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KIM [24]
2 years ago
12

How many total atoms does each element have? Plz help!

Chemistry
1 answer:
VLD [36.1K]2 years ago
7 0

Answer:

14 Calcium (4x3+2)

2 Phosphurus (2)

9 Oxygen (4x2+1)

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What is the value of δg°' (or, to put it another way, the cost) when 2nadp+ and 2h2o are converted to 2nadph plus 2h+ plus o2?
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According to the information in your study unit which of the following locations require the highest illuminance
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7 0
3 years ago
A certain first-order reaction has a half-life of 25.2 s at 20°C. What is the value of the rate constant k at 60°C if the activa
DochEvi [55]

Answer:

t

(

2

)

1/2

=

85.25 s

Notice how you're given the half-life (for one temperature), a second temperature, and the activation energy. The key to doing this problem is recognizing that:

the half-life for a first-order reaction is related to its rate constant.

the rate constant changes at different temperatures.

Go here for a derivation of the half-life of a first-order reaction. You should find that:

t

1/2

=

ln

2

k

Therefore, if we label each rate constant, we have:

k

1

=

ln

2

t

(

1

)

1/2

k

2

=

ln

2

t

(

2

)

1/2

Recall that the activation energy can be found in the Arrhenius equation:

k

=

A

e

−

E

a

/

R

T

where:

A

is the frequency factor, i.e. it is proportional to the number of collisions occurring over time.

E

a

is the activation energy in

kJ/mol

.

R

=

0.008314472 kJ/mol

⋅

K

is the universal gas constant. Make sure you get the units correct on this!

T

is the temperature in

K

(not

∘

C

).

Now, we can derive the Arrhenius equation in its two-point form. Given:

k

2

=

A

e

−

E

a

/

R

T

2

k

1

=

A

e

−

E

a

/

R

T

1

we can divide these:

k

2

k

1

=

e

−

E

a

/

R

T

2

e

−

E

a

/

R

T

1

Take the

ln

of both sides:

ln

(

k

2

k

1

)

=

ln

(

e

−

E

a

/

R

T

2

e

−

E

a

/

R

T

1

)

=

ln

(

e

−

E

a

/

R

T

2

)

−

ln

(

e

−

E

a

/

R

T

1

)

=

−

E

a

R

T

2

−

(

−

E

a

R

T

1

)

=

−

E

a

R

[

1

T

2

−

1

T

1

]

Now if we plug in the rate constants in terms of the half-lives, we have:

ln

⎛

⎜

⎝

ln

2

/

t

(

2

)

1/2

ln

2

/

t

(

1

)

1/2

⎞

⎟

⎠

=

−

E

a

R

[

1

T

2

−

1

T

1

]

This gives us a new expression relating the half-lives to the temperature:

⇒

ln

⎛

⎜

⎝

t

(

1

)

1/2

t

(

2

)

1/2

⎞

⎟

⎠

=

−

E

a

R

[

1

T

2

−

1

T

1

]

Now, we can solve for the new half-life,

t

(

2

)

1/2

, at the new temperature,

40

∘

C

. First, convert the temperatures to

K

:

T

1

=

25

+

273.15

=

298.15 K

T

2

=

40

+

273.15

=

313.15 K

Finally, plug in and solve. We should recall that

ln

(

a

b

)

=

−

ln

(

b

a

)

, so the negative cancels out if we flip the

ln

argument.

⇒

ln

⎛

⎜

⎝

t

(

2

)

1/2

t

(

1

)

1/2

⎞

⎟

⎠

=

E

a

R

[

1

T

2

−

1

T

1

]

⇒

ln

⎛

⎜

⎝

t

(

2

)

1/2

400 s

⎞

⎟

⎠

=

80 kJ/mol

0.008314472 kJ/mol

⋅

K

[

1

313.15 K

−

1

298.15 K

]

=

(

9621.78 K

)

(

−

1.607

×

10

−

4

K

−

1

)

=

−

1.546

Now, exponentiate both sides to get:

t

(

2

)

1/2

400 s

=

e

−

1.546

⇒

t

(

2

)

1/2

=

(

400 s

)

(

e

−

1.546

)

=

85.25 s

This should make sense, physically. From the Arrhenius equation, the higher

T

2

is, the more negative the

[

1

T

2

−

1

T

1

]

term, which means the larger the right hand side of the equation is.

The larger the right hand side gets, the larger

k

2

is, relative to

k

1

(i.e. if

ln

(

k

2

k

1

)

is very large,

k

2

>>

k

1

). Therefore, higher temperatures means larger rate constants.

Furthermore, the rate constant is proportional to the rate of reaction

r

(

t

)

in the rate law. Therefore...

The higher the rate constant, the faster the reaction, and thus the shorter its half-life should be.

Explanation:

Sorry just go here https://socratic.org/questions/588d14f211ef6b4912374c92#370588

3 0
2 years ago
Gases become more soluble in liquids as the temperature
Oksi-84 [34.3K]
 gases become more soluble in liquids as the temperature gets higher
<span />
3 0
2 years ago
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