The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.

Answers

Answer 1

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

Half-life at 629K = 252.4min

Answer 2

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature  

   

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

Find more about activation energy here:

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I hope it helps you!


Related Questions

. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:

Answers

Answer:

pH of solution B is 5

Explanation:

A weak acid, HA, is in equilibrium with water as follows:

HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)

Where Ka (10^-pKa = 1x10⁻⁹) is:

Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

Where concentrations of this species are equilibrium concentrations

As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:

[HA] = 0.1M - X

[A⁻] = X

[H₃O⁺] = X

Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.

Replacing in Ka expression:

1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

1x10⁻⁹ = [X] [X] / [0.1 - X]

1x10⁻¹⁰ - 1x10⁻⁹X = X²

1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0

Solving for X:

X = -0.00001 → False solution, there is no negative concentrations.

X = 1x10⁻⁵ → Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 1x10⁻⁵M

And pH = -log[H₃O⁺]

pH = 5

pH of solution B is 5

Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O

Answers

Answer:

0.423 m.

Explanation:

The following data were obtained from the question:

Mass of glycine (NH2CH2COOH) = 141.5 g

Mass of water = 4.456 kg

Molality =.?

Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.

This is illustrated below:

Mass of glycine (NH2CH2COOH) = 141.5 g

Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol

Mole of glycine (NH2CH2COOH) =.?

Mole = mass /Molar mass

Mole of glycine (NH2CH2COOH) = 141.5/75

Mole of glycine (NH2CH2COOH) = 1.887 moles

Finally, we shall determine the molality of the solution as follow:

Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:

Molality = mole / mass (kg) of water

With the above formula, we can obtain the molality of the solution as follow:

Mole of glycine (NH2CH2COOH) = 1.887 moles

Mass of water = 4.456 kg

Molality =.?

Molality = mole /mass (kg) of water

Molality =1.887/4.456

Molality = 0.423 m

Therefore, the molality of the solution is 0.423 m

What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond

Answers

Answer:

A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable

Explanation:

243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.​

Answers

Answer:

A. The atomic mass is 243 amu, and the atomic number is 95.

Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.

Answers

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

At a constant temperature, a sample of a gas in a balloon that originally had a volume of 5.00 L and pressure of 626 torr has its volume changed to 6.72 L. Calculate the new pressure in torr.

Answers

Answer:

466 torr

Explanation:

Step 1: Given data

Initial pressure (P₁): 626 torrInitial volume (V₁): 5.00 LFinal pressure (P₂): ?Final volume (V₂): 6.72 LConstant temperature

Step 2: Calculate the final pressure

Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁ / V₂

P₂ = 626 torr × 5.00 L / 6.72 L

P₂ = 466 torr

For each bond, show the direction of polarity by selecting the correct partial charges. _________ Si-P _________ _________ Si-Cl _________ _________ Cl-P _________ The most polar bond is _______

Answers

Answer:

Siδ⁺ -- Pδ⁻⁻

Clδ⁻⁻ -- Pδ⁺

Siδ⁺ -- Clδ⁻⁻

Of the mentioned bonds the most polar bond is Si -- Cl

The polarity of the bond primarily relies upon the electronegativity difference between the two atoms that forms the bond. Therefore, if the electronegativity difference between the two atoms that forms the bond is more the bond will be more polar, and if it is less then the bond will be less polar. The electronegativity of the atoms mentioned is Si = 1.8 , P = 2.1 and Cl = 3.00.  

Therefore, the Si - Cl atoms exhibit more electronegativity difference, thus, the Si - Cl bond will be the most polar bond.  

whts the ph of po4 9.78

Answers

Answer:

4.22

Explanation:

We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].

If the pOH of a solution is given, one may obtain the pH of such solution from the formula;

pH + pOH =14

Hence we can write;

pH = 14-pOH

pH = 14 - 9.78 = 4.22

Hence the pH of the solution is 4.22.

Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b. Glu-Ala-Phe-Gly-Ala-Tyr by chymotrypsin

Answers

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

b. Glu-Ala-Phe + Gly-Ala-Tyr

Explanation:

In this case, we have to remember which peptidic bonds can break each protease:

-) Trypsin

It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.

-) Chymotrypsin

It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.

With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).

In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.

What is the formula of a compound if a sample of the compound contains 0.492 mol X, 0.197 mol Y, and 0.295 mol Z?

Answers

Answer:

X₅Y₂Z₃

Explanation:

The formula of a compound is determined as the whole number ratio between moles of each element present in the molecule.

The molecule is made from X, Y and Z. To fin the ratio we will divide the given moles in the moles of Y (0.197 moles), because is the element with the low number of moles.

