In which of the following is light energy converted to electrical energy
2 answers:
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Answer:
0.4 M
Explanation:
The process that takes place in an aqueous K₂HPO₄ solution is:
- K₂HPO₄ → 2K⁺ + HPO₄⁻²
First we <u>calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution</u>:
- 200 mL * 0.2 M = 40 mmol K₂HPO₄
Then we <u>convert K₂HPO₄ moles into K⁺ moles</u>, using the <em>stoichiometric coefficients</em> of the reaction above:
- 40 mmol K₂HPO₄ *
= 80 mmol K⁺
Finally we <em>divide the number of K⁺ moles by the volume</em>, to <u>calculate the molarity</u>:
- 80 mmol K⁺ / 200 mL = 0.4 M
Two months later 13.8 milligrams of the barium-131 still be radioactive.
<h3>How is the decay rate of a radioactive substance expressed ? </h3>It is expressed as:
where,
A = Amount remaining
A₀ = Initial Amount
t = time
T = Half life
Here
A₀ = 0.50g
t = 2 months = 60 days
T = 11.6 days
Now put the values in above expression we get
= 0.50 × 0.0277
= 0.0138 g
= 13.8 mg [1 mg = 1000 g]
Thus from the above conclusion we can say that Two months later 13.8 milligrams of the barium-131 still be radioactive.
Learn more about the Radioactive here: brainly.com/question/2320811
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Question: Suppose that 0.50 grams of ban that 0.50 grams of barium-131 are administered orally to a patient. Approximately many milligrams of the barium would still be radioactive two months later? The half-life of barium-131 is 11.6 days.