The cation of the salt is sodium ion, and the anion is aurothiosulfate ion. Based on the chemical formula of the salt, what must the charge be of the aurothiosulfate ion [Au(S2O3)2 n- ]

Answers

Answer 1

Answer:

3-

Explanation:

Sodium aurothiosulfate is a salt with the formula Na₃Au(S₂O₃)₂. The cation of the salt is sodium ion, and the anion is aurothiosulfate ion. We can determine the charge of the aurothiosulfate ion, considering that the sum of the positive and negative charges must be equal to the charge of the compound, which is zero.

3 × Na⁺ + 1 × Au(S₂O₃)₂ⁿ⁻ = 0

3 × +1 + 1 × Au(S₂O₃)₂ⁿ⁻ = 0

Au(S₂O₃)₂ⁿ⁻ = 3-


Related Questions

What is the mass of a piece of iron if its density is 1.98 g/mL and its volume is 2.45 mL?
0.80 g
4.858
1.248
5.998
2.71 g

Answers

Answer:

4.858 g

Explanation:

Start with the formula

density = [tex]\frac{mass}{volume}[/tex]

density = 1.98 g/mL

volume = 2.45 mL

mass = ??

rearrange the formula to solve for mass

(density) x (volume) = mass

Add in the substitutes and solve for mass

1.98 g/mL x 2.45 mL = 4.858 g

How do I do this? What are the answers to the 5 questions shown?

Answers

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

The standard enthalpies of combustion of fumaric acid and maleic acid (to form carbon dioxide and water) are - 1336.0 kJ moJ-1 and - 1359.2 kJ moJ-1, respectively. Calculate the enthalpy of the following isomerization process:

maleic acid ----> fumaric acid

Answers

Answer:

Explanation:

maleic acid ⇒ fumaric acid

ΔHreaction = ΔHproduct - ΔHreactant

ΔHproduct = -1336.0 kJ mol⁻¹

ΔHreactant = - 1359.2 kJ mol⁻¹.

ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)

=   1359.2 kJ mol⁻¹   -1336.0 kJ mol⁻¹

= 23.2 kJ mol⁻¹ .

Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.

A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. What is the empirical formula of the compound? a. C2H5NO b. CH3NO c. C3H9N2O2 d. C4HN3O4 e. C4H13N3O3

Answers

Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

Moles N -Molar mass: 14.01g/mol-

0.420g N * (1mol/14.01g) = 0.0300 moles N

Moles O -Molar mass: 16g/mol-

0.480g O * (1mol/16g) = 0.0300 moles O

Moles C -Molar mass: 12.01g/mol-

0.540g C * (1mol/12.01g) = 0.0450 moles C

Moles H -Molar mass: 1.0g/mol-

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

c. C3H9N2O2

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

Answers

Answer:

c- ability to undergo chemical reaction

What is the molality of a glucose solution prepared by dissolving 16.7 g of glucose, C6H12O6, in 133.6 g of water

Answers

Answer:

0.696 m

Explanation:

We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:

Mass of C₆H₁₂O₆ = 16.7 g

Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)

= 72 + 12 + 96

= 180 g/mol

Mole of C₆H₁₂O₆ =?

Mole = mass / molar mass

Mole of C₆H₁₂O₆ = 16.7 / 180

Mole of C₆H₁₂O₆ = 0.093 mole

Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

133.6 g = 133.6 g × 1 Kg / 1000 g

133.6 g = 0.1336 Kg

Thus, 133.6 g is equivalent to 0.1336 Kg.

Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:

Mole of C₆H₁₂O₆ = 0.093 mole

Mass of water = 0.1336 Kg

Molality =?

Molality = mole / mass of water (in Kg)

Molality = 0.093 / 0.1336

Molality = 0.696 m

Therefore, the molality of the solution is 0.696 m

A dehydration reaction starting with 3.0 g cyclohexanol produces 1.9 g cyclohexene. Calculate the theoretical yield for this reaction. Report your answer with two significant figures.

