Explanation:

There will occur a straight line for the curve of Ln[(CH_{3})_{3}CCl] Vs. time. Since, it is a first order reaction therefore, the formula of its half-life will be as follows.

Half-life =

or, k =

=

So, rate of the reaction will be as follows.

Rate =

=

= 0.000473

=

Thus, we can conclude that the instantaneous rate of given reaction is .

A second-order reaction has a half-life of 12 s when the initial concentration of reactant is 0.98 M. The rate constant for this reaction is __________ M-1s-1.

The rate constant for the reaction will be "0.085 M⁻¹s⁻¹".

The given values:

- Half life = 12 s
- Initial concentration, = 0.98 M

As we know the formula,

→

By substituting the given values, we get

Thus the above answer is appropriate.

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Answer:

0.085 (Ms)⁻¹

Explanation:

Half life = 12 s

is the initial concentration = 0.98 M

Half life expression for second order kinetic is:

So,

k = 0.085 (Ms)⁻¹

The rate constant for this reaction is 0.085 (Ms)⁻¹ .

An enzyme catalyzes a reaction with a K m of 6.50 mM and a V max of 4.25 mM ⋅ s − 1 . Calculate the reaction velocity, v 0 , for each substrate concentration.

The given question is incomplete. The complete question is as follows.

An enzyme catalyzes a reaction with a K m of 6.50 mM and a V max of 4.25 . Calculate the reaction velocity, v 0 , for each substrate concentration.

(a) 4.00 mM, (b) 6.50 mM

Explanation:

It is given that the enzyme catalyzes a reaction with a of 6.50 mM and a of 4.25 .

So, we will calculate the reaction velocity (), for the following substrate concentrations as follows.

V =

Putting the given values into the above formula as follows.

V =

a) It is given that [S] = 4 mM

So, V =

V =

V = 1.619 mM/s

Hence, the reaction velocity (), for substrate concentration 4 mM is 1.619 mM/s.

b) It is given that [S] = 6.5 mM

So, V =

V =

V = 2.125 mM/s

Hence, the reaction velocity (), for substrate concentration 6.5 mM is 1.619 mM/s.

Classify each of these reactions. Pb ( NO 3 ) 2 ( aq ) + 2 KCl ( aq ) ⟶ PbCl 2 ( s ) + 2 KNO 3 ( aq ) C 2 H 4 ( g ) + 3 O 2 ( g ) ⟶ 2 CO 2 ( g ) + 2 H 2 O ( l ) Cu ( s ) + FeCl 2 ( aq ) ⟶ Fe ( s ) + CuCl 2 ( aq ) 2 KOH ( aq ) + H 2 SO 4 ( aq ) ⟶ K 2 SO 4 ( aq ) + 2 H 2 O ( l )

Answer:

For a: The reaction is a double displacement reaction.

For b: The reaction is a combustion reaction.

For c: The reaction is a single displacement reaction.

For d: The reaction is a neutralization reaction.

Explanation:

For the given equations:

Equation 1:

This reaction is an example of double displacement reaction because here exchange of ions takes place.

Equation 2:

This reaction is an example of combustion reaction because here hydrocarbon is reacting with oxygen gas to produce carbon dioxide gas and water molecule

Equation 3:

This reaction is an example of single displacement reaction because here more reactive element displaces a less reactive element.

Equation 4:

This reaction is an example of neutralization reaction because here a base reacts with an acid to produce a salt and water molecule

Compute diffusion coefficients for the interdiffusion of carbon in both (a) α-iron (BCC) and (b) γ-iron (FCC) at 1000˚C. Assume that D0 for the interdiffusion of carbon in α-iron and in γ-iron are 1.1 × 10-6 and 2.3 × 10-5 m2/s, respectively, and that Qd are 80 and 148 kJ/mol, respectively.

Explanation:

According to Ficks first law, the relation between diffusion flux and diffusion coefficient is as follows.

J =

where, D = function of temperature

D =

where, = pre-exponential independent of temperature

(a) Substituting the given values into the above formula, we will calculate the value of D as follows.

D =

=

=

=

=

Therefore, diffusion coefficient for the inter-diffusion of carbon in -iron is .

(b) For -iron we will calculate the value of D as follows.

D =

=

=

=

=

Therefore, diffusion coefficient for the inter-diffusion of carbon in -iron is .

