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serg [7]
3 years ago
8

Write the nuclear equation to describe the alpha decay of 243 95 Am

Chemistry
2 answers:
motikmotik3 years ago
7 0

Answer:

Here's what I get.

Explanation:

Your unbalanced nuclear equation is:

_{95}^{243}\text{Am} \longrightarrow \, _{x}^{y}\text{Z} +\, _{2}^{4}\text{He}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

  95 = x + 2, so x =   95 - 2 =    93

243 = y + 4, so y = 243 - 4 = 239

Element 93 is neptunium, so the nuclear equation becomes

_{95}^{243}\text{Am} \longrightarrow \, _{93}^{239}\text{Np} + \, _{2}^{4}\text{He}  

scoundrel [369]3 years ago
7 0

<u>Answer:</u> The nuclear reaction is written below.

<u>Explanation:</u>

Alpha decay is defined as the decay process in which alpha particle is released. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units and a mass of 4 units.

The released alpha particle is also known as helium nucleus.

In this decay process, the atomic number of the atom decreases by 2 units and the mass number decreases by 4 units.

The chemical equation for alpha decay process follows:

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

The chemical equation for alpha decay of _{95}^{243}\textrm{Am} follows:

_{95}^{243}\textrm{Am}\rightarrow _{93}^{239}\textrm{Np}+_{2}^{4}\alpha

Hence, the nuclear reaction is written above.

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Read 2 more answers
At a given temperature, K = 1.3x10^-2 for the reaction:
son4ous [18]

Answer:

a) 0.11

b)76.9

c) 8.8

d) 1.7*10^-4

Explanation:

Step 1: Data given

K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 2: Formula of K

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K = [C]^c *[D]^d  / [A]^a * [B]^b

K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]

Step 3:

a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

1/2N2 + 3/2H2(g) ⇌ NH3(g)    =>K' =  \sqrt{K}

K' = \sqrt{1.3*10^-2} = 0.11

<em>b. 2NH3(g) ⇌ N2(g) + 3H2(g)</em>

N2(g) + 3H2(g) ⇌ 2NH3

2NH3(g) ⇌ N2(g) + 3H2(g)    =>K' = 1/K

K' = 1/(1.3*10^-2) = 76.9

c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)    

=>K' = \frac{1}{\sqrt{K} }

K' = \frac{1}{\sqrt{1.3*10^-2} }

K' = 8.8

d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

2N2(g) + 6H2(g) ⇌ 4NH3(g)

K' = K²

K' = (1.3*10^-2)²

K' = 1.7 *10 ^-4

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