Answers

Answer:

77%

Explanation:

First we convert 3.0 g of cyclohexanol (C₆H₁₂O) to moles, using its molar mass:

Molar mass of C₆H₁₂O = 100.158 g/mol3.0 g ÷ 100.158 g/mol = 0.030 mol

Then we convert 1.9 g of cyclohexene (C₆H₁₀) to moles, using its molar mass:

Molar mass of C₆H₁₀ = 82.143 g/mol1.9 g ÷ 82.143 g/mol = 0.023 mol

Finally we calculate the theoretical yield:

0.023 mol / 0.030 mol * 100% = 77%

The Tollen's test is the reaction of aldehydes with silver(l) ions in basic solution to form silver metal and a carboxylate. 2 Ag+ + + 3 OH- - HR 2 Ag +_ W + 2 H2O ÃR Which species is being oxidized in the reaction? Choose... Which species is being reduced in the reaction? Choose... Which species is the visual indicator of a positive test? v Choose... Carboxylate ion Aldehyde Silver metal Water Silver(1) ion Hydroxide ion

Answers

Answer:

Explanation:

When Tollen's test is done by aldehyde , silver ion is converted into silver which forms a layer which looks like a mirror.

Ag⁺ + e = Ag

It is a reduction process where silver(1) ion is reduced to metallic silver.

Aldehyde is oxidized to carboxylate ion.

CH₃CHO + 2 OH⁻ = CH₃COO⁻ + H₂O + H⁺ + 2e

Visual indicator is silver metal which forms silver mirror at the bottom of test tube .

name a factor tht affects the value of electron affinity​

Answers

Answer:

Atomic sizeNuclear chargesymmetry of the electronic configuration
Various factors that affect electron affinity are atomic size, nuclear charge and the symmetry of the electronic configuration. Atomic size: With increase in the atomic size, the distance between the nucleus and the incoming electron also increases.

For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb

Answers

Answer:

11.12 → pH

Explanation:

This is a titration of a weak base and a strong acid.

In the first step we did not add any acid, so our solution is totally ammonia.

Equation of neutralization is:

NH₃ + HCl → NH₄Cl

Equilibrium for ammonia is:

NH₃ + H₂O ⇄  NH₄⁺  +  OH⁻      Kb = 1.8×10⁻⁵

Initially we have 50 mL . 0.10M = 5 mmoles of ammonia

Our molar concentration is 0.1 M

X amount has reacted.

In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.

Expression for Kb is : x² / (0.1 - x)  = 1.8×10⁻⁵

As Kb is so small, we can avoid the x to solve a quadratic equation.

1.8×10⁻⁵ = x² / 0.1

1.8×10⁻⁵  .  0.1 = x²

1.8×10⁻⁶ = x²

√1.8×10⁻⁶ = x → 1.34×10⁻³

That's the value for [OH⁻] so:

1×10⁻¹⁴ = [OH⁻] . [H⁺]

1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺]7.45×10⁻¹²

- log [H⁺] = pH

- log 7.45×10⁻¹² = 11.12 → pH

Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?

Answers

Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

[A] = 11.9g remains after 48.2 days

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. Decide, in each case, whether or not an insoluble precipitate is formed.

a. K2S and NH4Cl
b. CaCl2 and NH4CO3
c. Li2S and MnBr2
d. Ba(NO3)2 and Ag2SO4
e. RbCO3 and NaCl

Answers

Answer:

a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :

There are no insoluble precipitate forms.

b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :                        

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.

                                          KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex]  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.

                                        [tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex]  ⇒ soluble.

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c)

Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.

                                        [tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex]  ⇒ insoluble.

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d)

Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.

                                     

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.

                                        [tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex]  ⇒ soluble.

There are no insoluble precipitates forms.

A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide

Answers

Answer:

MoO2

Explanation:

The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.

