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vaieri [72.5K]
3 years ago
12

Predicting nuclear stability is important when determining whether or not a nuclear reaction can take place spontaneously.Use th

e following set of guidelines to predict the nuclear stability of the elements listed below.
1. Stable nuclei have a high neutron-to-proton ratio, greater than 1.25 beyond atomic number 40 and continuing to rise at higher atomic numbers.
2. The majority of stable nuclei have even numbers of neutrons and protons. Odd-even and even-odd combinations can be stable but stability is less likely.
3. Nuclei that have the magic numbers of 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82 or 126 neutrons are more stable than nuclei that do not contain these numbers.
4. Nuclei that have atomic numbers greater than 83 are unstable.Nuclei are considered stable if one or more criteria are met.
Chemistry
1 answer:
Nitella [24]3 years ago
8 0

Answer:

a) unstable

b) unstable

c) unstable

d) Stable

Explanation:

a)

^{212}_{84}PO

The given isotope , atomic number is greater than 83.(84>83).

Therefore, it is unstable.

b)

^{30}_{15}P

The given isotope  has even number of protons(15) and neutrons(15).

Therefore, it is unstable.

c)

^{137}_{55}Cs

The given isotope  has even number of protons(15).

Therefore, it is unstable.

d)

^{118}_{50}Sn

The given isotope have even number of neutrons ( 68) an d the proton ratio is \frac{68}{50}= 1.36.It is more than 1.25.

Therefore, it is stable.

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3 0
3 years ago
Read 2 more answers
2.56 g of hydrogen reacts completely with 20.32 g of oxygen<br> to form X g of water. X = g
Brilliant_brown [7]

Answer:

Mass of water produced is 22.86 g.

Explanation:

Given data:

Mass of hydrogen = 2.56 g

Mass of oxygen = 20.32 g

Mass of water = ?

Solution:

Chemical equation:

2H₂ + O₂   →  2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 20.32 g/ 32 g/mol

Number of moles = 0.635 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 2.56 g/ 2 g/mol

Number of moles = 1.28 mol

Now we will compare the moles of water with oxygen and hydrogen.

                    O₂            :            H₂O

                     1              :             2

                  0.635        ;            2×0.635 =  1.27

                   H₂             :              H₂O

                    2              :              2

                 1.28            :           1.28

The number of  moles of water produced by oxygen are less thus it will be limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 1.27 × 18 g/mol

Mass = 22.86 g

 

4 0
4 years ago
Read 2 more answers
How many moles of oxygen are there in 3 moles of Zn(NO3)2? (2 points)
muminat

there are 6 moles of oxygen in Zn(No3)2

4 0
2 years ago
The element whose neutral isolated atoms in the ground state have three half filled orbital is
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One would be phosporous whose configuration is 1s2 2s2 2p6 3s2 3p3
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3 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
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