Answer:
"The active site of the enzyme has a complementary rigid structure" belongs to the key and lock system
"The conformation of the enzyme changes when it binds to the substrate so that the active site conforms to the substrate." belongs to the induced fit system.
"The substrate binds to the enzyme at the active site, forming an enzyme-substrate complex" belongs to both, that is, the key and lock system and the induced fit system.
"The substrate binds to the enzyme through non-covalent interactions" can belong to both enzyme systems.
Explanation:
Enzymatic key and lock systems bear this name because the enzyme at its site of union with the substrate has an ideal shape so that its fit is perfect, similar to a headbreaker, so once they are joined they are not It can bind another substrate to the enzyme, since they are generally associated with strong chemical bonds.
The shape of the enzyme's active site is a negative of what the shape of the substrate would be.
On the other hand, in the mechanism or enzyme system of induced adjustment, the enzyme has an active site that is where it binds with the substrate and another site where another chemical component binds, which when this chemical component binds this enzyme changes its morphology and becomes "active" to bond with your substrate.
This happens a lot in the inactive enzymes that are usually activated in digestive processes since the fact that these enzymes are constantly active would be dangerous, therefore the body takes the induced enzyme system as a control mechanism, where a molecule or chemical compound induces change morphological of an enzyme by means of the allosteric union so that it joins its substrate and catalyzes or analyzes it, depending on the enzymatic character of the enzyme.
The mathematics of combining quantum theory with wave motion of atomic particles is known as _____.
Combining quantum theory with wave motion of atomic particles is: Wave Mechanics
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid = 1.9 x 10-5.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M
Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M
GIven that :
Ka = [tex]1.9 x 10^{-5}[/tex]
Thus; it's pKa = 4.72
[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]
[tex]pH =4.72 + log(1.1376)[/tex]
[tex]pH =4.72 + 0.05598[/tex]
[tex]pH =4.77598[/tex]
pH ≅ 4.80
The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.
How we calculate the pH?pH of the given solution will be used by using the following equation:
pH = pKa + log[conjugate base] / [weak acid]
Given chemical reaction will be represented as:
HN₃ + OH⁻ → N₃⁻ + H₂O
Moles will be calculated as:
n = M×V, where
M = molarity
V = volume
Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375 mol
Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol
From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:
0.00375 - 0.001995 = 0.001755 mol
And 0.001995 mol of N₃⁻ is preduced by the reaction.
Total volume of the solution = 0.025 + 0.0133 = 0.0383 L
To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:
[N₃⁻] = 0.001995 mol / 0.0383 L = 0.0521 M
[HN₃] = 0.001755 mol / 0.0383 L = 0.0458 M
Ka for HN₃ = 1.9 × 10⁻⁵
pKa = -log( 1.9 × 10⁻⁵ ) = 4.72
On putting all these values on the above equation, we get
pH = 4.72 + log (0.0521) / (0.0458)
pH = 4.80
Hence, pH of the solution is 4.80.
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A 3.00-g sample of an alloy (containing only Pb and Sn) was dissolved in nitric acid (HNO3). Sulfuric acid was added to this solution, which precipitated 2.93 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (molar mass of PbSO4 = 303.3 g/mol)
Answer:
33.3% of Sn in the sample
Explanation:
The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.
Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:
2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.
As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:
9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺
As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g
And the percentage of Sn in the sample is:
1.00g / 3.00g ₓ 100 =
33.3% of Sn in the sampleLewis structure of methyl metcaptain
Answer:
The lewis structure of the compounds can be drawn by making the skeleton of the molecule first. Then the different atoms are arranged and the electrons are arranged in their bonding pattern. The lone pair of the atoms, which are not involved in the bonding are represented by the dots.
So the lewis structures of both the compound methyl mercaptan has been in the attached image:
Spelling of methyl metcaptain is wrong, the correct spelling is methyl mercaptan.
Answer:
Methyl mercaptan is also known as Methanethiol with the chemical formula CH3SH and it is an organosulfur compound.
