List some of the important biological functions of carbohydrates
1 answer:
Answer:
Carbohydrates are the biological molecules having a general emperical formula .
There are various important biological functions of carbohydrates. Carbohydrates serve as structural components, component of coenzymes, and backbone of RNA. Carbohydrates provide and store energy, and also plays important role in blood clotting, development, immune system, and preventing pathogenesis.
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Answer:
9.64g/mL
Explanation:
Given parameters:
Mass of the metal = 106g
Volume of cylinder = 50mL
Volume difference = 31mL - 20mL = 11mL
Unknown:
Density of the metal = ?
Solution:
To find the density of the metal, we use;
Density =
Density = = 9.64g/mL
Lewis Structure is drawn in following steps,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,
Answer:
i. Molar mass of glucose = 180 g/mol
ii. Amount of glucose = 0.5 mole
Explanation:
<em>The volume of the glucose solution to be prepared</em> = 500
<em>Molarity of the glucose solution to be prepared</em> = 1 M
i. Molar mass of glucose () = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
ii.<em> mole = molarity x volume</em>. Hence;
amount (in moles) of the glucose solution to be prepared
= 1 x 500/1000 = 0.5 mole