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Brilliant_brown [7]
3 years ago
7

_H2 + _02 → _H20 Need help

Chemistry
1 answer:
Dominik [7]3 years ago
8 0

Answer:

h + O2

  • cross over valencies
  • h20
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A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm
VARVARA [1.3K]

<u>Answer:</u> The nuclear equations for the given process is written below.

<u>Explanation:</u>

The chemical equation for the bombardment of neutron to U-238 isotope follows:

_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

The chemical equation for the first beta decay process of _{92}^{239}\textrm{U} follows:

_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta

The chemical equation for the second beta decay process of _{93}^{239}\textrm{Np} follows:

_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta

Hence, the nuclear equations for the given process is written above.

6 0
3 years ago
What is the entropy of a solid based on
NeX [460]
The answer is:

The arrangement of the Atoms
4 0
4 years ago
CH3COOH  CH3COO– + H+
Oxana [17]

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

6 0
2 years ago
For each solute, identify the better solvent: water or carbon tetrachloride
Olin [163]
Do you mean which one out of water and carbon tetrachloride
if so it will be water
3 0
4 years ago
_____ were common ocean scavengers long ago, and they are frequently found in fossilized form.
quester [9]

Answer:

a. trilobites.

Explanation:

6 0
4 years ago
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