Answers
2,2-dimethyl-4-propyloctane has SIX secondary carbons.
Explanation:
Secondary Carbon:
In organic compounds secondary carbon is that carbon which is bonded to two carbon atoms and two hydrogen atoms. Below Red arrows indicate secondary carbons.
Primary Carbon:
Primary carbon is that carbon which is further attached to one carbon atom and three hydrogen atoms. Below in structure Blue arrows indicate primary carbons.
Tertiary Carbon:
These are those carbons which are bonded to to at least three carbon atoms. Below Green arrows indicate tertiary carbons.
The molecule 2,2-dimethyl-4-propyloctane has five secondary carbons which are those bonded to two other carbon atoms.
Explanation:The hydrocarbon 2,2-dimethyl-4-propyloctane belongs to the class of alkanes, characterized by single bonds between carbon atoms. This molecule consists of an 8-carbon chain (octane), with three side branches: two methyl groups (-CH3) on the second carbon and a propyl group (-CH2-CH2-CH3) on the fourth carbon. When identifying secondary carbons, which are carbon atoms bonded to two other carbon atoms, it becomes evident that 2,2-dimethyl-4-propyloctane contains five secondary carbons.
The two carbons on either end of the main chain don't fit the criteria of secondary carbons as they are bonded to only one other carbon atom. The five secondary carbons are as follows: two from the main chain (excluding the ends); two found at the ends of the propyl branch; and one where the methyl branches connect to the main chain.
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Related Questions
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days __________?
a.0.835
b.0.268
c.0.422
d.0.440
e.0.499
Answers
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
0.499 mg will remain after 13.0 days. So, the correct answer is (E).
What is Half life?The time which is required for a quantity to reduce to half of its initial value is defined as Half life. This is commonly used in nuclear physics to for the description of unstable atoms undergo radioactive decay or how much time stable atoms survive.
Formula for Half Life,
t1/2 = ln 2 / λ
where,
t1/2 is half-life
λ is radioactive decay constant
For given information,
here, t1/2 = 8.04 days
so, 8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
As per Radioactive decay law,
[tex]N_t = N_o[/tex]e∧(-λt)
where,
[tex]N_t[/tex] is amount of compound at t time,
[tex]N_o[/tex] is amount of compound at t = 0 time
t is time taken to decay
λ is radioactive decay constant.
[tex]N_o[/tex] = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
[tex]N_t[/tex]= 1.53 mg e∧{(-0.693/8.04 days) x 13.0 days)}
[tex]N_t[/tex] = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
Thus, 0.499 mg will remain after 13.0 days. So, the correct answer is (E).
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In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be needed in the reaction?
1.) 18 g
2.) 54 g
3.) 72 g
4.) 90 g
Answers
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol
18g C*1mol C/12 g C = 1.5 mol C
C + O2 → CO2
from reaction 1 mol 1 mol 1 mol
from problem 1.5 mol 1.5 mol 1.5 mol
1.5 mol O2*32 g O2/1 mol O2 = 48 g O2
In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount of O2 that was added for the reaction (probably people who wrote this problem wanted this)
we need (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is 2.) 54 g.
Trans-2-butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm−1); however, ir spectroscopy is still helpful in identifying the presence of the double bond. identify the other signal that would indicate the presence of a c=c bond.
Answers
1) 3010-3100 cm⁻¹:
As in trans-2-Butene a hydrogen atoms ate attached to sp² hybridized carbon, therefore the stretching of =C-H (C-H) bond will give a peak of medium intensity in the range of 3010-3100 cm⁻¹.
2) 675-1000 cm⁻¹:
Another peak which is given by the bending of =C-H (C-H) bond with strong intensity will appear in the range of 675-1000 cm⁻¹.
Trans-2-butene's C=C double bond can be indicated by IR absorption signals of C-H stretching near 3100 cm⁻¹ for the =CH₂ group, 3020 cm⁻¹ for the –CH= group and C-H out-of-plane bending near 915 cm⁻¹ and 1000 cm⁻¹
To indicate the presence of a C=C double bond in trans-2-butene despite the absence of a distinct signal in the double bond region of the infrared spectrum (1600–1850 cm−1), one would look for other characteristic IR absorption signals. These include the C-H stretching vibrations associated with alkenes, found around 3100 cm⁻¹ for the =CH2 group and around 3020 cm⁻¹ for the –CH= group. Additionally, the out-of-plane bending vibrations of the C-H bonds at the terminal =CH2 group occur near 915 cm⁻¹ and those of the –CH= grouping near 1000 cm⁻¹. These peaks would be indicative of the alkenic C-H bonds as opposed to alkane C-H stretches, which occur at slightly different frequencies.
