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Feliz [49]
3 years ago
7

How much energy is released when a proton combines with a deuterium nucleus to produce 3He?

Physics
1 answer:
slava [35]3 years ago
6 0

Answer:

E =  7.99 *10^{-13} J

Explanation:

the given reaction is

_1 H^1 + _1 H^2 = _2 He^3

we know that energy is given asE = \Delta mc^2E = (m_1 H^1 + m_1 H^2  - _2 He^3)c^2

where

m_1 H^1 is mass of proton = 1.672622 *10^{-27}

m_1 H^2 is mass of deuterium = 3.344494 *10^{-27}

m_2 H^3 is mass of He = 5.008234 *10^{-27}

E = [1.672622 *10^{-27} + 3.344494 *10^{-27} - 5.008234 *10^{-27} ] *(3*10^8)^2

E =  7.99 *10^{-13} J

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{\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}

0.72 liters

Explanation:

\sf Volume \: before \: heated(V_1) = 0.5l \\  \\  \sf Temperture \: before \: heated(T_1) = 20{ \degree} C = 20 + 273.15 = 293.15K \\  \\  \sf Volume \: after \: heated(V_2) =  ? \\  \\  \sf Temperature \: after \: heated(T_2) = 150{ \degree}C = 150 + 273.15 = 423.15K

\textsf{By using Charles's law}  \\  \\  \tt{ \frac{V_1}{T_1}  =  \frac{V_2}{T_2} }

\sf Now \: sustituting \: the \: values

\sf  \longmapsto \frac{0.5}{293.15}  =  \frac{V_2}{423.15}  \\  \\  \\  \sf \longmapsto  \frac{0.5 \times 423.15}{293.15}  = V_2 \\  \\  \\  \sf  \longmapsto  \frac{211.575}{293.15}  = V_2 \\  \\  \\  \boxed {\longmapsto 0.72l = V_2}

4 0
2 years ago
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Answer:

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4 years ago
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swat32

Answer:

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Explanation:

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3 years ago
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Answer:

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3 years ago
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