Nêu và giải thích hiện tượng trong các thí nghiệm sau

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago

A horse falls into a mudhole 5 m deep and becomes stuck. A truck uses a force of 7580 N to lift the horse out of the mudhole. Ho

vovikov84 [41]

I think the answer is B but i could be wrong

3 0

3 years ago

Liquids do not have the energy to overcome the ___ and move to the gaseous state.

Ugo [173]

Answer - Inter-molecular attractions

Explanation-

As we know everything around us is made up of matter that means everything has molecules as their basic structure. The state of anything is decided by the spaces between the molecules.

The state of the objects that have strong inter-molecular attractions a solid and gradually the lesser will be in state of liquid and gas. The attraction between the molecules is overcome only when a certain amount of energy is provided from outside.

4 0

3 years ago

suppose you use a pitcher to to estimate 1.25 qt of water ( experimental volume) . You later measured the volume more precisely

saul85 [17]

The percent error in the initial measurement would be 6.47%

<h3>Percent Error</h3>

Percent error in any measurement is mathematically given as:

absolute error/true measurement x 100%

In this case;

True measurement = 1.34 qt

Absolute Error = 1.34 - 1.25

= 0.09 qt

Percent error = 0.09/1.34 x 100%

= 6.47%

More on percent error can be found here: brainly.com/question/3105259?referrer=searchResults

6 0

3 years ago

When 0.300 g of a diprotic acid was titrated with 0.100 M LiOH, 40.0 mL of the LiOH solution was needed to reach the second equi

Iteru [2.4K]

Answer:

H₂C₄H₄O₆

Explanation:

The clue of this question is to find the molar mass of the diprotic acid and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.

The procedure is:

Find the number of moles of the base: LiOH
Use stoichionetry to infere the number of moles of the acid.
Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.
Compare and conclude.

Solution:

1. Number of moles of the base, LiOH:

M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.

2. Stoichiometry:

Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.

Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.

3. Molar mass of the acid:

molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol

4. Molar mass of the possible diprotic acids:

a. H₂Se: 2×1.008 g/mol + 78.96 g/mol = 80.976 g/mol

b. H₂Te: 2×1.008 g/mol + 127.6 g/mol = 129.616 g/mol

c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol

d. H₂C₄H₄O₆ = 6×1.008 g/mol + 4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.

Conclusion: since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.

6 0

3 years ago