Why do all humans have similar features? A. They have the same number and kinds of chromosomes. B. They have identical alleles. C. They have different alleles. D. Their genes do not allow for any variation.
Answers
Fe2o3 (molar mass = 159.7 g/mol) reacts with co (molar mass = 28.0 g/mol) according to the equation fe2o3(s) + 3 co(g) → 3 co2(g) + 2 fe(s). when 352 g fe2o3 reacts with excess co, how much fe (iron) is produced? 1. 4.41 mol fe 2. 0.907 mol fe 3. 147 g fe 4. 25.1 mol fe
Answers
Fe₂O₃ + 3CO ---> 3CO₂ + 2Fe
Fe₂O₃ reacts with excess CO, which makes Fe₂O₃ the limiting reactant. the amount of products formed depends on the amount of limiting reactant present.
Number of moles of Fe₂O₃ reacted - 352 g / 159.7 g/mol = 2.204 mol
stoichiometry of Fe₂O₃ to Fe is 1:2
number of Fe moles formed - 2 x 2.20 mol = 4.408 mol
therefore number of Fe moles formed - 4.41 mol
answer is 1. 4.41 mol Fe
When 352 g of Fe₂O₃ reacts with excess CO, it produces 4.41 mol of Fe.
To determine how much Fe (iron) is produced when 352 g of Fe₂O₃ reacts with excess CO, follow these steps:
Calculate moles of Fe₂O₃:
The molar mass of Fe₂O₃ is 159.7 g/mol.
Moles of Fe₂O₃ = 352 g / 159.7 g/mol = 2.20 mol Fe₂O₃.
Determine the mole ratio of Fe₂O₃ to Fe using the balanced equation:
The balanced equation is Fe₂O₃ + 3CO → 3CO₂ + 2Fe.
1 mol of Fe₂O₃ produces 2 mol of Fe.
Calculate moles of Fe produced:
Moles of Fe = 2.20 mol Fe₂O₃ * 2 mol Fe / 1 mol Fe₂O₃ = 4.40 mol Fe.
Convert moles of Fe to grams:
The molar mass of Fe is 55.85 g/mol.
Mass of Fe = 4.40 mol * 55.85 g/mol = 245.74 g of Fe.
The correct answer is option 1: 4.41 mol Fe.
Which metal can replace Cr in Cr2O3
Answers
Answer:
Aluminium.
Explanation:
Hello,
In this case, the most appropriated metal turns out being the Aluminum since it is higher in the activity series than nickel. Don't forget that activity series is an empirical tool used to predict products in displacement reactions and reactivity of metals with water and acids in replacement reactions and ore extraction.
Best regards.
Choose the molecule or compound that exhibits dispersion forces as its strongest intermolecular force.choose the molecule or compound that exhibits dispersion forces as its strongest intermolecular force.cohbrcl2naclall of these have intermolecular forces stronger than dispersion.
Answers
Cl₂ exhibits dispersion forces as its strongest intermolecular force.
Explanation:
Chlorine molecule is a non-polar molecule because the electronegativity difference both chlorine atoms is zero. Such molecules are also called as symmetrical molecules due to symmetry in their electron density.
So, Cl₂ molecules can not show either Hydrogen bond interactions due to unavailability of partial positive hydrogen nor they can show dipole-dipole interactions due to unavailability of dipole.
Therefore, such non-polar compounds generate instantaneous dipole which on approaching next non-polar molecule induces dipole in it. Hence, develops Dispersion forces.
Chlorine (Cl2) exhibits dispersion forces as its strongest intermolecular force. HBr exhibits dipole-dipole forces, and Co and NaCl show stronger forms of bonding (metallic and ionic, respectively). Larger, heavier atoms or molecules generally display stronger dispersion forces due to their greater number of electrons.
Explanation:The molecule or compound that exhibits dispersion forces as its strongest intermolecular force among Co, HBr, Cl2, and NaCl is Cl2. Dispersion forces are the only type of intermolecular force that occurs between noble gas atoms and nonpolar molecules such as Cl2. They arise as a result of fluctuations in electron density in an atom or molecule, that create temporary regions of positive and negative charge. These forces can attract nearby oppositely charged regions in neighboring molecules.
