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elena-14-01-66 [18.8K]
2 years ago
14

The half-life of 131I is 80 days. How much of a 500.0 mg sample remains after 3 half lives?

Chemistry
1 answer:
I am Lyosha [343]2 years ago
6 0

Answer:

62.5 mg

Explanation:

Just multiply the original amount by  1/2   three times:

500 mg  x  1/2  x  1/2  x  1/2 = 62.5 mg

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What is the scientific notation of 36000x10 ex.10?
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Answer:

3.6 times 10^4

Explanation:

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3 years ago
Martha ran 420 m, and Cole ran 60 m less than Martha. The distance Kate ran is right in the middle of Martha and Cole's distance
prohojiy [21]

Answer:

390 m

Explanation:

Martha ran 420 m

Cole ran 60 m less than Martha = 420 m - 60 m = 360 m

The distance Kate ran is right in the middle of Martha and Cole's distances

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8 0
2 years ago
Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

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3 years ago
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