Which one of the following is not matches the organelle with its function
Answer:
rip there isnt a photo
Explanation:
i do know a lot about cells tho lol
what is the IUPAC name of 2NaOH(s)
Answer:
NaoH= sodium hydroxide
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
what are the properety of covalent bond
Explanation:
1. boiling and melting point
2. electrical conductivity
3. Bond strength
4. bond length
A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.
Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.
Answer:
10.77%
Explanation:
Molar mass of Cu = mass deposited/number of moles of Cu
Molar mass of Cu = 0.4391 g/6.238x10^-3 moles
Molar mass of Cu = 70.391 g/mol
%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100
%error = 10.77%
What is the energy of a photon emitted with a wavelength of 654 nm?
O A. 3.04 x 10^-19 J
O B. 1.01 * 10^-27 J
O C. 1.30 x 10^-22 J
O D. 4.33 * 10^-22 J
Answer:
A. 3.04×10^-19J
Explanation:
Hope this will help you.
Calculate the pH of each solution.
A. 0.18 M CH3NH2
B. 0.18 M CH3NH3Cl
C. a mixture of 0.18 M CH3NH2 and 0.18 M CH3NH3Cl
Answer:
See Explanations
Explanation:
pH =-log[H₃O⁺] = -log[H⁺]
pOH = -log[OH⁻]
For weak acids [H⁺] = SqrRt(Ka·[Acid])
For weak bases [OH⁻] = SqrRt(Kb·[Base])
pH + pOH = 14
__________________________________________
A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95
CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;
[OH⁻] = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M
=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05
=> pH = 14 - pOH = 14 - 2.05 = 11.95.
*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.
___________________________________________________
B. Given 0.18M CH₃NH₃Cl
In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴
Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.
Hydrolysis Reaction of Methylammonium Ion:
CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻
Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹ Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.
*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.
C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl
Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl
In Water ...
=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl
=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻
=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻
-----------------------------------------------------------
Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)
=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹
----------------------------------------------------------
From the 0.36M CH₃NH₃⁺
=> CH₃NH₃⁺ + H₂O ⇄ CH₃NH₄⁺ + OH⁻
C(eq) 0.36M ---- x x (<= at equilibrium after mixing)
Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)
=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M
=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process
=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻
--------------------------------------------------------
Cl⁻ + H₂O => No Rxn (Cl⁻ does not react with H₂O)The 0.1926M in OH⁻ => [H⁺] = Kw/[OH⁻] = (10⁻¹⁴/0.1926)M = 5.192 x 10⁻¹⁴M in H₃O⁺ ions (= H⁺ ions) ...∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix
The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and
CH₃NH₃Cl and the mixture give the following approximate values;
A. The pH value of the 0.18 M CH₃NH₂ is 11.93
B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69
C. The pH value of the mixture is 10.644
Which method can be used to calculate the pH values?A. 0.18 M CH₃NH₂
The solution is presented as follows;
CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻
Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we
have;
The number of moles of CH₃NH₂ remaining = 0.18 - x
Which gives;
[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]
[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴
Therefore;
[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]
4.167 × 10⁻⁴ × (0.18 - x) = x²
4.167 × 10⁻⁴ × (0.18 - x) - x² = 0
Which gives;
x = [OH⁻] = 8.455 × 10⁻³
pH = 14 + log[OH⁻]
Which gives;
pH = 14 + log(8.455 × 10⁻³) ≈ 11.93
B. 0.18 M CH₃NH₃Cl
The solution is presented as follows;
CH₃NH₃⁺ → CH₃NH₂ + H⁺
Let x represent the number of moles of CH₃NH₂ and H⁺ produced,
respectively, we have;
The number of moles of CH₃NH₃⁺ remaining = 0.18 - x
Which gives;
[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]
Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹
Therefore;
[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]
2.27 × 10⁻¹¹ × (0.18 - x) = x²
2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0
Which gives;
x = [H⁺] ≈ 2.02 × 10⁻⁶
pH = -log[H⁺]
Which gives;
pH = -log(2.02 × 10⁻⁶) ≈ 5.69
C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;
Based on the Henderson-Hasselbalch equation, we have;
[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]
Which gives;
[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]
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5. Calcule las concentraciones cuando se alcanza el equilibrio si partimos de unas concentraciones iniciales [A]=[B]=1M ; [C]=[D]=0M y una constante de equilibrio de 5.
Las concentraciones en el equilibrio para la reacción química presentada son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
Consideremos la siguiente reacción química genérica:
A + B ⇄ C + D
Para calcular las concentraciones en el equilibrio, debemos construir una Tabla ICE. Cada fila representa una instancia (Inicial, Cambio, Equilibrio) y la completamos con la concentración o cambio de concentración ("x" para concentraciones desconocidas). Como inicialmente no hay productos, la reacción se desplazará hacia la derecha para alcanzar el equilibrio.