In the provided list, Hydrogen Bromide (HBr) exhibits dipole-dipole forces which are stronger than dispersion forces because it's a polar molecule. Cobalt (Co) is a metal and exhibits metallic bonding which is stronger than the dispersion forces. Sodium chloride (NaCl) forms an ionic compound and hence shows ionic bonding, which is much stronger than the dispersion forces.
Molecules with larger and heavier atoms, like chlorine (Cl2) , generally exhibit stronger dispersion forces compared to the smaller and lighter atoms, because of the presence of more electrons that can create temporary poles. At room temperature, Cl2 is a gas, but its dispersion forces allow it to condense into a liquid or solid under the right conditions.
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The ability to attract an electron for bonding is called:
A. Electrostatic attraction
B. Electronegativity
C. Delocalized electron attraction
D. Polar bonding
Answers
The ability to attract an electron for bonding is called Electronegativity. Hence, option C is correct.
What is an atom?An atom consists of a central nucleus that is usually surrounded by one or more electrons.
Electronegativity is a measure of an atom's ability to attract shared electrons to itself.
The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity.
Hence, option C is correct.
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Write the lewis structure for mgi2. draw the lewis dot structure for mgi2. include all lone pairs of electrons.
Answers
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,
The Lewis structure for MgI2 shows 3 lone pairs of electrons on each I atom and no lone pairs on the Mg atom.
Explanation:The Lewis structure for MgI2 can be drawn by following a few steps:
Calculate the total number of valence electrons for MgI2.Put the least electronegative atom, Mg, in the center and put the I atoms around it.Connect the central atom (Mg) to each of the I atoms using a single bond. This will account for 2 electrons.Place the remaining electrons as lone pairs around the I atoms.If any electrons remain, place them as lone pairs around the central Mg atom.The final Lewis structure for MgI2 will have 3 lone pairs of electrons on each I atom and no lone pairs on the Mg atom.
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which part of the blood is correctly paired with its function
Answers
A stock bottle of concentrated hydrochloric acid is a 37% (w/w) hcl solution. what is the molarity of this solution? how many moles of hcl are in a 10 ml sample of this concentrated acid? the density of 37% hcl is 1.19 g/ml.
Answers
therefore number of moles of HCl in 1000g - 370 g / 36.5 g/mol = 10.1 mol
the density of solution is 1.19 g/ml
volume of 1 mL weighs - 1.19 g
then 10 mL would weigh - 11.9 g
so if 1000 g contains - 10.1 mol
then 11.9 g contains - 10.1 mol / 1000 g x 11.9 g = 0.120 mol
therefore number of HCl moles in 10 mL is 0.120 mol
On a piece of paper, calculate the number of moles of kool-aid® powder needed to make 100 ml of a 0.1 m solution.
Answers
Molarity = Moles / Volume of Solution ----- (1)
Data Given:
Molarity = 0.1 mol.L⁻¹
Volume = 100 mL = 0.1 L
Mole = ?
Solving Eq. 1 for Moles,
Moles = Molarity × Volume of Solution
Putting Values,
Moles = 0.1 mol.L⁻¹ × 0.1 L
Moles = 0.01 moles
When the above equation is balanced the coefficients in order are
Answers
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + Al(OH)₃
Ca are balance on both sides,
There are 2 Al at left side and one at right so, multiply Al(OH)₃ by 2 to balance,
So,
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + 2 Al(OH)₃
Now, Ca and Al are balanced, now balance SO₄, which is 3 at left hand side and one at right hand side, so multiply CaSO₄ on right side by 3, so,
Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
Again Ca got imbalance, so multiply Ca(OH)₂ by 3 to balance Ca, So,
3 Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
The Equation is balance now with respect to every element.
Result:
The Ratio is 3 : 1 : 3 : 2, so, Option-G is correct.
What is the molecular geometry around each carbon atom in a saturated hydrocarbon?
Answers
Molecular geometry around each carbon atom in a saturated hydrocarbon is Tetrahedral.
