The right answer is indeed letter c
Answer:
The answer to your question is the number 4. 1.53175 L
Explanation:
Data
Volume 1 = V1 = 557 ml
Concentration 1 = C1 = 11 M
Volume 2 = V2 = ?
Concentration 2 = C2 = 4 M
Formula
Volume 1 x Concentration 1 = Volume 2 x Concentration 2
or V1 x C1 = V2 x C2
Solve for V2 V2 = V1 x C1 / C2
-Substitution
V2 = (557)(11) / 4
-Simplification
V2 = 6127 / 4
-Result
V2 = 1531.75 ml or 1.53175 L
Conclusion
557 ml of 11 M must be diluted to 1.53175 L
Okay thanks for reminding me
Answer:
12 mmilligrams of Po-218 was the mass of the original starting material
Explanation:
The half-life of a radioactive material is the time taken for half the amount ofnthe original material present in a radioactive material to decay or disintegrate.
After each half-life, half the original material present at the start remains.
For the radioactive polonium-218 having a half-life of 3.04 minutes, it means that if 1 g is the starting material, after 3.04 minutes, 1/2 g will be remaining; after, 6.08 minutes 1/2 of 1/2 which is 1/4 of the starting material will be remaining; and after 9.12 minutes, 1/2 of 1/4 = 1/8 g will be remaining.
From the question, number of half-lives undergone after 9.12 minutes = 9.12/3.04 = 3 half-lives.
After 3 half-lives, 1/8 of the original material is remaining.
1/8 = 1.50 mg
The original mass of the sample at the start = 1.50 mg × 8 = 12 mg
Therefore, 12 milligrams of Po-218 was the mass of the original starting material.