After decaying for 48 hours, one-sixteenth (1/16) of the original mass of a radioisotope sample remains unchanged. What is the h
1 answer:
The half-life of this radioisotope : 12 hr
<h3>Further explanation </h3>The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually radioactive elements have an unstable atomic nucleus.
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
t=48 hr
The half-life :
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Answer:
Explanation:
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In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:
Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:
Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:
Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:
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The equation for this question could be →
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so for 6.75 moles of *3 moles of O_{2}/2 moles of
= 10.125