Answer : The Lewis-dot structure of
is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, 
As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in
= 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :





Hence, the Lewis-dot structure of
is shown below.
For Iron:

For Oxygen:

These are the two chemical symbols for the two elements found in Iron Oxide.
Answer:
The answer is b
Explanation:
have common properties and are good conductors of heat and electricity. They reflect light and are malleable, and ductile.
Answer:
c. decarboxylation of an a-keto acid.
Explanation:
Decarboxylation refers to the removal of the carboxyl group from a carboxylic acid and thus releasing carbon dioxide. Decarboxylases are enzymes that speed up the removal of the carboxyl group from acids. These reactants could be amino acids, alpha-keto acids, and beta-keto acids. Biotin is known to catalyze the decarboxylation of malonyl CoA to acetyl CoA during fatty acid synthesis.
Malonyl CoA is converted to acetyl CoA after decarboxylation assisted by biotin also known as Vitamin H. Alpha keto acids are involved in fatty acids synthesis and Malonyl CoA is an alpha-keto acid because the keto group is located in the first carbon near the carboxylic acid group. Keto acids have both a carboxyl group and a ketone group.
Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.