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NISA [10]
3 years ago
6

What happens ????????

Chemistry
2 answers:
Lemur [1.5K]3 years ago
6 0

The air pressure changes

ivolga24 [154]3 years ago
5 0
Correct me if I wrong but I think it's "c"
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4 years ago
Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C
Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, E = \triangle M * 931.5

\triangle M = \sum M_{product} - \sum M_{reactant}

Mass of 1H, M_{H} = 1.007823

Mass of 2H, M_{2H} = 2.0141u

Mass of 3H, M_{3H} = 3.016 u

Mass of Helium, M_{4He} = 4.002602u

Mass of Beryllium, M_{7Be} = 7.01693 u

Mass of neutron, M_{n} = 1.008664 u

a) 2H + 3H \rightarrow 4He + n

\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u

Energy released,

E = -0.01884 * 931.5\\E = -17.55 Mev

Energy released = 17.55 MeV

b) 4He + 4He \rightarrow 7Be + n

\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} +  M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u

Energy released,

E = 0.02039 * 931.5\\E = 18.99 Mev

c) 2H + 2H \rightarrow 3 He + n

\triangle M = (M_{3He} + M_{n} ) - (M_{2H} +  M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u

Energy released,

E = -0.003536 * 931.5\\E = -3.29 Mev

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) 2H + 2H \rightarrow 3H + 1H

\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} +  M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u

E = -0.00435 * 931.5\\E =-4.075 Mev

E = 4.075 MeV ( Energy is released)

5 0
3 years ago
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