A + B ⇄ C + D
I 1 1 0 0
C -x -x +x +x
E 1-x 1-x x x
La constante de equilibrio, Kc, es:
[tex]Kc = 5 = \frac{[C][D]}{[A][B]} = \frac{x^{2} }{(1-x)^{2} } \\\sqrt{5} = x/1-x\\x = 0.69[/tex]
Las concentraciones en el equilibrio son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
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1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.
The structures are shown in the image attached.
A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.
Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.
I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2) and tetrahydrofuran (image 3).
All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.
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State the different radiations emitted by radioactive elements.
When comparing Be2 and H2:
I. Be2 is more stable because it contains both bonding and antibonding valence electrons.
II. H2 has a higher bond order than Be2.
III. H2 is more stable because it only contains 1s electrons.
IV. H2 is more stable because it is diamagnetic, whereas Be2 is paramagnetic
a. II,III,IV
b.II,III
c.III only
d.I,II
e.III,IV.
Answer:
The answer is "Option b".
Explanation:
H2 does have bond energy of 1, while Be2 has a covalent bond of zero. Be2 has eight electrons, each of which dwells in a distinct orbital. As just a result, four of them are linked molecular orbitals and two are antibonding molecular orbitals, respectively. As just a result, this molecule is unstable. This chemical orbital, with a bond order of 1, has just two electrons. As a result, it is a very solid substance. H2's bond length is higher than Be2's. Since it only has one electron, H2 is more stable than that of other compounds.
For an atoms electrons, how many energy sublevels are present in the principal energy level n = 4?
A. 4
B. 9
C. 10
D. 16
E. 32
Answer:
by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32
What is the molarity of a solution that contains 0.75 mol Naci in 3.0 L of solution? Select one: O a. 4.0 M O b. 2.3 M O d. 3.8 M O d. 0.25 M Clear my choice
Answer:
[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]
The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.
moles of solute = 0.75 mol NaCl liters of solution = 3.0 LSubstitute these values into the formula.
[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]
[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]
Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.
[tex]molarity= 0.25 \ M \[/tex]
The molarity of the solution is 0.25 Molar and Choice D is correct.
Consider the chemical reaction: N2 3H2 yields 2NH3. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2.5 x 10-1 M, calculate the reaction quotient (Q) and determine which way the chemical system would shift by comparing the value of Q to K.
In this equilibrium, the chemical system will shift to the right in order to produce more NH₃.
The equilibrium constant of a reaction is defined as:
"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".
The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.
To solve this question we need this additional information:
For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:
[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
[tex]K = \frac{[NH_3]^2}{[H_2]^3[N_2]}[/tex]
Where [] are concentrations in equilibrium.
And Q, is:
[tex]Q = \frac{[NH_3]^2}{[H_2]^3[N_2]}[/tex]
Where actual concentrations are:
[NH₃] = 1.0x10⁻⁴M
[N₂] = 4.0M
[H₂] = 2.5x10⁻¹M
Replacing:
Q = 1.6x10⁻⁷
As Q < K,
The chemical system will shift to the right in order to produce more NH₃
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Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5
B)5,5
C)6,5
D)7,5
E)9,5
20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
The empirical formula is OsO₄ :
Explanation:
Osmium oxide contains osmium and oxygen only.
Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:
Mass of compound = 2.89 g
Mass of Os = 2.16 g
Mass of O =?Mass of O = (Mass of compound) – (Mass of Os)
Mass of O = 2.89 – 2.16
Mass of O = 0.73 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Os = 2.16 g
Mass of O = 0.73 g
Empirical formula =..?Os = 2.16 g
O = 0.73 g
Divide by their molar mass of
Os = 2.16 / 190 = 0.011
O = 0.73 / 16 = 0.046
Divide by the smallest
Os = 0.011 / 0.011 = 1
O = 0.046 / 0.011 = 4
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PLEASE HELP ME
a)
b)
c)
or d)?
Answer:
D / 15.0 g
Explanation:
3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions
.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationWhich subshells are found in each of the following shells
electron subshell - M shell
Answer:
3
Explanation:
The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.
