If the element with atomic number 60 and atomic mass 211 decays by beta minus emission. What is the atomic mass of the decay product?

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Answer 1

The atomic mass of the decay product with atomic number 60 and atomic mass 211 decays by beta minus emission is 211.

The atomic number of a beta-minus decayed element increases by 1 and the atomic mass remains the same. For instance, if element Z decays by beta-minus decay, the resulting element would be Z + 1, and the atomic mass would be unchanged. Therefore, the atomic mass of the decay product is 211.

Beta minus decay (β− decay) is a type of radioactive decay in which an unstable nucleus converts into a stable nucleus by converting a neutron into a proton. The atomic number of the element increases by 1 in β− decay, while the atomic mass of the element remains unchanged.

Hence, the atomic mass of the decay product is the same as the atomic mass of the initial nucleus, which is 211.

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Related Questions

An ammonia solution has a pH of 11.30, what is the H3O+ concentration in this solution

A.) 5.0 x 10 ^ -23 M
B.) 2.0 x 10 ^ -9 M
C.) 5.0 x 10 ^ -12 M
D.) 2.0 x 10 ^ 11 M

Answers

The H₃O+ concentration in the ammonia solution with a pH of 11.30 is approximately option C.) 5.0 x [tex]10 ^-^1^2[/tex] M.

Ammonia (NH₃) is a weak base that can undergo a reaction with water to produce hydroxide ions (OH-) and ammonium ions (NH₄+). In this reaction, water acts as an acid, donating a proton (H+) to the ammonia molecule.

The pH scale is a logarithmic scale that measures the concentration of H₃O+ ions in a solution. It is defined as the negative logarithm (base 10) of the H₃O+ concentration. Therefore, to find the H₃O+ concentration, we need to convert the given pH value to a concentration.

Given that the pH of the ammonia solution is 11.30, we can use the formula pH = -log[H₃O+] to find the concentration of H₃O+. Rearranging the equation, we have [H₃O+] = [tex]10^(^-^p^H^)[/tex].

Substituting the given pH value into the equation, we get [H₃O+] = [tex]10^(^-^1^1^.^3^0^)[/tex]. Calculating this value yields approximately 5.0 x [tex]10^(^-^1^2^)[/tex] M.

Therefore, the correct answer is: C.) 5.0 x [tex]10 ^-^1^2 M[/tex]

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