Answer:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.
Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin. Palmitate oxidation however, does not involve carboxylation.
Explanation:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.
Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme. The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.
Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.
The simplest carboxylic acid is called *
O Formaldehye
O formic acid
acetic acid
O
acetone
under the same conditions carbon (iv) oxide,propane and nitrogen (i) oxide diffuse at the same rate.Explain
Answer:
Rate of diffusion is same .
Explanation:
As we know that Rate of the diffusion is directly proportional to the [tex]\frac{1}{\sqrt{M} }[/tex] .They have same mass if there is same rate and similar condition therefore the mass of carbon (iv) oxide,propane and nitrogen (i) oxide will be similar.
The mass is directly proportional to the Rate of the diffusion.Therefore the rate of diffusion is similar in all carbon (iv) oxide,propane and nitrogen (i) oxide .A student mixes baking soda and vinegar in a glass. Are there any new substances created from this mixture?
Answer:
Explanation:
1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.
Yes, there are new substances created from this mixture.
Given the following Fischer projection: Fischer projection for an entantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right. draw the perspective formula of the molecule. Be sure to correctly indicate stereochemistry in your answer.
Answer:
Explanation:
Stereoisomers are two or more atoms that have the same bonding order of atoms but there is a difference spatial arrangement of the atoms in space.
A plane of symmetry divides a molecule into two equal halves.
A chiral stereoisomer are not superimposed on a mirror image , Hence they do not posses a plane of symmetry.
As a result to that. these non-superimposable mirror images are said to be Enantiomers.
However, a Fischer Projection emanates from a two - dimensional figure which is used for presenting a three - dimensional organic molecules.
From the given question;
Fischer projection for an enantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right.
we can sketch the way the enantiomer of 2-bromo-2,3-dihydroxypropanal can be seen like the one shown below:
CH₂OH
|
|
|
Br -------------|----------------OH
|
|
|
CHO
The objective of this question is to drawn the perspective formula of the molecule.
So , from the attached file below; we can see the perspective formula of the molecule in a well structured 3-D format.
a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a buffer composed of HC2H3O2 and C2H3O2 - ? Write a chemical equation for that reaction.c) What happens when HBr is added to this buffer? Write a chemical equation for that reaction.
Answer:
a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻
b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
Explanation:
a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:
HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺
C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻
Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.
b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.
HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.
C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
Which of the following would be more reactive than magnesium (Mg)?
A. Calcium (Ca)
B. Potassium (K)
C. Argon (Ar)
D. Beryllium (Be)
Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium
Explanation:
I hope it will help you
Answer: B. Potassium(K)
Explanation:
A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The specific heat capacity of
wood is 1.716 joules/gram degree Celsius. What is the mass of the piece of wood?
ОА. 16 g
OB. 29 g
ОC. 36 g
OD. 61 g
Answer:
35.578g or 36g if you round
Explanation:
Q=mc ∆∅ where ∅ is temperature difference
1160= m x 1.716 x (42-23)
m = 1160/ 1.716 x19
m=35.578g
m = 36g to nearest whole number
Answer: C. 36 g
Explanation: I got this right on Edmentum.
Of Sr or Ba , the element with the higher first ionization energy is
Answer:
Sr
Explanation:
Sr has an ionization of 550 whereas Ba has an ionization of 503
25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311 M NaOH solution. What is the concentration of the H2SO4 solution
Answer:
Concentration of the H₂SO₄ solution is 0.0737 M
Explanation:
Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:
H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O
From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.
mole ratio of acid to base, nA/nB = 1:2
Concentration of the base, Cb = 0.1311 M
Volume of base, Vb, = 28.11 mL
Concentration of acid, Ca = ?
Volume of acid, Va + 25.0 mL
Using the formula, CaVa/CbVb = nA/nB
making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB
substituting the values into the equation
Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M
Therefore, concentration of the H₂SO₄ solution is 0.0737 M
Enter an abbreviated electron configuration for magnesium: Express your answer in complete form, in order of increasing energy. For example, [He]2s22p2 would be entered as [He]2s^22p^2.
Answer:
[Ne]3s²
Explanation:
Mg
1s2 2s2 2p6 3s2 or [Ne]3s²
Abbreviated electronic configuration of magnesium is [Ne]3 s² and in complete form it is 1 s² 2 s² 2 p⁶ 3 s².
What is electronic configuration?Electronic configuration is defined as the distribution of electrons which are present in an atom or molecule in atomic or molecular orbitals.It describes how each electron moves independently in an orbital.
Knowledge of electronic configuration is necessary for understanding the structure of periodic table.It helps in understanding the chemical properties of elements.
Elements undergo chemical reactions in order to achieve stability. Main group elements obey the octet rule in their electronic configuration while the transition elements follow the 18 electron rule. Noble elements have valence shell complete in ground state and hence are said to be stable.
Learn more about electronic configuration,here:
https://brainly.com/question/13497372
#SPJ5
4. Which of the following statements explains the cause of lanthanide contraction?
A. All lanthanides and actinides are radioactive
B. Protons exhibit a stronger pull on outer f orbitals
C. The d orbitals in lanthanides have unpair electrons
D. The d orbitals in actinides have paired electrons
Answer:
B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS
Explanation:
Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.
Answer:
B) Protons exhibit a stronger pull on outer f orbitals than on d orbitals.
Explanation:
Rank the following transitions in a hydrogen atom in order of increasing wavelength of electromagnetic radiation that could produce them. Answer this question without doing any calculations. Explain the order.
n=2 to n=4
n=6 to n=8
n=10 to n=12
n=14 to n=16
Answer:
n=2 to n=4 < n=6 to n=8 < n=10 to n=12 < n=14 to n=16
Explanation:
According to Neils Bohr, electrons in an atom are found in specified energy levels. Transitions are possible from one energy level to another when the electron receives sufficient energy usually in the form of a photon of electromagnetic radiation of appropriate frequency and wavelength. The energy of this photon corresponds to the energy difference between the two energy levels. Thus the higher the energy difference between energy levels, the greater the energy of the photon required to cause the transition and the shorter the wavelength of the photon.
High energy photons have a very short wavelength. It should be noted that as n increases, the energy of successive energy levels decreases and transitions between them now occurs at longer wavelengths. Hence, the highest energy and shortest wavelength of photons are required for transition involving lower values of n because such electrons are closer to the nucleus and are more tightly bound to it than electrons found at a greater distance from the nucleus.
Hence transition involving electrons at higher energy levels occur at a longer wavelength compared to transition involving electrons closer to the nucleus. This is the basis for the arrangement of wavelengths required to effect the various electronic transitions shown in the answer.