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SIZIF [17.4K]
3 years ago
5

Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. Which of

the following statements about containers A and B is true? Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. Which of the following statements about containers A and B is true? The rate of decay of atoms (half-life) in container A is greater (or longer) than the rate of decay of atoms (half-life) in container B. The rate of decay of atoms (half-life) in container B is greater (or longer) than the rate of decay of atoms (half-life) in container A. The rate of decay of atoms (half-life) in container B is the same as the rate of decay of atoms (half-life) in container A.
Chemistry
2 answers:
Angelina_Jolie [31]3 years ago
6 0

Answer:

b

Explanation:

Sloan [31]3 years ago
3 0

Answer:

B

Explanation:

This is due to the differences in the concentrations of the radioactive isotopes in the two containers of the same volume. In container A, with a higher density of atoms, there is a higher chance of a neutron released in the decay of one atom, hitting another atom and splitting it. This is in comparison to container B that has fewer atoms that are widely dispersed due to the lower density. Atoms in cointainer A will therefore decay faster (shorter half-life) than atoms in container B.

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Tetraphosphorus decaiodide is represented by which formula? A) P3I10 B) P4I10 C) P3I8 D) P4I8
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I need help pleaseee
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Answer: So, If we mix 40.0 g of NaOH with enough distilled water to make 500 mL, we will get a 2.00 M NaOH solution.

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5 0
2 years ago
If the aluminum block is initially at 25 ∘C∘C, what is the final temperature of the block after the evaporation of the alcohol?
Effectus [21]

Answer:

Final temperature of aluminum block = 12.1°C

<em>Note: The question is not complete. The complete question is given below:</em>

<em>If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol Suppose that 1.12 g of rubbing alcohol (C3H8O) evaporates from a 73.0 g aluminum block.</em>

Explanation:

Heat lost by aluminum block = heat required for vaporization of alcohol

Heat required to vaporize ethanol, H = mass of alcohol * heat of vaporization of alcohol

Mass of alcohol = 1.12 g; molar mass of rubbing alcohol = 60 g/mol

Heat of vaporization = (45.4 kJ/mol)/ 60 g/mol =  0.75666 kJ/g

H = 1.12 g × 0.7566 kJ/g

H = 0.8474 kJ = 847.4 J

Heat lost by aluminum block, Q = -(mass × specific heat capacity × temperature change)

Mass of aluminum block = 73.0 g

Specific heat capacity of aluminum = 0.900 J/g

Temperature change = (Final temperature, T - 25)

Q = -73.0 g × 0.900 J/g × (T - 25)

Q = -65.7 J × (T - 25°C)

Since, Heat lost by aluminum block = heat for vaporization of alcohol

-65.7 J × (T - 25°C) = 847.4 J

T - 25 = 847.4/-65.7

T - 25 = -12.9

T = -12.9 + 25

T = 12.1°C

5 0
3 years ago
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