The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239.9g/mol-

12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo

Mass Mo -95.95g/mol-:

0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo

Mass oxygen in the oxide:

13.197 - 9.895g = 3.302g Oxygen

Moles oxygen -Molar mass: 16g/mol-:

3.302g Oxygen * (1mol / 16g) = 0.206 moles O

Now, the ratio of moles O / moles Mo is:

0.206 moles O / 0.1031 moles Mo = 2

That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:

MoO2

11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride

Answers

Explanation:

Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.

OR

Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3

what are the properety of covalent bond​

Answers

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.

what is the difference between 25ml and 25.00ml​

Answers

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 type of bonding is occuring in the compound below?

A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar

Answers

Answer:

(B). it's metallic bonding

A gas at 273K temperature has a pressure of 590 MM Hg. What will be the pressure if you change the temperature to 273K? 

Answers

Explanation:

here's the answer to your question

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)

Answers

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.

Learn more about stoichiometry at: https://brainly.com/question/9743981

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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

Answers

6.5< x < 8.5 hope this helps

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.

Answers

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.

https://brainly.com/question/9988658

tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine

Answers

.36 g of caffeine for this problem. 2% of 18g is 0.36g

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.

Answers

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%

here is the question

Answers

Answer:

1. Nitrate ions, NaNO3 - Sodium nitrate.

2. Sulphide ions, K2S - Potassium sulphide.

3. Sulphate ions, CaSO4 - Calcium sulphate.

4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.

5. Carbonate ions, CaCO3 - Calcium carbonate.

6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.

7. Phosphite ions, PH3 - Hydrogen phosphite.

8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).

9. Ethanoate ions, CH3COONa - Sodium ethanoate.

10. Methanoate ions, HCOONa - Sodium methanoate.

11. Fluoride ions, HF - Hydrogen fluoride.

12. Chloride ions, KCl - Potassium chloride.

13. Bromide ions, HBr - Hydrogen bromide.

14. Iodide ions, NaI - Sodium iodide.

15. Phosphate ions, K3PO3 - potassium phosphate.

A major component of gasoline is octane (C8H8). When liquid octane is burned in air it reacts with oxygen (O2) gas to produce "0.050 mol" carbon dioxide gas and water vapor. Calculate the moles of octane needed to produce of carbon dioxide.

Answers

Answer:

0.0063 mol

Explanation:

Step 1: Write the balanced combustion equation

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(g)

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of C₈H₁₈ to CO₂ is 1:8.

Step 3: Calculate the number of moles of C₈H₁₈ needed to produce 0.050 moles of CO₂

0.050 mol CO₂ × 1 mol C₈H₁₈/8 mol CO₂ = 0.0063 mol C₈H₁₈

calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume​

Answers

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

Which subshells are found in each of the following shells
electron subshell - M shell

Answers

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.

Please help thank you

Answers

Answer:

[tex]K=1.7x10^{-3}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:

[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]

Next, we plug in the given concentrations on the data table to obtain:

[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]

Regards!

Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?

Answers

cyclohexane will distill off first as it will have lower boiling point compared to ortho xylene which has higher molecular mass

Melanie has completed the analysis of her data for the reaction of KMnO4 with malonic acid and data for a reaction of KMnO4 with tartaric acid. She compared the activation energies, Ea, she calculated for the two reactions and found the Ea for the malonic acid reaction to be greater than the Ea for the tartaric acid reaction.

Required:
What does this mean about the magnitude of the rate constant, k, and the rate of the reaction?

Answers

Answer:

See explanation

Explanation:

The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;

k= Ae^-Ea/RT

Where;

k= rate constant

A= pre-exponential factor

Ea=activation energy

R= gas constant

T= temperature

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.

Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

The study of chemistry and bonds is called chemistry. There are two types of elements metal and nonmetals.

The correct answer is mentioned below.

What is the Arrhenius equation?The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation

The equation is as follows:-

[tex]k= Ae^{-Ea/RT[/tex] Where;

k= rate constantA= pre-exponential factorEa=activation energyR= gas constantT= temperature

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have very high activation energy are markedly slow. Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

Hence, the correct answer is mentioned above.

For more information about the equation, refer to the link:-

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