Find how many milliliters of NaOH should be used to reach the half-equivalence point during the titration of 20.00 mL 0.888 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54×10–5) with 0.425 M NaOH solution. Enter 2 decimal places.

Answer: The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

Explanation:

The chemical equation for the dissociation of butanoic acid follows:

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the negative value because concentration cannot be negative

To calculate the volume of base, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is butanoic acid

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

Consider the following multistep reaction: A+B→AB(slow) A+AB→A2B(fast)–––––––––––––––––––– 2A+B→A2B(overall) Based on this mechanism, determine the rate law for the overall reaction. Express your answer in standard MasteringChemistry format. For example, if the rate law is k[A]3[B]2 type k*[A]^3*[B]^2. View Available Hint(s)

Answer : The rate law for the overall reaction is,

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : (slow)

Step 2 : (fast)

(overall)

The rate law expression for overall reaction should be in terms of A and B.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

The expression of rate law for this reaction will be,

Hence, the rate law for the overall reaction is,

Write the Ka expression for an aqueous solution of hypochlorous acid: (Note that either the numerator or denominator may contain more than one chemical species. Enter the complete numerator in the top box and the complete denominator in the bottom box. Remember to write the hydronium ion out as H3O+ , and not as H+) Ka= _______________

Answer:

Ka = [H3O+] [ClO-] / [HOCl]

Explanation:

First let us generate the dissociation equation of hypochlorous acid(HOCl) in water this given below:

HOCl + H2O H3O+ + ClO-

Now we can write the Ka as follows:

Ka = [H3O+] [ClO-] /[HOCl]

If a solution containing 53.99 g 53.99 g of mercury(II) nitrate is allowed to react completely with a solution containing 9.718 g 9.718 g of sodium sulfide, how many grams of solid precipitate will form?

Answer:

28.97 g

Explanation:

The formula for the calculation of moles is shown below:

For mercury(II) nitrate

Mass = 53.99 g

Molar mass = 324.7 g/mol

The formula for the calculation of moles is shown below:

Thus,

For sodium sulfide

Mass = 9.718 g

Molar mass = 78.0452 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the given reaction:

1 mole of mercury(II) nitrate reacts with 1 mole of sodium sulfide

So,

0.16627 mole of mercury(II) nitrate reacts with 0.16627 mole of sodium sulfide

Moles of sodium sulfide = 0.16627 moles

Available moles of sodium sulfide = 0.12451 moles

Limiting reagent is the one which is present in small amount. Thus, sodium sulfide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of sodium sulfide produces 1 mole of mercury sulfide (precipitate)

So,

0.12451 mole of sodium sulfide produces 0.12451 mole of mercury sulfide (precipitate)

Moles of mercury sulfide = 0.12451 mole

Molar mass of mercury sulfide = 232.66 g/mol

Mass of mercury sulfide = Moles × Molar mass = 0.12451 × 232.66 g = 28.97 g

28.97 g of solid precipitate will form.

If a reaction mixture contains 3 moles of O2(g) with 4 moles of H2(g), which reactant is the limiting reactant and what is the mass (in grams) remaining of the reactant in excess?

Answer:

Limiting reactant: H₂

Mass in grams remaining of the reactant in excess: 32 g

Explanation:

Reaction for the excersise is: 2H₂(g) + O₂(g) → 2H₂O (l)

Ratio is 2:1. We make these rules of three:

2 moles of hydrogen need to react with 1 mol of oxygen

Then 4 moles of H₂ will react with (4 .1) / 2 = 2 moles of O₂

We have 3 moles of O₂ and we need 2 moles, so the oxygen is my reagent in excess. 1 mol of O₂ remains (32 g) with no reaction.

Limiting reactant is the hydrogen, but let's verify:

1 mol of oxygen needs 2 moles of H₂ to react

Therefore, 3 moles of O₂ will react with (3 . 2) / 1 = 6 moles of H₂

We need 6 moles of H₂, but we have only 4 moles, not enough H₂

Kf(water) = -1.86 °C/m In a laboratory experiment, students synthesized a new compound and found that when 13.39 grams of the compound were dissolved in 202.1 grams of water, the solution began to freeze at -2.050 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?

Answer:

The molecular weight for the compound is 60.1 g/mol

Explanation:

We need to determine the molality of solute to find out the molar mass of it.

