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olesya19994013
@olesya19994013
August 2022
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Определите среднюю кинетическую энергию молекул водорода при 127 градусах Цельсия.
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mg58
Е = 3кТ/2
к — постоянная Больцмана = 1,38х10⁻²³Дж/К
Т = 273+127= 400°К
Е = 3·1,38·10⁻²³·4·10²/2 =
8,28·10⁻²¹ Дж
.
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Answers & Comments
к — постоянная Больцмана = 1,38х10⁻²³Дж/К
Т = 273+127= 400°К
Е = 3·1,38·10⁻²³·4·10²/2 = 8,28·10⁻²¹ Дж.