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bearhunter [10]
2 years ago
9

What is the nuclear equation for decay of Am - 241

Chemistry
1 answer:
adelina 88 [10]2 years ago
3 0

Americium is a radioactive element .

  • It goes through alpha decay during nuclear reactions .

The nuclear reaction.is given by

\rm {}^{241}_{95}Am\longrightarrow {}^{237}_{93}Np+{}^4_2\alpha

  • Np stands for Nuptunium

Here alpha can be written as He too

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The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0
vodomira [7]
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
5 0
3 years ago
This system has an equilibrium constant of 0.105 at 472°C: N2(g) + 3H2(g) ↔ 2NH3(g)
Kamila [148]

Answer:

What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?

A. 0.0625

How does Q compare to Keq?

B. Q < Keq

Explanation:

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4 years ago
What mineral has the least hardness according to the scale?
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3 years ago
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Dennis_Churaev [7]

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4 0
3 years ago
Read 2 more answers
A heat energy of 645 J is applied to a sample of glass with a mass of 28.4 g. Its temperature increases from –11.6 ∞C to 15.5 ∞C
Anika [276]
The heat that is required to raise the temperature of an object is calculated through the equation,
                        heat = mass x specific heat x (T2 - T1)
Specific heat is therefore calculated through the equation below,
                                specific heat = heat / (mass x (T2 - T1))
Substituting,
                                specific heat = 645 J / ((28.4 g)(15.5 - - 11.6))
The value of specific heat from above equation is 0.838 J/g°C. 
5 0
3 years ago
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