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3 years ago

Elaborate on the difference in natural occurrences between fission and fusion reactions.

1 answer:

5 0

Fission is the division of one atom into two, and fusion is the combination of two lighter atoms into larger one. They are opposing processes and therefore very different.

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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent

elixir [45]

Answer : The initial rate for a reaction will be $3.8\times 10^{-4}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

$A+B+C\rightarrow P$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1$

$k=7.6\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1$

$\text{Rate}=3.8\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.8\times 10^{-3}Ms^{-1}$

6 0

3 years ago

Choose the options below that are true.

Assoli18 [71]

Answer:

The options (A) -The rate law for a given reaction can be determined from a knowledge of the rate-determining step in that reaction's mechanism. and (C) -The rate laws of bimolecular elementary reactions are second order overall ,is true.

Explanation:

(A) -The rate law can only be calculated from the reaction's slowest or rate-determining phase, according to the first sentence.

(B) -The second statement is not entirely right, since we cannot evaluate an accurate rate law by simply looking at the net equation. It must be decided by experimentation.

(D)-The fourth argument is incorrect. We must track the rates of and elementary phase that is following the reaction in order to determine the rate.

Therefore , the first and third statement is true.

5 0

2 years ago

Help pleaaasee??????????? can u tell me whether it's physical or chemical change

Karo-lina-s [1.5K]

2) chemical
3) chemical
4) physical
5)chemical
6)chemical
7)physical
8)physical
9)physical / chemical
100% sure

5 0

3 years ago

How do you know when a phase change happens?

VladimirAG [237]

Answer:

phase for what?

Explanation:

4 0

3 years ago

What are the answers for this question Don't just answer for points

Kisachek [45]

Answer:

i cant see it very much i dont think no one can

Explanation:

5 0

3 years ago

Read 2 more answers