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kumpel [21]
3 years ago
5

Which of the following correctly represents the transmutation in which a curium-242 nucleus is bombarded with an alpha particle

to produce a californium-245 nucleus?^242_96 Cm(^1_0 n, ^4_2 He)^245_98 Cf^242_96 Cm(^4_2 He, ^1_1 p)^245_98 Cf^242_96 Cm(^4_2 He, 2^1_1 p)^245_98 Cf^242_96 Cm(^4_2 He, ^1_0 n)^245_98 Cf^242_96 Cm(^4_2 He, ^1_-1 e)^245_98 Cf
Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
6 0

<u>Answer:</u> The chemical equation is written below.

<u>Explanation:</u>

Transmutation is defined as the process in which one chemical isotope gets converted to another chemical isotope. The number of protons or neutrons in the isotope gets changed.

The chemical equation for the reaction of curium-242 nucleus with alpha particle (helium nucleus) follows:

_{96}^{242}\textrm{Cm}+_4^2\textrm{He}\rightarrow _{98}^{245}\textrm{Cf}+_0^1\textrm{n}

The product formed in the nuclear reaction are californium-245 nucleus and a neutron particle.

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Common brass is a copper and zinc alloy containing 37.0% zinc by mass and having a density of 8.48g/cm3. A fitting composed of c
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First, we calculate the mass of the sample:

mass = density x volume
mass = 8.48 x 112.5
mass = 954 grams

Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.

Zinc mass = 0.37 x 954
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Zinc moles = 352.98 / 65
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When solutions of barium chloride and lithium sulfate are mixed, the spectator ions in the resulting reaction are?
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Cobalt-63 has a half-life of 5.3 years. If a pellet that has been in storage for 15.9 years contains 40.0g of Cobalt-63, how muc
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Answer:

320 g  

Step-by-step explanation:

The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of               Fraction         Mass

half-lives   t/yr   Remaining   Remaining/g

      0        0              1

      1         5.3           ½

     2        10.6           ¼

     3        15.9           ⅛                 40.0

     4        21.2           ¹/₁₆

We see that 40.0 g remain after three half-lives.

This is one-eighth of the original mass.

The mass of the original sample was 8 × 40 g = 320 g

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