If the half-life of a radioactive substance is 10 days, what fraction of the original substance will remain after 30 days? MCQ with Answer and Explanation
If the half-life of a radioactive substance is 10 days, what fraction of the original substance will remain after 30 days?
A. 1/2
B. 1/16
C. 1/4
D. 1/8
Answer: Option D
Solution (By JKSSB Mock Tests)
The number of half-lives elapsed is 30 days / 10 days = 3. The fraction remaining is (1/2)ⁿ, where n is the number of half-lives. So, (1/2)³ = 1/8. After 10 days: 1/2; after 20 days: 1/4; after 30 days: 1/8 of the original amount remains.
Explanation:
Minamata disease is a neurological syndrome caused by severe mercury poisoning. It was first discovered in Minamata city, Japan, due to the release of methylmercury in industrial wastewater from a chemical factory, which bioaccumulated in fish and shellfish consumed by the local population.
Explanation:
CH₃CH₂OH is ethyl alcohol or ethanol. Methanol is CH₃OH. Propanol C₃H₇OH. Butanol C₄H₉OH. Ethanol is the active ingredient in alcoholic beverages and used as biofuel.
Explanation:
A Lewis acid is a chemical species that can accept a pair of electrons to form a covalent bond. BF3 has an incomplete octet on the boron atom (only 6 electrons), making it an electron-deficient molecule and a strong Lewis acid. NH3, H2O, and OH- have lone pairs and act as Lewis bases.
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