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Vesna [10]
3 years ago
14

Californium-252 is bombarded with a boron-10 nucleus to produce another nuclide and six neutrons. what nuclide forms?

Chemistry
2 answers:
Llana [10]3 years ago
6 0
²⁵²Cf₉₈ + ¹⁰B₅ → ²⁵⁶Lr₁₀₃ + 6 ¹n₀ 
The new nuclide formed = Lawrencium - 256 = ²⁵⁶Lr₁₀₃
Yanka [14]3 years ago
5 0

Answer: the nuclide is lawrencium-103


Using the isotopes notation it can be written as ²⁵⁶ ₁₀₃ Lr.


The number 256 is a superscript to the left of the symbol of the element (Lr) and the number 103 is a subscript to the left of the symbol


Explanation:


1) To determine the nuclide formed you have to rely in a mass and particles balance.


2) californium - 252 is the isotope has mass number 252, atomic number 98 (you obtain this information in a periodic table, remember that it is the number of protons).


The number of neutrons is determined from the formula


mass number = atomic number + number of neutrons ⇒ number of neutrons = mass number - atomic number


∴252 - 98 = 154 neutrons


3) boron - 10 nucleus has mass number 10, and atomic number 5 (from periodic table)


⇒ number of neutrons = 10 - 5 = 5.


4) Now the mass balance leads to:


mass number of californium-252 + mass number of boron-10 = unknown mass number + mass number of mass number of 6 neutrons


⇒ 252 + 10 = x + 6 ⇒ x = 252 + 10 - 6 = 256


5) Since neutrons do not change the atomic masses, number of protons, you can do this balance based on the atomic numbers (i.e. the same that number of protons):


atomic number of californium-252 + atomic number of boron - 10 = unknown atomic number


98 + 5 = A


A = 103


The element with atomic number 103 is lawrencium (Lr)


Calculate the number of neutrons: mass number - atomic number = 256 - 103 = 153.


Therefore, the new isotope is lawrencium (Lr) whith 153 neutrons.


Using the notation for isotopes the equation that show the reaction is:


²⁵² ₉₈ Cf + ¹⁰ ₅ B → ²⁵⁶ ₁₀₃ Lr + 6 ¹₀n

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