For lewis structure of methyl mercaptan (CH3SH), there are total 14 valence electrons. Four hydrogen atoms has one valence electron each, carbon has four valence electrons and sulfur has six valence electrons. Carbon form one bond with three hydrogen atoms by sharing one electron with each, carbon form one single bond with sulfur atom by sharing one electron with it and sulfur form one single bond with hydrogen. Sulfur left with four unpair electrons.
g Which statement is incorrect regarding oxidation? Oxidation is a "gain" of electrons. Oxidation is the combination with O atoms. Oxidation is an increase in oxidation state. Oxidation is always accompanied by reduction. none of these
Answer:
The incorrect statement from the options is OXIDATION IS A "GAIN" OF ELECTRONS
Explanation:
Oxidation in a redox reaction is the loss of electrons. It is also the increase in the oxidation states of an atom or ion or atoms in a molecule. A redox reaction is a type of chemical reaction in which there is a transfer of electrons from an atom or ion to another resulting in a change in oxidation states of the substances involved. The reducing agent in the reaction is undergoes oxidation by losing electrons while the oxidating agent is reduced that is it gains electrons at the end of the reaction. The atom or ion from which electron is lost is said to be oxidized while the other atom or ion involved in the reaction is reduced.
Oxidation is also the combination with O atoms and it is always accompanied by reduction because oxidation forms a half of the whole redox reaction. A substance cannot be oxidized except it has reduced another substance by losing electrons to it.
Hypochlorous acid is formed in situ by reaction of aq. sodium hypochlorite solution with acetic acid. Draw balanced chemical equations to show the formation of hypochlorous acid and protonated hypochlorous acid.
Answer:
NaClO + CH₃COOH ----> HClO + CH3CO- + Na
Explanation:
This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction
A solid white substance A is heated strongly in the absence of air. It decomposes to form a new white substance B and a gas C. The gas has exactly the same properties as the product obtained when carbon is burned in an excess of oxygen. Based on these observations, can we determine whether solids A and B and the gas C are elements or compounds?
Answer:
A, B and C are compounds
Explanation:
First of all, I need to establish that when carbon is burnt in excess oxygen, carbon dioxide is obtained as shown by this equation; C(s) + O2(g) ----> CO2(g).
Looking at the presentation in the question, A was said to be heated strongly and it decomposed to B and C. Only a compound can decompose when heated. Elements can not decompose on heating. Secondly, compounds usually decompose to give the same compounds that combined to form them. Compounds hardly decompose into their constituent elements.
Again from the information provided, the compound A is a white solid. This is likely to be CaCO3. It decomposes to give another white solid. This may be CaO and the gas was identified as CO2.
Hence;
CaCO3(s)--------> CaO(s) + CO2(g)
Cual es la diferencia entre agua pesada y agua ligera a) el agua pesada contiene mas minerales que el agua ligera b) el agua ligera es liquida mientras el agua pesada es solida c) el agua ligera es agua purificada y el agua pesada es agua contaminada d) el agua pesada contiene mas elementos estearato de sodio
Answer:
d) El agua pesada contiene mas elementos
Explanation:
La diferencia fundamental entre el agua pesada y el agua ligera es que la primera tiene una proporción mayor de deuterio que la segunda. El deuterio es un ión del hidrógeno que tiene un peso atómico mayor que el hidrógeno común y corriente. Por ende, la opción D ofrece la mejor aproximación.
Answer:
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Explanation:ki
(a) How many stereoisomers are possible for 4-methyl-1,2-cyclohexanediol? ___ (b) Name the stereoisomers formed by oxidation of (S)-4-methylcyclohexene with osmium tetroxide. If there is only one stereoisomer formed, leave the second space blank. Isomer #1: Isomer #2: (c) Is the product formed in step (b) optically active? _____
Answer:
See explanation
Explanation:
For the first part of the question, we have to check the chiral carbons in 4-methyl-1,2-cyclohexanediol. In this case carbons, 1 and 2 are chiral, if we have 2 chiral carbons we will have 4 isomers. We have to remember that formula 2^n in which "n" is the number of chiral carbons, so:
2^n = 2^2 = 4 isomers
And the isomers that we can have are:
1) (1R,2S)-4-methylcyclohexane-1,2-diol
2) (1S,2S)-4-methylcyclohexane-1,2-diol
3) (1S,2S)-4-methylcyclohexane-1,2-diol
4) (1S,2R)-4-methylcyclohexane-1,2-diol
See figure 1
For the second part of the question, we have to remember that the oxidation with [tex]OsO_4[/tex] is a syn addition. In other words, the "OHs" are added in the same plane. In this case, we have the methyl group with a wedge bond, so the "OH" groups will have a dashed bond due to the steric hindrance. Due to this we only can have 1 isomer ((1S,2R,4S)-4-methylcyclohexane-1,2-diol). Finally, on this molecule, we dont have any symmetry planes (this characteristic will cancel out the optical activity), so the product of this reaction has optical activity.