Explanation:
In saturated hydrocarbons (-CH₂-) the central atom (carbon) is bonded to either three or two hydrogen atoms and one or two carbon atoms. So, the central atom is having four electron pairs and all pairs are bonding pairs and lacks any lone pair of electron. According to Valence Shell Electron Pair Repulsion (VSEPR) Theory the central atom with four bonding pair electrons and zero lone pair electrons will attain a tetrahedral geometry with bond angles of 109°.
the molecular egometry around each carbon atom in a staurated hydro carbon is tetrahydral
Determine the kinds of intermolecular forces that are present in each element or compound. o2 nbr3 ch4 hf
Answers
NB₃ (polar molecule so it contain dipole-dipole interaction in addition to dispersion force)
CH₄ (Non polar compound and has no hydrogen bond so only dispersion is present)
HF (It can form hydrogen bond because it is polar and contain F which has high electronegativity in addition to dipole-dipole and dispersion)
O2 experiences London dispersion forces, NBr3 experiences dipole-dipole interaction, CH4 experiences London dispersion forces, and HF experiences hydrogen bonding due to the presence of a hydrogen atom bonded to a highly electronegative atom.
Explanation:The given molecules show different types of intermolecular forces. Oxygen gas (O2) is a non-polar molecule and it interacts through London dispersion forces, the weakest form of forces. Nitrogen tribromide (NBr3) has polar bonds, so this compound's main intermolecular force would be dipole-dipole interaction. Methane (CH4) is non-polar, so the intermolecular forces are due to London dispersion forces. Hydrogen Fluoride (HF) experiences hydrogen bonding because it contains a hydrogen atom bonded to a more electronegative atom, fluorine.
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Which formula represents an organic compound?cah2c4h8h2o2p2o5?
Answers
C₄H₈ is hydrocarbon, organic compound consisting only of carbon and hydrogen.
CaH₂ (calcium hydride) is alkaline earth hydride, inorganic compound.
H₂O₂(hydrogen peroxide) is pale blue, clear, inorganic liquid.
P₂O₅ (phosphorus pentoxide) is white, crystalline solid inorganic compound.
One mole of air at 310 k confined in a cylinder under a heavy piston occupies a volume of 4.9 l. the heat capacity of air under constant volume is 5 2 r. find the new volume of the gas if 3.9 kj is transferred to the air. answer in units of l
Answers
An empty 10 ml volumetric flask weighs 10.2634 g. when filled to the mark with distilled water and weighed again in the air at 20 oc, the mass is 20.2144 g. what is the true volume of the flask at 20 oc?
Answers
Explanation:
It is known that density of water at [tex]20^{o}C[/tex] is 998,29 [tex]kg/m^{3}[/tex] or 0.998 g/ml (as 1 [tex]kg/m^{3}[/tex] = 0.001 g/ml).
Therefore, calculate mass of water as follows.
Mass of water = Final mass - initial mass
= 20.2144 g - 10.2634 g
= 9.951 g
As we known that density is the amount of mass present in a unit volume.
Mathematically, Density = [tex]\frac{mass}{volume}[/tex]
Therefore, putting the given values into the above formula we will calculate the volume as follows.
Density = [tex]\frac{mass}{volume}[/tex]
0.998 g/ml = [tex]\frac{9.951 g}{Volume}[/tex]
Volume = 9.97 ml
Thus, we can conclude that volume of the flask at [tex]20^{o}C[/tex] is 9.97 ml.
The true volume of the flask at 20°C is 8.30145 mL.
The true volume of the flask can be calculated using the difference in mass before and after adding water.
Weigh the empty flask: 10.2634 gWeigh the filled flask: 20.2144 gSubtract the mass of the empty flask from the mass of the filled flask to get the mass of the water added: 20.2144 g - 10.2634 g = 9.9510 gSince 1 mL of water weighs 1 g, the volume of water added is 8.30145 mLTherefore, the true volume of the flask at 20°C is 8.30145 mL.
Correct question is: An empty 10 ml volumetric flask weighs 10.2634 g. when filled to the mark with distilled water and weighed again in the air at 20°C , the mass is 20.2144 g. What is the true volume of the flask at 20°C ?
Chemistry to make tomato soup, you add one can of water to the condensed soup. why is this dilution?
Answers
Pentane (c5h12) undergoes combustion with excess oxygen to produce water and carbon dioxide. how many liters of water are produced when 50.5 grams of pentane combustion with excess oxygen at stp?
Answers
What is the mass in grams of na 2 co 3 (s) (molar mass = 105.98 g/mol) that, when dissolved in 125 ml of water, will make a 0.15 m aqueous solution?
