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il63 [147K]
3 years ago
12

What are reactants of nuclear fission

Chemistry
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Answer:

When a large, fissile atomic nucleus such as uranium-235 or plutonium-239 absorbs a neutron, it may undergo nuclear fission. The nucleus splits into two or more lighter nuclei, releasing kinetic energy, gamma radiation, and free neutrons.

Explanation:

Hope this helps. :)

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State the oxidation number of S in <br><img src="https://tex.z-dn.net/?f=H_%7B2%7DSO_%7B3%7D" id="TexFormula1" title="H_{2}SO_{3
Nezavi [6.7K]

Taking into account the definition of oxidation number, the oxidation numbers of S in H₂SO₃ is 4.

<h3>Definition of oxidation number</h3>

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

<h3>Oxidation number determination</h3>

To determine the oxidation state of different elements it is necessary to know that:

  • The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where is –1.
  • The oxidation number of oxygen in a compound is –2, except in peroxides, where it is –1.

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

<h3>Oxidation numbers of S</h3>

In this case, the oxidation numbers of S in H₂SO₃ is calculated as:

2× (+1) + oxidation numbers of S + 3×(-2)= 0

2 + oxidation numbers of S -6= 0

oxidation numbers of S -4= 0

<u><em>oxidation numbers of S= 4</em></u>

Finally, the oxidation numbers of S in H₂SO₃ is 4.

Learn more about the oxidation number:

brainly.com/question/8990767

brainly.com/question/6498977

5 0
2 years ago
Read 2 more answers
A solution of HCl has mc012-1.jpg = 0.01 M. What is the pH of this solution?
xxMikexx [17]
PH = -log10 [H+]. So anwer 2 pH
4 0
3 years ago
Read 2 more answers
Can anyone please help me with my chemistry homework? It’s about molarity. I can’t message anyone here because I’m new to brainl
GREYUIT [131]

Answer:

ok sure

Explanation:

4 0
3 years ago
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
LiRa [457]

Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
3 years ago
What does the atomic number of each atom represent?
saw5 [17]
It's the number of Protons of that atom..
6 0
2 years ago
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