X = 0.492 moles / 0.197 moles = 2.5

Y = 0.197 moles / 0.197 moles = 1

Z = 0.295 moles / 0.197 moles = 1.5

But, as the formula is given just with whole numbers, if we multiply each number twice:

X = 2.5*2 = 5

Y = 1*2 = 2

Z = 1.5*2 = 3

The formula is:

X₅Y₂Z₃

What will be formed when 2,2,3-trimethylcyclohexanone reacts with hydroxylamine?

Answers

Answer:

Following are the solution to this equation:

Explanation:

In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:

In the given question "Option (iii)" is correct, which is defined in the attachment file.

When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.

which of the following is an example of precipitate a. liquid evaporating into gas b. a solid form from a frozen liquid c. a chunky solid form from 2 liquids combining d. a liquid solution that contains 2 substances

Answers

Answer:

B and C

Explanation:

The google definition of precipitate is "a solid formed by a change in a solution, often due to a chemical reaction or change in temperature that decreases solubility of a solid". So in this case, the "solid formed from a frozen liquid" and "a chunky solid form from 2 liquids combining" are both examples of a precipitate.

The correct answer is C, a chunky solid form from 2 liquids combining.

In the laboratory, we usually mix two chemicals. When we mix the chemicals, we expect them to interact in one way or another. The interaction of those chemicals is known as a chemical reaction.

When we mix two liquid chemicals and they interact with each other to yield a solid product, we say that a chemical reaction has occurred and that such chemical reaction has produced a precipitate.

https://brainly.com/question/24276721

A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but addition of (SO4)2- ion resulted in a precipitate. Which cation is present

Answers

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ

Answers

Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ

Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

Enthalpy is defined as internal heat existent in the system. It is calculated as:

[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]

Using Enthalpy Formation Table:

[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]

[tex]\Delta H^{0} = 62,6 kJ[/tex]

Entropy is the degree of disorder in the system. It is found by:

[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]

Calculating:

[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]

[tex]\Delta S^{0} = -354.1J[/tex]

And so, Gibbs Free energy will be:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]

[tex]\Delta G^{0} = 168121.8 J[/tex]

Rounding to the nearest kJ:

[tex]\Delta G^{0}[/tex] = 168.12 kJ

An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (answer must be in 3 sig figs)

Answers

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Thats all i know

How do covalent bonds form? A Sharing valence electrons between atoms. B Donating and receiving valence electrons between atoms. C Opposite slight charges attract each other between compounds. D Scientists are still not sure how they form.

Answers

Answer:

A. Sharing valence electrons between atoms.

Explanation:

This is the definition of a covalent bond. Option B describes ionic bonds, Option C describes intermolecular forces, and Option D is wrong because then there wouldn't be any mention of them in our high school chemistry textbooks :).  

What is the density of a 10 kg mass that occupies 5 liters?
( pls need help)

Answers

Answer: d=2000 g/L

Explanation:

Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.

[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]

Now that we have grams, we can divide to get density.

[tex]d=\frac{10000g}{5 L}[/tex]

d=2000g/L

g Does a reaction occur when aqueous solutions of barium hydroxide and aluminum sulfate are combined

Answers

Answer:

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

Explanation:

Yes! A reactiin occurs between barium hydroxide and auminium sulphate.

barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.

The reaction is given by the equation below;

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide

Answers

Answer:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Explanation:

Hello,

In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Best regards.

The new hybrid car can get 51.5 km/gal. It has a top speed of 40000.00 cm/min and is 4m long. How fast can the car go in m/hr?


Answers

Answer:

The anawer of this question is 0.024 m/h

Explanation:

Other explanations of the question are additional.

An atom of 108Te has a mass of 107.929550 amu. Calculate the binding energy per MOLE in kJ. Use the values: mass of 1H atom

Answers

Answer:

The binding energy = 8.64972649×10¹⁰ kJ/mole

Explanation:

Given that:

An atom of 108Te has a mass of 107.929550 amu.

In a 108 Te atom, there are 52 protons and 56 neutrons

where;

mass of proton= 1.007825 amu

mass of neutron= 1.008665 amu

Similarly; The atomic number of Te = 52

the mass of 52 protons = 52 ×  1.007825  amu

the mass of 52 protons = 52.4069 amu

the mass of 56 neutrons = 56 ×  1.008665 amu

the mass of 56 neutrons = 56.48524 amu

The total mass can now be = the mass of 52 protons + the mass of 56 neutrons

The total mass = 52.4069 amu +  56.48524 amu

The total mass = 108.89214  amu

Recall  : it is given that An atom of 108Te has a mass of 107.929550 amu.