Answers
molarity of sodium carbonate is 0.15 M
this means that 0.15 moles of sodium carbonate are dissolved in 1 L of solution
if 1 L contains - 0.15 mol
then 125 mL should contain - 0.15 mol/L x 0.125 L = 0.019 mol
mass of sodium carbonate in 125 mL - 0.019 mol x 105.98 g/mol = 2.0 g
2.0 g of sodium carbonate is required to make a 0.15 M solution
Answer:
[tex]m_{Na_2CO_3}=2.0gNa_2CO_3[/tex]
Explanation:
Hello,
In this case, we consider the equation defining molarity, in order to compute the mass that is present into 125 mL of an aqueous 0.15M of sodium carbonate (solute), Na₂CO₃ as shown below:
[tex]M=\frac{n_{solute}}{V_{solution}}[/tex]
Now, since the unknown is the mass which comes from the moles, by solving for it and subsequently using its molar mass, one obtains:
[tex]m_{Na_2CO_3}=125mL*\frac{1L}{1000mL}*0.15\frac{mol{Na_2CO_3}}{L}*\frac{105.98gNa_2CO_3}{1mol{Na_2CO_3}} \\m_{Na_2CO_3}=2.0gNa_2CO_3[/tex]
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If 1495 J of heat is needed to raise the temperature of a 319 g sample of a metal from 55.0°C to 66.0°C, what is the specific heat capacity of the metal?
Answers
Q(heat)= MC delta T
Q= 1495 j
c= specific heat capacity =?
M(mass)=319 g
delta T = change in temperature = 66-55 =11 c
by making c the subject of the formula
C=Q /M delta T
c= 1495 j/ 319 g x 11 c =0.426 j/g/c
a gas occupies 3.5 l at standard presure. find the volume of the gas when the pressure is 1140 mm hg
Answers
For the initial and final state of pressure and volume for a given gas at given temperature is as,
P₁ V₁ = P₂ V₂ --------- (1)
Data Given;
P₁ = 760 mmHg (standard Pressure)
P₂ = 1140 mmHg
V₁ = 3.5 L
V₂ = ?
Solving equation 1 for V₂,
V₂ = P₁ V₁ / P₂
Putting values,
V₂ = (760 mmHg × 3.5 L) ÷ 1140 mmHg
V₂ = 2.33 L
Result:
Hence, with increase in pressure from 760 mmHg to 1140 mmHg the volume has decreased from 3.5 l to 2.33 L.
The question concerns Boyle's Law in physics, which states that pressure and volume of a gas are inversely related at a constant temperature. Using the provided initial conditions and the final pressure, the final volume of the gas can be calculated using Boyle's Law to be approximately 2.33 L.
Explanation:The subject matter of this question falls under the concept of gases in Physics, especially where Boyle's Law is applicable. Boyle's Law states that the pressure (P) and volume (V) of a gas have an inverse relationship when the temperature is kept constant. In other words, an increase in pressure causes a decrease in volume and vice versa.
To solve the student's question, we'll use the equation of Boyle's Law: P1*V1 = P2*V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
In this case, the standard pressure (P1), which is generally 760 mm Hg, and the initial volume (V1) is 3.5 L. The final pressure, P2, is given as 1140 mm Hg, and we need to find the final volume (V2).
Substituting the values we know into the Boyle's Law equation: (760mm Hg * 3.5 L) = (1140 mm Hg * V2) Solving for 'V2' gives us a final volume of approximately 2.33 L. Therefore, the volume of the gas when the pressure is 1140 mm Hg will be about 2.33 L.
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Question 7 unsaved liquid water can store more heat energy than an equal amount of any other naturally occurring substance because liquid water question 7 options: 1) covers 71% of earth's surface 2) has its greatest density at 4°c 3) has the higher specific heat 4) can be changed into a solid or a gas
Answers
___________
a substance with a higher specific heat will take more energy to raise its temperature ?
True or False
Answers
Answer:
True
Explanation:
what is the density of a smaple if it weight 3.5 grams and has a volume of 2cm^3
Answers
density=mass/volume.
d=3.5g/2cm^3
d=1.75 g/cm^3
The density of the sample is 1.75 g/cm^3.
Which alkanealkene pair would be formed by a disproportionation reaction of the two radicals produced by the most energetically favored homolytic bond cleavage in the molecule shown?
Answers
Write the overall equation for the conversion of pyruvate to acetyl coa. express your answer as a chemical equation.