See figure 2
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Draw the structure 2 butylbutane
Answer:
please look at the picture below.
Explanation:
What would have happened to your % Transmittance reading and to your calculations of Keq if the spectrophotometer had been set at 520 nm rather than 447nm
Answer:
On the off chance that the wavelength(λ) maximum worth has been changed to 520 nm from 470 nm on the spectrophotometer, less light would be absorbed and in this way %T would be higher than the one found at 470 nm. The wavelength utilized at 520 nm isn't adequate for the excitation and consequently lesser light is absorbed by the arrangement.
Explanation:
A spectrophotometer is an analytical equipment used to quantitatively gauge the transmission(passage) or impression of visible light, UV light or infrared light through a medium.
Transmittance (τ) is the ratio of the brilliant or luminous flux at a given wavelength that is transmitted to that of the incident radiation.
where, Keq is the equilibrium constant.
On the off chance that the wavelength(λ) has been changed to 520 nm from 470 nm on the spectrophotometer, less light would be absorbed and in this way %T would be higher than the one found at 470 nm.
What happens to Transmittance?A spectrophotometer is an analytical equipment used to quantitatively gauge the transmission(passage) or impression of visible light, UV light or infrared light through a medium. Transmittance (τ) is the ratio of the brilliant or luminous flux at a given wavelength that is transmitted to that of the incident radiation. The wavelength utilized at 520 nm isn't adequate for the excitation and consequently lesser light is absorbed by the arrangement. As the concentration goes up, more radiation is absorbed and the absorbance goes up. Therefore, the absorbance is directly proportional to the concentration.
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Write the empirical formula
Answer:
[tex]Pb(CO_{3})_{2} \\Pb(NO_{3})_{4} \\FeCO_{3}\\Fe(NO_{3})_{2}[/tex]
Explanation:
[tex]Pb^{4+}(CO_{3}^{2-})_{2} --->Pb(CO_{3})_{2} \\Pb^{4+} (NO_{3}^{-})_{4} --->Pb(NO_{3})_{4} \\Fe^{2+} CO_{3}^{2-} --->FeCO_{3}\\Fe^{2+} (NO_{3}^{-})_{2}--->Fe(NO_{3})_{2}[/tex]
Which correctly lists three characteristics of minerals?
solid, crystal structure, definite chemical composition
organic, crystal structure, definite chemical composition
human-made, solid, organic
crystal structure, definite chemical composition, human-made
Answer:a
Explanation:
The three characteristics of minerals are that they are solid, have definite crystal structure and definite chemical composition.
What are minerals?Minerals are defined as a chemical compound which has a well -defined composition and possesses a specific crystal structure.It occurs naturally in the pure form.
If a compound occurs naturally in different crystal structure then each structure is considered as a different mineral.The chemical composition of a mineral varies depending on the presence of small impurities which are present in small quantities.
Some minerals can have variable proportions of two or more chemical elements which occupy equivalent position in the crystal structure.It may also have variable composition which is split into separate species.
Physical properties of minerals include color,streak, luster,specific gravity and cleavage.
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why homoannular diene in uv spectrometry have higher wavlenth than hetroannular diene?
Answer:
Homoanular dienes have a greater base value than heteroanular dienes
Explanation:
Woodward in 1945 gave a set of rules relating the wavelength of maximum absorption to the structure of a compound. These rules were modified by Fieser in 1959. These sets of rules describe the absorption of organic molecules in the UV region of the electromagnetic spectrum.
Each system of diene or triene has a given fixed value at which maximum absorption is expected to occur according to Woodward rules. This given fixed value is called the base or parent value. If the two double bonds are trans to each other, the diene is said to be transoid. If the two double bonds belong to different rings, the system is said to be heteroanular and the base value in each case is 215nm. If the double bonds are cis to each other (cisoid), or the two double bonds are in the same ring (homoanular), then the base value is 253nm.
Since λmax = base value + ∑ substituent contributions + ∑ other contributions, if the other contributions are not very significant, homoanular diene will have a greater λmax because of its larger base value compared to heteroanular diene. This correlates well with the fact that conjugated systems absorb at a longer wavelength.
which process is used to produce gases from solutions of salts dissolved in water or another liquid?
A.Electrolysis
B.Metallic bonding
C.Ionic bonding
D. Polar covalent bonding
Answer:
A.Electrolysis
Explanation:
A.Electrolysis
For example, electrolysis of solution of NaCl in water gives H2 and O2.
g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empirical formula of the compound?
Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O