1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
madreJ [45]
3 years ago
10

The rearrangement of methyl isonitrile (CH3NC) to acetonitrile (CH3NC) is a first-order reaction and has a rate constant of 5.11

x10-5s-1 at 472k . If the initial concentration of CH3NC is 3.00×10-2M :
(A) What is the half-life (in hours) of this reaction?
t1/2=________________hr
(B) How many hours will it take for the concentration of methyl isonitrile to drop to 6.25% of its initial value?
t=____________________hr
Chemistry
1 answer:
Jobisdone [24]3 years ago
6 0

Explanation:

Reaction is first order.

Rate Constant (k) = 5.11x10-5s-1

Temperature = 472k

Initial Concentration = 3.00×10-2M

(A) What is the half-life (in hours) of this reaction?

Formular for half life of a first order reaction is;

t1/2 = 0.693 / k

t1/2 = 0.693 / (5.11x10-5)

t1/2 = 0.1386 x 10 ^ 5 s

Upon converting to hours by dividing the value by 3600;

t1/2 = (0.1386 x 10 ^ 5) / 3600 = 3.85 hours

(B) How many hours will it take for the concentration of methyl isonitrile to drop to 6.25% of its initial value?

6.25% of initial concentration;

(6.25 / 100) * 3.00×10-2 = 0.1875 x 10 ^ -2M

ln[A] = ln[A]o − kt

[A] = Final Concentration

[A]o = Initial Concentration

upon making t subject of interest;

t = (ln[A]o - ln[A] ) / k

Inserting the values;

t = [  In(3.00×10-2) - In(0.1875 x 10 ^ -2) ] / 3.85

t = 2.7726 / 3.85

t = 0.72 hours

You might be interested in
Balance the equation AlCl3 + H2SO4 yields Al2(SO4)3 +HCl
Natalija [7]

Answer:

2AlCl3 + 3H2SO4 → Al2(SO4)3 + 6HCl

Explanation:

3 0
3 years ago
1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
a. For the reaction 2H2 + 02→ 2H20, how many moles of water can be produced from 6.0 mol of oxygen? 2.0 mol b. 6.0 mol 12 mol d.
pickupchik [31]

Explanation:

from the equation 1 mole of O2 will give 2 moles of H2O then 6.0 moles of O2 will give x

6.0*2 moles/ 1 mole

= 12 moles

this implies that, 6.0 moles of O2 will give = 12 moles of water

3 0
3 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
3 years ago
1. The answer to the research question is the...
katovenus [111]

Answer:

1. A claim

2. B evidence

8 0
3 years ago
Other questions:
  • How could you make plastic sink and steel float
    11·1 answer
  • 3
    13·1 answer
  • Element X
    7·1 answer
  • What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4
    11·1 answer
  • The ability to conduct electricity in the solid state is a characteristic of metallic bonding. This characteristic is best expla
    6·1 answer
  • How could you separate a mixture of rocks and sand
    13·1 answer
  • If an atom has a mass number of 25 and an atomic number of 11 how many neutrons are there in this atom?
    7·1 answer
  • How to find specific heat capacity of a liquid
    11·1 answer
  • A chemistry student needs of acetone for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student dis
    8·1 answer
  • Please help me I will give you the brain thing with extra points if you help me, please. I need to get this right. 3/10
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!