After 184.5 years, onky 0.625 grams of an original 20 gram sample of a radioactive isotope is left. Find the half life
1 answer:
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The question is incomplete. Complete question is attached below.
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Correct Answer: <em>Option 1) 2-pentene</em>
Reason:
Following are the IUPAC rules for naming the compound
1) Select the longest carbon chain. In present case longest carbon chain has 5 carbon atom. Hence, it is a pentane derivative.
2) In case of alkene, replace 'e' of alkane by 'ene'
3) Give lowest number to function group. In present case, it is double bond.
Applying above rules, the IUPAC name of compound is 2-pentene
..........................................................................................................................
Correct Answer: <em>Option 1) 2-pentene</em>
Reason:
Following are the IUPAC rules for naming the compound
1) Select the longest carbon chain. In present case longest carbon chain has 5 carbon atom. Hence, it is a pentane derivative.
2) In case of alkene, replace 'e' of alkane by 'ene'
3) Give lowest number to function group. In present case, it is double bond.
Applying above rules, the IUPAC name of compound is 2-pentene
<u>Answer:</u> The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.
<u>Explanation:</u>
We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.
To calculate the volume of KBr, we use the formula:
Mass of the solution = 100 grams
Density of KBr solution = 1.03g/mL
Volume of the solution = ? mL
Putting values in above equation, we get:
Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.
In 97.08 mL of solution, mass of KBr present is 8.3 grams.
So, 41.2 mL of solution will contain = of KBr.
Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.