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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

Are inner planets rockier or more like gas??

Eduardwww [97]

Inner planets are rockier and outer planets are gaseous. Hope this helps! Can I have brainliest please

4 0

3 years ago

Read 2 more answers

According to Hund's Rule, if you were filling the three orbitals at the 2p energy level, you would first put an electron in the

GenaCL600 [577]

Answer:

True.

Explanation:

The Hund's Rule states that all orbitals must be singled occupied before any orbital is doubly occupied, and all the electrons at the singly occupied orbitals have the same spin number. By doing that, the electrons filled the lowest energy orbitals first.

The 2p level has 3 orbitals: 2px, 2py, and 2pz. So, when filling it, first put an electron in the 2px, then in the 2py, then and the 2pz (all with the same spin). After that, the remains electrons can be paired up.

5 0

3 years ago

Use the given half reactions to "construct" an electrolytic cell. Zn^2+ + 2 e^--------->Zn E°cell = -0.76 V Cu^2+ + 2 e^-----

Neko [114]

Answer:

See explanation and image attached

Explanation:

The standard cell potential at 298 K is given by;

E°cathode - E°anode

Hence;

E°cell = 0.34 V - (-0.76 V)

E°cell = 0.34 V + 0.76 V

E°cell = 1.1 V

To reduce Zn^2+ to Zn then Zn must be the cathode, hence;

E°cell = (-0.76 V) - 0.34 V

E°cell = -1.1 V

5 0

3 years ago

If you have a 4.6 L of gas in a piston at a pressure of 1.2 atm and compress the gas unit it's volume is 2.6 L, what will the ne

klio [65]

Answer:

<h3>The answer is 2.12 atm</h3>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new pressure

$P_2 = \frac{P_1V_1}{V_2} \\$

So we have

$P_2 = \frac{4.6 \times 1.2}{2.6} = \frac{5.52}{2.6} \\ = 2.1230769...$

We have the final answer as

Hope this helps you

8 0

3 years ago