Therefore, the mass defect will be = 108.89214  amu - 107.929550 amu

the mass defect = 0.96259amu

where 1 amu = 1.66× 10⁻²⁷ kg

Therefore, 0.96259amu = (0.96259  × 1.66× 10⁻²⁷) kg

= 1.5978994 × 10⁻²⁷kg

The binding energy = mass defect × (speed of light)²

where;

speed of light c = 2.99792 × 10⁸ m/s

The binding energy = 1.5978994 × 10⁻²⁷kg  ×  2.99792 × 10⁸ m/s

The binding energy = 1.43611597 × 10⁻¹⁰  J

The binding energy =  1.43611597 × 10⁻¹³ kJ/atom

since 1 mole = 6.023 × 10²³ atom (avogadro's constant)

Then;

The binding energy = ( 1.43611597 × 10⁻¹³ )× (6.023 × 10²³)  kJ/mole

The binding energy = 8.64972649×10¹⁰ kJ/mole

1500 L has how many significants figures

Answers

Answer:

It has 2

Explanation:

The significant figures are 1 and 5!

Hope this helps:)

Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(

Answers

Answer:

The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].

The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].

Explanation:

The oxidation states of atoms in a compound should add up to zero.

Ag₂O

There are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:

[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].

SO₂

Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:

[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.

Therefore:

[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the oxidation state of [tex]\rm S[/tex] here:

[tex]\text{Oxidation state of $\rm S$} = 4[/tex].

Consider the bond dissociation energies listed below in kcal/mol. CH3-Br 70 CH3CH2-Br 68 (CH3)2CH-Br 68 (CH3)3C-Br 65 These data show that the carbon-bromine bond is weakest when bromine is bound to a __________.

Answers

Answer:

The answer is "Tertiary carbon".

Explanation:

Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is=  68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.

The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.

The rate constant for the decay of a radioactive element is 3.68 × 10⁻³ day⁻¹. What is the half-life of this element?

Answers

Answer:

half life=0.693/rate constant =188.3

The half-life of this element is 188.32 years

The formula for calculating the half-life of an element is expressed according to the equation:

[tex]t_{1/2}=\frac{ln 2}{\lambda}[/tex]

[tex]\lambda[/tex] is the decay constantt1/2 is the half-life

Given the following parameters:

The rate constant for the decay = 3.68 × 10⁻³ day⁻¹.

Substitute into the formula for calculating the half-life as shown:

[tex]3.68\times 10^{-3}=\frac{0.693}{\lambda}\\ 0.00368=\frac{0.693}{\lambda}\\\lambda=\frac{0.693}{0.00368}\\\lambda = 188.32 years[/tex]

Hence the half-life of this element is 188.32 years

Learn more on half-life here: https://brainly.com/question/23438944

A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculate the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol. He next tried the same experiment with 100 ml of 1.0 M HCl and 100 ml of 1.0 M NaOH. The calculated molar enthalpy change of reaction for his second trial was:

Answers

Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.

Explanation:-

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]

Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]

Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]

Given for second trial:

[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]

[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]

0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release  heat = 54 kJ

0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release  heat  =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]

Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.

How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−

Answers

The given question is incomplete, the complete question is:

How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures

Answer:

The correct answer is spontaneous at all the temperatures.

Explanation:

Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS

When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol

= -126000 J/mol, it is negative

ΔS = 146 J/K/mol, it is positive

Now, ΔG = ΔH-TΔS

= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.  


Zeros laced at the end of the significant number are...

Answers

Answer:

Zeros located at the end of significant figures are significant.

Explanation:

Hope it will help :)

Chemistry
What is a chemical reaction

Answers

Answer:

A process that involves rearrangement

Explanation:

A chemical reaction is the process that involves rearrangement of the molecular or ironic structure of a substance, as a distinct from a change in physical form or a nuclear reaction.

Answer:

Explanation:

Chemistry

The chemical reaction H2(g) + ½ O2(g) → H2O(l) describes the formation of water from its elements.

The reaction between iron and sulfur to form iron(II) sulfide is another chemical reaction, represented by the chemical equation:

8 Fe + S8 → 8 FeS

What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0×10–14), with a concentration of hydroxide ions of 2.21×10–6 M? answer options: A) 2.8×10–8 M B) 4.52 ×10–9 M C) 1.6×10–9 M D) 3.1×10–6 M

Answers

Answer:

ITS NOT D. ITS B. 4.52x10^-9 M

Explanation:

Answer:

4.52 ×10–9 M

Explanation:

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