Answers
CH3COO-COO- + NAD+ + HS-COA = ch3-COO-S -COA +NADH +CO2
The oxidation of pyruvate led to a conversation of NAD+ to NADH and produces acetyl COA and CO2
Final answer:
The conversion of pyruvate to acetyl CoA is represented by the equation: C3H3O3- + CoA-SH + NAD+ → CH3-C-S-CoA + CO2 + NADH + H+. This step catalyzed by the pyruvate dehydrogenase complex involves oxidative decarboxylation and connects glycolysis to the Krebs cycle.
Explanation:
The conversion of pyruvate to acetyl CoA is a critical step that links glycolysis, the initial phase of glucose catabolism, to the Krebs cycle, where further energy extraction occurs. The overall chemical equation representing the conversion of pyruvate (a three-carbon molecule) to acetyl CoA (a two-carbon molecule) is:
C3H3O3- (pyruvate) + CoA-SH + NAD+ → CH3-C-S-CoA (acetyl CoA) + CO2 + NADH + H+
This process, known as pyruvate oxidation, is catalyzed by the multi-enzyme complex pyruvate dehydrogenase. It involves the oxidative decarboxylation of pyruvate, resulting in one molecule of carbon dioxide and the reduction of NAD+ to NADH. The transferred acetyl group is then bound to Coenzyme A, forming acetyl CoA, which subsequently enters the Krebs cycle within the mitochondria.
Now, imagine a cow who eats grass (which is mostly non- digestible cellulose) all day long. How does a cow get any energy out of that cellulose?
Answers
Cows can derive energy from cellulose-rich grass due to the presence of symbiotic bacteria in their digestive system, which secrete an enzyme called cellulase, breaking down cellulose into usable glucose monomers. Their multiple-chambered stomach and appendix further facilitate cellulose digestion. This unique adaptation allows cows and other ruminants to utilize cellulose as an energy source.
Explanation:Cows and other herbivores are able to derive energy from cellulose-rich grass due to symbiotic bacteria that reside in their rumen, a part of their digestive system. These bacteria secrete an enzyme called cellulase which breaks down cellulose into glucose monomers, providing a viable energy source for the animal. Herbivores, such as cows, buffalos, and horses, possess a multiple-chambered stomach where cellulose digestion mostly occurs, and also an appendix where bacteria help in further breaking down cellulose.
Cellulose is composed of glucose monomers packed tightly as extended long chains, giving it its high tensile strength and rigidity, an important attribute for plant cells. It's interesting to note that every other glucose monomer in cellulose is flipped over, which adds to its complexity and indigestibility in many animals including humans. However, the unique digestive system of ruminants, rooms for cellulase-secreting bacteria such as those in the rumen and appendix, enable them to utilize cellulose as an energy source.
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What is the ph of a 0.45 m solution of aniline (c6h5nh2)? (pkb 9.40)?
Answers
The pH of a 0.45 M solution of aniline (C6H5NH2) is 9.4.
Explanation:The pH of a 0.45 M solution of aniline (C6H5NH2) can be calculated using the given pKb value of 9.40.
To find the pH, we need to find the pOH first using the pKb value:
pOH = 14 - pKb = 14 - 9.40 = 4.6
Then, we can use the equation pH + pOH = 14 to find the pH:
pH = 14 - pOH = 14 - 4.6 = 9.4
Therefore, the pH of the solution is 9.4.
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The pH of a 0.45 M solution of aniline, given that its pKb is 9.40, is calculated to be approximately 10.99 after taking into account the equilibrium constant Kb for aniline and the concentration of OH- ions introduced on the ionization of aniline.
Explanation:To calculate the pH of a 0.45 M solution of aniline (C6H5NH2), we need to take into account its pKb value, 9.40. Aniline is a weak base, and we can use the formula pOH = pKb - log[(base)/(acid)]. First, we have to calculate the equilibrium constant Kb for aniline which is Kw/Ka = 10^-14 / Ka. We know Ka equals 10^-5.6, hence Kb for aniline = 10^-14 / 10^-5.6 = 4.3 × 10^-10.
Next, using an approximation approach and the ICE table, we calculate the concentration of OH- introduced by aniline when it ionizes, and find it to be roughly 9.7 × 10-4 M. The pOH of the solution is then -log(9.7 × 10-4) which equals 3.01. To get the pH, we subtract this from 14 (since pH + pOH = 14), yielding a pH for the solution of approximately 10.99.
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