How much of 45 grams of pu 234 remains after 27 hours if it’s half life is 4.98 hours

Chemistry

1 answer:

nadya68 [22]3 years ago

6 0

Answer:

1.07 g

Explanation:

Half-life of Pu-234 = 4.98 hours

Initially present = 45 g

mass remains after 27 hours = ?

Solution:

Formula

mass remains = 1/ 2ⁿ (original mass) ……… (1)

Where “n” is the number of half lives

To find "n" for 27 hours

n = time passed / half-life . . . . . . . .(2)

put values in equation 2

n = 27 hr / 4.98 hr

n = 5.4

Mass after 27 hr

Put values in equation 1

mass remains = 1/ 2ⁿ (original mass)

mass remains = 1/ 2^5.4 (45 g)

mass remains = 1/ 42.2 (45 g)

mass remains = 0.0237 x 45 g

mass remains = 1.07 g

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BlackZzzverrR [31]

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3 0

1 year ago

Please begging you guys someone help me

BaLLatris [955]

Answer:

See Explanation

Explanation:

10) From the options provided for this question, gamma particle is the most energetic. Recall that gamma rays are high energy electromagnetic radiation which are capable of causing a high degree of ionization in matter.

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$U\frac{235}{92} + n\frac{1}{0}---> I\frac{138}{53} + Y\frac{95}{39} +3n \frac{1}{0}$

Hence

a = 92, b= 95, c= 53

12) In positron emission, a proton is transformed into a neutron. The mass number of the daughter nucleus is the same as its parent but the atomic number decreases by 1.

Hence;

$Th\frac{231}{90} -----> e\frac{1}{0} +Ac \frac{231}{89}$