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3 years ago

Which is an example of a highly unstable isotope that is often used in fission reactions?

2 answers:

4 0

The right answer to this question is U-235. It is an isotope of Uranium and it is an unsteady heavy metal used in fission reactions because it can run long chains of reactions. I hope this helps.

Marta_Voda [28]3 years ago

4 0

Answer:

C. U-235

Explanation:

A fission reaction is one in which an unstable radioisotope breaks down into smaller nuclei when bombarded with energetic particles.

The highly unstable isotope that is often used in fission reactions is U-235. When bombarded with neutrons that atom breaks down to form krypton and barium along with the release of a large amount of energy. The reaction is given as:

₉₂U²³⁵ + ₀n¹ → ₃₆Kr⁸⁹ + ₅₆Ba¹⁴⁴ + 3₀n¹ + 210 Mev energy

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Water has a higher boiling point than acetone does. Which of the following statements about water and acetone in the liquid stat

Semenov [28]

Answer:

Water's boiling point is higher than acetone's one due to the stronger intermolecular forces it has in liquid phase.

Explanation:

Hello.

In this case, since no options are given we can infer from the statement that due to water's higher boiling point than acetone we can conclude that when they are in liquid state, water has stronger intermolecular forces which allow its particles to be held in a stronger way in comparison to the acetone's molecules, for that reason, more energy will be required in order to separate them and promote the boiling process, which is attained via increasing the temperature. Besides, less energy will be required for the separation of the acetone's molecules in order to boil it when liquid, therefore, a lower temperature is required.

In such a way, we can sum up that water's boiling point is higher than acetone's one due to the stronger intermolecular forces it has in liquid phase.

Regards.

7 0

3 years ago

What conclusion can she make from her experiment?

pishuonlain [190]

Answer:

hii samyy

Explanation:

nice killua profile pic :))

5 0

3 years ago

Read 2 more answers

Suppose that a catalyst lowers the activation barrier of a reaction from 125kJ/mol to 55kJ/mol125⁢kJ/mol to 55kJ/mol. By what fa

sineoko [7]

Answer:

The factor of increasing reaction rate is 1,85x10¹².

Explanation:

Using arrhenius formula:

$k = A e^\frac{-E_{a}}{RT}$

Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K

Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:

k = 1,25x10⁻²²

When activation energy is 55kJ/mol:

k = 2,31x10⁻¹⁰

Thus, the factor of increasing reaction rate is:

2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>

I hope it helps!

8 0

3 years ago

What is the oh- in a solution with a poh of 5.71

Rudik [331]

Answer:- The hydroxide ion concentration of the solution is $1.95*10^-^6$ .

Solution:- The formula used to calculate pOH from hydroxide ion is:

$pOH=-log[OH^-]$

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

$[OH^-]=10^-^p^O^H$