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Karo-lina-s [1.5K]
3 years ago
7

Which equation represents a transuranium​

Chemistry
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.

2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.

The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.

This are some of the transuranium elements:

Neptunio - 93

Plutonium - 94

Americium - 95

Curium - 96

Berkelium - 97

Californium - 98

Einstenium - 99

And so all the known elements (the last one is the 118).

3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.

4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:

0

 β, which means 0 mass and charge - 1.

-1

5) This is, then, an example of a β decay equation for one transuranium element:

239              239            0

     Np    →         Pu   +      β

 93                94            -1

As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.

Explanation:

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Answer:

y8y g8c g u to give give g gu tu tvtc8 ug fu t 7f to fu t7 to to tutorials

Explanation:

v8y to c g. fu uf gu t8 g8. g8 u g 8 g gi gi gi g gi if fu%:/"

5 0
2 years ago
What is the application of chemistry​
vlabodo [156]

Answer:

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3 0
3 years ago
1. In regards to curly hair, which of the following statements
Kamila [148]

Answer:

A and D are the answers

Explanation:

the other two don't make sense

3 0
3 years ago
The question is in the attached word document.
Alekssandra [29.7K]

Answer:i read it Explanation:

3 0
2 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
2 years ago
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