The skeletal structure of an organic compound is an abbreviated representation of its molecular structure, they are quick and easy to draw.
For example, the following image shows the skeletal structure of a compound:
The peaks represent the carbons. We must remember that carbon can have a maximum of 4 bonds.
Now, I will show you how is the structure of this specific compound:
This is ternary alcohol, called 2-methyl-2-butanol. If you see carefully, you will notice that each carbon has 4 bonds. The functional groups present will be OH. The skeletal structure will be:
Hey there!
The best way to balance chemical equations is to first start by balancing polyatomic ions such as OH and SO₄.
Next, balance other elements, but save elements that are by themselves for last, such as H₂ or Fe. Once you balance everything else you can do the ones by themselves, it's much easier.
Hope this helps!
Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm
Yes mitochondria does make necrssary chemicals for the cell therefore the answer to your question is yes
Answer:
saturated fats
Explanation:
bcoz it is solid at room temperature and does not have double bonds between carbons
