answer.
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
Tcecarenko [31]
2 years ago
13

Suppose that a 1.0-L reaction vessel initially contains 1.0 mol of O3 and 1.0 mol of O2. What fraction of the O3 will have react

ed when the rate falls to one-half of its initial value?
Chemistry
1 answer:
NemiM [27]2 years ago
8 0

Answer:

Explanation:

given reaction

                        2O₃        ⇄      3O₂

 initial                1                      1

    final           1-x                   1+ 1.5x

rate of reaction = k  [ O₃ ]² / [ O₂ ]

initial rate = k . 1² / 1 = k

final rate = k (1-x )² / (1+ 1.5x) = k/2

(1-x )² / (1+ 1.5x) = 1/2

2x² - 5.5 x + 1 = 0

x = 0.196

So fraction of ozone reacted = .196

You might be interested in
Sodium thiosulfate (Na2S2O3), photographer’s
vova2212 [387]

3Na2S2O3 + AgBr ------>Na5[Ag(S2O3) 3] +NaBr

from equation 3 mol 1 mol

given x mol 0.10 mol


x= (3*0.10)/1=0.30 mol Na2S2O3


Answer: 0.30 mol Na2S2O3

3 0
2 years ago
Read 2 more answers
Imagine that you are given the mass spectra of these two compounds, but the spectra are missing the compound names.
12345 [234]

The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment

Answer:

Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z

Explanation:

The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.

The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.

Compound 2 (2,5-dimethylhexane) structure shows that the cleavage  of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.

5 0
2 years ago
A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of
allochka39001 [22]

Answer:

The answer to your question is   C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

                       44 g of CO₂ ---------------  12 g of C

                        52.8 g          ---------------  x

                        x = (52.8 x 12)/44

                        x = 633.6/44

                        x = 14.4 g of C

                        12 g of C ------------------  1 mol

                        14.4 g of C ---------------   x

                         x = (14.4 x 1)/(12)

                         x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

                         18 g of H₂O ---------------  2 g of H

                         21.6 g of H₂O -------------  x

                          x = (21.6 x 2) / 18

                         x = 2.4 g of H

                         1 g of H -------------------- 1 mol of H

                         2.4 g of H -----------------  x

                          x = (2.4 x 1)/1

                          x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

                           = 6.4 g

                         16 g of O ----------------  1 mol

                          6.4 g of O --------------  x

                          x = (6.4 x 1)/16

                          x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

                                C₃H₆O

8 0
2 years ago
Convert 338 L at 63.0 atm to its new volume at standard pressure.
taurus [48]

The new volume at standard pressure of 1 atm is 21294 liters.

Explanation:

Data given:

Initial volume of the gas V1 = 338 liters

initial pressure on the gas P1 = 63 atm

standard pressure as P2 = 1 atm

Final volume at standard pressure V2 =?

The data given shows that Boyle's law equation is to used:

P1V1 = P2V2

rearranging the equation to calculate V2,

V2 = \frac{P1V1}{P2}

Putting the values in the equation:

V2 = \frac{338X63}{1}

     = 21294 L

as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.

7 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
2 years ago
Other questions:
  • The highest elevation Mt. Zembat in Alaska 40 years ago was measured at 7600 feet. Today the highest elevation is 7598 feet. Wha
    11·1 answer
  • You are asked to go into the lab and prepare an acetic acid - sodium acetate buffer solution with a ph of 4.00  0.02. what mola
    12·1 answer
  • The lab procedure involves several factors, listed below. Some were variable and some were constant. Label each factor below V f
    8·2 answers
  • Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. the initial temperature of the gas is 27.0c
    7·1 answer
  • KClKCl has a lattice energy of −701 kJ/mol.−701 kJ/mol. Consider a generic salt, ABAB , where A2+A2+ has the same radius as K+,K
    13·1 answer
  • Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra
    15·1 answer
  • Old aspirin exposed to moisture often smells like acetic acid (vinegar). When aspirin is heated in boiling water, it decomposes
    8·1 answer
  • Consider the following hypothetical reaction: 2 P + Q → 2 R + S The following mechanism is proposed for this reaction: P + P Q
    11·1 answer
  • Given six molecules, identify the molecules with polar bonds and the molecules that are polar.CCl4, CH3Cl, H20, CO2, O2
    6·1 answer
  • What percentage of the 20.0 g sample does sugar (sucrose) represent?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!