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Andrej [43]
3 years ago
6

The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The

decays proceed through a series of alpha-particle and beta-particle emissions.A.How many of alpha-particle emissions are involved in this series?B.How many of beta-particle emissions are involved in this series?
Chemistry
1 answer:
dsp733 years ago
3 0

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

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How many moles of H2O2 are needed to react with 1.07 moles of N2H4?
I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄  +   2H₂O₂       →   N₂ +  4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

                          N₂H₄     :      H₂O₂  

                            1           :        2

                            1.07      :         2×1.07 = 2.14 mol

                   

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Diol oxidation without C-C bond breakage ....which products are formed??
babunello [35]
If its oxidised, water and carbon dioxide is formed.
4 0
3 years ago
Discussion: (24 points)
lorasvet [3.4K]
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6 0
3 years ago
What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3
snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

ΔG°=-16.7 kJ/mol

Required

the free energy change

Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol

8 0
3 years ago
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