We apply the colligative property of freezing point depression:

ΔT = Kf . m . i

If the compound was also found to be nonvolatile and a non-electrolyte,

i = 1.

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

0°C - (-2.05°C) = 1.86°C/m . m

2.05°C / 1.86m/°C = m → 1.10 mol/kg

To determine the moles of solute we used, we can multiply molality by the mass of solvent in kg → 202.1 g . 1kg/1000g = 0.2021 kg

1.10 mol/kg . 0.2021kg = 0.223 moles

Molar mass→ g/mol → 13.39 g / 0.223 mol = 60.1 g/mol

Provide structures for the products obtained by the reaction of propanal under five conditions outlined below. 1. Sodium Borohydride followed by work up. [A] 2. Methyl lithium followed by work up. [B] 3. Isopropyl magnesium bromide followed by work up. [C] 4. Methyl triphenylphosphonium bromide and a base. [D] 5. Two equivalents of methanol in the presence of an acid catalyst. [E] 6. Zinc amalgam and hydrochloric acid. [F]

Hahahha na shsshsnsnana hahahaha ave has hshshsnsnsbsn

The unsaturated hydrocarbon butadiene (C4H6) dimerizes to 4-vinylcyclohexan (C8H12).When data collected in studies of the kinetics of this reaction were plotted against reaction time, plots of [C4H6] or ln[C4H6] produced curved lines, but the plot of 1/[C4H6] was linear.a) What is the rate law for the reaction?b) How many half-lives will it take for the [C4H6] to decrease to 3.1% of its original concentration? The relationship of [A]t/[A]o = 0.50^n, where n = the number of half-lives.

Explanation:

From the information of the graph plots, the reaction is a second order reaction.

a) What is the rate law for the reaction?

Rate = k [C4H6]²

The superscript 2 signifies it is a second order reaction.

b) How many half-lives will it take for the [C4H6] to decrease to 3.1% of its original concentration?

Using the formular below;

[A]t / [A]o = 0.50^n

[A]t = 100%

[A]o = 3.1%

100 / 3.1 = 0.50^n

32 = 0.50^n

2^5 = 2^-n

n = 5

It would take five half lives.

The equilibrium constant for the reaction below, at a given temperature is 45.6. If the equilibrium concentrations of F2 and BrF3 are 1.24 x 10-1 M and 1.99 x 10-1 M respectively, calculate the equilibrium concentration of Br2. (4)

This is an incomplete question, here is a complete question.

The given chemical reaction is:

The equilibrium constant for the reaction below, at a given temperature is 45.6. If the equilibrium concentrations of F₂ and BrF₃ are 1.24 × 10⁻¹ M and 1.99 × 10⁻¹ M respectively, calculate the equilibrium concentration of Br₂.

Answer : The equilibrium concentration of Br₂ is, 0.0428 M

Explanation : Given,

Concentration of at equilibrium =

Concentration of at equilibrium =

Equilibrium constant = 45.6

The given chemical reaction is:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:

Thus, the equilibrium concentration of Br₂ is, 0.0428 M

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be the mass (in mg) remaining after t years. Then we know the following. y ( t ) = y ( 0 ) e k t = ⋅ e k t Since the half-life is 30 years, then y ( 30 ) = mg. Thus, e 30 k = . Therefore, k = . Now, remembering that ln x n = n ln x and that e ln z = z , we have y ( t ) = 180 e ( t / 30 ) ( − ln 2 ) = mg. How much of the sample remains after 70 years? After 70 years we have the following. y ( 70 ) = 180 ⋅ 2 = mg (Round your answer to two decimal places.) After how long will only 1 mg remain? To find the time at which only 1 mg remains, we must solve 1 = y ( t ) = 180 ( 2 − t / 30 ) , and so we get the following. t = − 30 log 2 ( ) Hence, we conclude the following. t = yr (Round your final answer to one decimal place.)

Explanation:

1) Initial mass of the Cesium-137== 180 mg

Mass of Cesium after time t = N

Formula used :

Half life of the cesium-137 = = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

= half life of the isotope

= rate constant

Now put all the given values in this formula, we get

Mass that remains after t years.

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

N = 1 mg

t = ?

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for it to decrease to 0.085 M. The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for it to decrease to 0.085 M. 10. 11 7.0 12 8.0

The time taken is 8 minutes.

