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svp [43]
2 years ago
13

Radioactive elements decay at a know rate known as half-life. Choose all of the correct statements concerning half-life. Carbon-

14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years.
Chemistry
2 answers:
Lera25 [3.4K]2 years ago
8 0

Answer:

-Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years.

-Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years.

Explanation:

9966 [12]2 years ago
6 0

Explanation:

Half life is simply the amount of time it takes for half of a substance to decompose.

Options;

- Carbon-14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. This is incorrect. After 5730 years, 15g of the sample ought to remain.

- Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. This is correct. 3 * 76000 = 228,000

- Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. This is incorrect. Mass after 3 half lives (27/9) = 9.5 (38 / 2 / 2)

- Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. This is incorrect. Mass after 4 half lives (6 / 1.5) = 2.5 gram (40 / 2 / 2 /2 / 2)

- Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years. This is incorrect. Original sampe = 240 gram. So after 2 half lives (105,000/52500), mass left = 60 (240 / 2 /2)

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AleksAgata [21]

Volume of tank = 6.9 m^{3} (given)

Since, 1 m = 3.28 ft

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1 m^{3} = 35.287 ft^{3}

For 6.9 m^{3}:

6.9\times 35.287 ft^{3} = 243.4803 ft^{3}

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

243.4803 ft^{3} \simeq 243.5 ft^{3}

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5 1
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Answer is: C. H₂, molecule of hydrogen, g is c<span>hemistry abbreviations or physical state symbol for gas.</span>
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6 0
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How many moles of koh are contained in 750. ml of 5.00 m koh solution?
Marat540 [252]
Molarity is one of  the method of expressing concentration of solution. Mathematically it is expressed as,
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Given: Molarity of solution = 5.00 M
Volume of solution = 750 ml = 0.750 l

∴ 5 = \frac{\text{number of moles}}{.750}
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4 0
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A scuba tank contains a mixture of oxygen (O2) and nitrogen (N2) gas. The oxygen has a partial pressure of PO2=5.62MPa. The tota
dmitriy555 [2]

Answer:

21.16 MPa

Explanation:

Partial pressure of oxygen = 5.62 MPa

Total gas pressure = 26.78 MPa

But

Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.

This implies that;

Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen

Hence partial pressure of nitrogen=

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Therefore;

Partial pressure of nitrogen= 26.78 - 5.62

Partial pressure of nitrogen = 21.16 MPa

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Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio
AlexFokin [52]

Answer:

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Explanation:

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The equilibrium constant (Keq) is given for the following expresion:

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We have:

(CH3OH)= ?

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So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:

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14.5 x (0.15 M) x (0.36)^{2} = (CH₃OH)

0.2818 M = (CH₃OH)

6 0
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