156 g/mol. It is an aldohexose that contains an aldehydic functional group. In its structure, there are six oxygen atoms, six carbon atoms, and twelve hydrogen atoms.
Therefore, the glucose molecule is composed of C, H, and O.
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<u>Answer:</u> The concentration of
required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of
, we use the equation:

Moles of
= 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:

For the given chemical equations:

![Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq.%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK_f%3D1.2%5Ctimes%2010%5E9)
Net equation: ![NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?](https://tex.z-dn.net/?f=NiC_2O_4%28s%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK%3D%3F)
To calculate the equilibrium constant, K for above equation, we get:

The expression for equilibrium constant of above equation is:
![K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC_2O_4%5E%7B2-%7D%5D%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%7D%7B%5BNiC_2O_4%5D%5BNH_3%5D%5E6%7D)
As,
is a solid, so its activity is taken as 1 and so for 
We are given:
![[[Ni(NH_3)_6]^{2+}]=0.016M](https://tex.z-dn.net/?f=%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%3D0.016M)
Putting values in above equations, we get:
![0.48=\frac{0.016}{[NH_3]^6}}](https://tex.z-dn.net/?f=0.48%3D%5Cfrac%7B0.016%7D%7B%5BNH_3%5D%5E6%7D%7D)
![[NH_3]=0.285M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.285M)
Hence, the concentration of
required will be 0.285 M.
Answer:
C.)One electron in each p orbital
Explanation:
In a P-sublevel with 3 electrons, they should be arranged with one electron going into each p-orbitals.
This is in accordance with the Hund's rule of maximum multiplicity.
The rule states that "electrons go into degenerate orbitals or sub-levels(p,d and f) singly before paring up".
Since the p-orbital is 3-fold degenerate with a capacity to accommodate a maximum number of 6 electrons, given 3 electrons, they will follow the Hund's rule in order to fill the orbitals.
So one electron will go in each p - orbitals easily.
Hehehehwgwgw. Be the hardest thing ever for a long day and I have a windows
Answer:- There are 32 valence electrons and it's tetrahedral in shape.
Explanations:- Atomic number of carbon is 6 and it's electron configuration is
. It has 4 electrons in the outer most shell means it has 4 valence electrons.
Atomic number of Br is 35 and it's electron configuration is
. It has 7 electrons in the outer most shell(2 in 4s and 5 in 4p) .
There is one C and four Br in the given compound. So, total number of valence electrons = 4+4(7) = 4+28 = 32
Four Br atoms are bonded to the central carbon atom and also there isn't any lone pair present on carbon. It makes it tetrahedral.