The rate of decay of a radioactive substance is given by:

λN

here N is the concentraion, λ is the decay constant. If inially the concentration is t time t = 0.

Then solving the above equation we get:

λ =

Now the half-life is defined as the time after which half of the initial concentration is left. The relation between half-life and decay constant is:

λ =

Given that = 13 min

λ = 0.693/13 = 0.0533

So, the time taken for a first-order reaction such that the concentration decreases from 0.13 M to 0.085M can be calculated by following equation:

λ =

t =

t = 8 min

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Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

Here

To find the time takes for it to decrease to 0.085 we use the below equation

Here , , [A₀] = 0.13 m and [ A] = 0.085 M

Therefore it takes 8.0 mins for it to decrease to 0.085 M

The half-life of carbon-14 is about 5,730 years. A tree fossil was estimated to have about 4.2 picograms of carbon-14 when it died. (A picogram is a trillionth of a gram.) The fossil now has about 0.5 picogram of carbon-14. About how many years ago did the tree die? Show your reasoning.

The number of half lives will be :17,190 years

Given:

Half life = 5,730 years

Initial amount = 4.2 picograms

Present amount = About 0.5 picograms

To find:

Number of half lives = ?

What does half-life tell?

The half life a material is simply the amount of time it takes for half of that material to decay.

4.2 --> 2.1 (First Half life)

2.1 --> 1.05 (Second Half life)

1.05 --> 0.525 (Third Half life)

It takes three half lives to reach it's present mass. This means number of years passed = 3 * 5,730 years = 17,190 years

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Answer:

17,190 years

Explanation:

Half life = 5,730 years

Initial amount = 4.2 picograms

Present amount = About 0.5 picograms

Number of half lives = ?

Before proceeding;

The half life a material is simply the amount of time it takes for half of that material to decay.

4.2 --> 2.1 (First Half life)

2.1 --> 1.05 (Second Half life)

1.05 --> 0.525 (Third Half life)

It takes three half lives to reach it's present mass. This means number of years passed = 3 x 5,730 years = 17,190 years

The half-life of Erbium-165 is 10.4 hours. After 24 hours a sample has been reduced to a mass of 2 mg. What was the initial mass of the sample, and how much will remain after 3 days?

Answer:

Explanation:

Given that:

Half life = 10.4 hours

Where, k is rate constant

So,

The rate constant, k = 0.06664 hour⁻¹

Time = 24 hours

Using integrated rate law for first order kinetics as:

Where,

is the concentration at time t = 2 mg

is the initial concentration = ?

So,

Now, time = 3 days = 3*24 hours = 72 hours ( 1 day = 24 hours)

Thus,

(20 pts) When 3.018 moles of chlorine reacts with excess aluminum, how many moles of aluminum chloride are formed?

2 Al + 3 Cl2 → 2 AICI3:

Answer:

Explanation:

2Al + 3Cl₂ → 2AlCl₃

n/mol: 3.018

The molar ratio is 2 mol AlCl₃:3 mol Cl₂.

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compute the atomic radius for Ti.

Answer : The atomic radius for Ti is,

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number

First we have to calculate the volume of HCP crystal structure.

Formula used :

.............(1)

where,

= density =

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

= Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

Now we have to calculate the atomic radius for Ti.

Formula used :

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

Now put all the given values in this formula, we get:

Therefore, the atomic radius for Ti is,

What would you expect to happen to the weight of a dialysis tubing bag containing 0% sugar solution after being placed into a beaker containing 50% sugar solution for one half hour?

Answer: The sugar will move from the 50 % concentration to the 0 % concentration.

Explanation:

Dialysis is an analytical technique which uses a semipermeable membrane in order to remove the impurities from the body.

This technique is used when the kidney failure inside the body takes place. The kidney inside the body perform function of filtration, when kidney failure takes place then the dialysis is used.

In this case the tubing bag containing 0% sugar is put into a sugar solution having 50% solution.

After sometime the sugar will move from 50 % to 0% bag because of the phenomenon known as osmosis.

There are 35 children for every 6 teachers. Using the same child to teacher ratio, how many teachers needed for 140 children?
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A second-order reaction has a half-life of 12 s when the initial concentration of reactant is 0.98 M. The rate constant for this reaction is __________ M-1s-1.
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