As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Hence, M shell contains s,p and d subshells.
calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume
Answer :
volume of a gas = weight * 22.4 l / gram molecular weight
volume of o2 = ?
weight given = 20.5 g
gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )
volume of oxygen = 20.5 * 22.4 / 32
volume of oxygen = 14.35 liters
Explanation:
hope this helps you
if wrong just correct me
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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Another method for creating a buffer, in situ, is to add an appropriate amount of a strong base, e.g., NaOH, to a weak acid OR add an appropriate amount of a strong acid, e.g., HNO3, to a weak base. As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide. Given this information, which of the following, when mixed with the appropriate amount of HCl, would create a buffer solution?
a. HNO3
b. HClO2
c. LiCl
d. NH3
Answer:
As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide.
When HClO2 is mixed with the appropriate amount of HCl it would create a buffer solution. That is option B.
Methods used to form buffer solutionA buffer solution is the solution that resists a change in pH of a solution when acid or base is added because it is made up of weak acid and the conjugate base or weak base and the conjugate acid.
The methods that can be used to form a buffer solution include:
Adding a strong base to a weak acid: For example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate.Adding a weak acid to a conjugate base: For example HCl is a strong acid which will react with a conjugate base such as HClO2.Although HCl is a strong acid, it can be converted to a weak acid through dilution with water. It is in this context that it can be used to form a buffer solution.
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What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.
Question Completion with Options:
O coarse...few...rapid
O fine...few...slow
O fine...multiple...rapid
O coarse...few...slow
O fine...multiple...slow
Answer:
The choice that best completes the sentence is:
O coarse...few...slow
Explanation:
Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow. This is because of the process that starts with recrystallization, recovery, and nucleation before growth can occur. While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.
which of the following is indicated by the ph value of a solution?
a- it's hydrogen ion concentration
b- its ammonium ion concentration
c- ability to undergo chemical reaction
d- its ratio of solute amount to solvent volume
Answer:
c- ability to undergo chemical reaction
6. Who stated that matter is not composed of particles
After careful consideration your answer is...
Leucippus and Democritus
*Hope I helped*
~Alanna~
Answer:
The first theories of matter were put forward by Empedocles in 450 BC, he proposed that all matter was composed of four elements - Earth, air, fire and water. Later, Leucippus and Democritus suggested matter was made up of tiny indestructible particles continuously moving in empty space.
Explanation:
Consider the following reaction:
Cr(NO3)3 (aq) + 2NaF (aq) --> 3NaNO3 (aq) + CrF3 (s)
If 21.0 grams of NaF are needed to precipitate all of the Cr+3 ions present in 0.125L of a solution of Cr(NO3)3, what is the molarity of the Cr(NO3)3 solution?
Your answer should be to 2 decimal places.
Answer:
2.01
Explanation:
First, let's convert grams to moles
(Na) 22.99 + (F) 18.998 = 41.988
Every mole of NaF is 41.988 grams
21/41.988 = 0.500143 moles of NaF
For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF
0.500143/2 = 0.250071
molarity = moles/liters
0.250071/0.125 = 2.0057 M
How many moles of Al2O3 can be formed from 10.0 g of Al?
Select an answer and submit. For keyboard navigation, use the up/down arrow keys to
select an answer.
Answer:
n Al= 10/27( mol)- >n Al2O 3 =5/27(mol)
Explanation:
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
a) potassium tetracyanonickelate(II)
b) sodium diamminedicarbonatoruthenate(III)
c) diamminedichloroplatinum(II)
Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
A reaction rate increases by a factor of 500. in the presence of a catalyst at 37oC. The activation energy of the original pathway is 106 kJ/mol. What is the activation energy of the new pathway, all other factor being equal
Answer:
[tex]E_2=999984KJ/mole[/tex]
Explanation:
From the question we are told that:
Factor [tex]dK=500[/tex]
Temperature [tex]T=37 C=310k[/tex]
Activation energy [tex]E=10^6kJ/mol[/tex]
Generally the Arhenius equation is mathematically given by
[tex]ln \frac{K_2}{K_1}=\frac{ E_1-E_2}{RT}[/tex]
Where
[tex]\frac{K_2}{K_1}=500[/tex]
[tex]ln 500=\frac{ 10^6-10^3-E_2}{8.314*310}[/tex]
[tex]E_2=999984KJ/mole[/tex]
The activation energy of the new reaction is 105.99 kJ/mol.
Using the Arrhenius equation;
ln(k2/k1) = -Ea2/RT2 + Ea1/RT1
Now, from the information in the question;
k2/k1 = 500
Ea = ?
R = 8.314 JKmol-1
T2 = 37oC + 273 = 310 K
T1 = 37oC + 273 = 310 K
Substituting values;
ln (500) =- Ea2 + Ea1
6.2 = -Ea2 + 106 × 10^3 J
Ea = 106 × 10^3 J - 6.2
Ea = 105.99 × 10^3 J or 105.99 kJ/mol
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