Chemistry, asked by parth4c, 5 days ago

Consider the following radioactive decay of 238 U 2380—+ "He- In a closed vessel of 11.2 ml 5.95 g of uranium is taken , then the total pressure at the end of 36 years at 0°C is (ty2 of 238U is 12 years) (1) 1 atm (2) 20 atm (3) 35 atm (4) 43.7 atm​

Answers

Answered by sumantraprajapati746
0

Answer:

43.7 atm

Explanation:

Consider the following radioactive decay of 238 U 2380—+ "He- In a closed vessel of 11.2 ml 5.95 g of uranium is taken , then the total pressure at the end of 36 years at 0°C is 43.7 atm

Answered by rohitsingh9014
0

Explanation:

14

N+

2

4

Hc→

8

17

O+

1

1

H

4

9

Be+

2

4

He→

6

12

C+

0

1

n

4

9

Be(p,α)

3

6

Li

15

30

P→

14

30

Si+

+1

0

e

1

3

H→

2

3

He+

−1

0

e

20

43

Ca(α,....)→

21

46

Sc+p(proton)

In these equations, the sum of the atomic numbers on LHS is equal to the sum of the atomic numbers on RHS.

Similarly, the sum of the mass numbers on LHS is equal to the sum of the mass numbers on RHS

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