radon-222, a highly radioactive gas with a half-life of 3.8 days that originates in the decay of234u (see the chart ofnuclides), may be present in uranium mines in dangerous concentrations if the mines are not properly ventilated. calculate the activity of 222rn in bq per metric ton of natural uranium.

Answers

Answer 1

The activity of 222Rn in bq per metric ton of natural uranium is dependent on the concentration of 222Rn and the decay constant of 222Rn.

Solution:

To calculate the activity, we need to know the concentration of 222Rn in the uranium mine. The activity of a radioactive substance is given by the equation:
Activity = concentration * decay constant.
The decay constant for 222Rn can be calculated using its half-life:
decay constant = ln(2) / half-life.

So, Once we have the decay constant, we can multiply it by the concentration of 222Rn to find the activity in bq per metric ton of natural uranium.

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Related Questions

A graduated cylinder contains 26 cm3 of water. an object with a mass of 21 grams and a volume of 15 cm3 is lowered into the water. what will the new water level be

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When the object with a volume of 15 cm3 is lowered into the water in the graduated cylinder, the new water level will be 11 cm3.

The new water level in the graduated cylinder can be determined by considering the principle of displacement. When the object is lowered into the water, it will displace an amount of water equal to its own volume.

Given that the object has a volume of 15 cm3, it will displace 15 cm3 of water. Since the initial volume of water in the graduated cylinder is 26 cm3, the new water level can be calculated by subtracting the volume of water displaced by the object from the initial volume of water.

Therefore, the new water level in the graduated cylinder will be 26 cm3 - 15 cm3 = 11 cm3.

To summarize, when the object with a volume of 15 cm3 is lowered into the water in the graduated cylinder, the new water level will be 11 cm3.

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The liquid dispensed from a burette is called ___________.

i. solute

ii. water

iii. titrant

iv. analyte

Answers

The liquid dispensed from a burette is called the titrant. A titrant is a solution with a known concentration that is added in a controlled manner to react with the analyte in a chemical analysis. The option C is correct.

The burette is a precise measuring instrument used in titrations to deliver the titrant.In a titration, the analyte is the substance being analyzed or tested. It reacts with the titrant to form a product, and the reaction is monitored to determine the concentration or amount of the analyte.

For example, in an acid-base titration, a solution of known concentration called the titrant is slowly added to the analyte solution until the reaction between the acid and base is complete. The burette allows for precise measurement of the volume of titrant added.The other options given are not accurate in this context. Solute refers to the substance being dissolved in a solvent, while water is a common solvent. Analyte, as mentioned earlier, is the substance being analyzed. The correct term for the liquid dispensed from a burette in a titration is the titrant.

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You prepare a stock solution that has a concentration of 2. 5 m. An aliquot with a volume of 10. 0 ml is removed from the solution. What is the concentration of the aliquot?.

Answers

The concentration of the aliquot is 2.5 M.

The concentration of a solution is defined as the amount of solute present per unit volume of the solution.

In this case, the stock solution has a concentration of 2.5 M (moles per liter).

An aliquot is a small portion or sample taken from a larger solution. In this scenario, an aliquot with a volume of 10.0 ml is removed from the stock solution.

Since the concentration of the stock solution is given in terms of moles per liter (M), the concentration of the aliquot will be the same as the concentration of the stock solution.

The concentration does not change when a specific volume is removed from the solution.

Therefore, the concentration of the aliquot is 2.5 M. It is important to note that the concentration remains the same regardless of the volume of the aliquot, as long as the proportion of solute to solvent remains constant.

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at constant temperature, a 144.0 ml sample of gas in a piston chamber has a pressure of 2.25 atm. calculate the pressure of the gas if this piston is pushed down hard so that the gas now has a volume of 36.0 ml.

Answers

The pressure of the gas would be 9.0 atm if the piston is pushed down hard to a volume of 36.0 ml.

To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume at constant temperature.

First, we need to set up the equation: P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Given that the initial volume (V1) is 144.0 ml and the initial pressure (P1) is 2.25 atm, and the final volume (V2) is 36.0 ml, we can plug in the values into the equation:

2.25 atm * 144.0 ml = P2 * 36.0 ml

Next, we can solve for P2 by dividing both sides of the equation by 36.0 ml:

2.25 atm * 144.0 ml / 36.0 ml = P2

P2 = 9.0 atm

Therefore, the pressure of the gas would be 9.0 atm if the piston is pushed down hard to a volume of 36.0 ml.

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a homogeneous solution contains copper(ii) ions (cu2 ), silver ions (ag ) and potassium ions (k ). you have sodium bromide (nabr) and sodium sulfide (na2s) available to use. what should you add and in what order to separate the three metal ions? ksp (sulfides) ksp (bromides) cus 6.0×10–37 cubr2 soluble ag2s 6.0×10–51 agbr 7.7×10–13 k2s soluble kbr soluble

Answers

To separate Cu2+, Ag+, and K+ from the homogeneous solution, add sodium sulfide (Na2S) first to precipitate CuS. Then add sodium bromide (NaBr) to precipitate AgBr. Finally, the remaining solution contains only K+.

To separate the copper (II), silver, and potassium ions from the homogeneous solution, you can employ the following procedure.

Firstly, add sodium sulfide (Na2S) to the solution, resulting in the formation of insoluble copper sulfide (CuS) precipitate due to its low solubility (Ksp = 6.0×10–37). By filtering the solution, the insoluble CuS precipitate can be separated.

Next, introduce sodium bromide (NaBr) to the filtrate, causing the formation of insoluble silver bromide (AgBr) precipitate due to its low solubility (Ksp = 7.7×10–13). By filtering the solution once again, the insoluble AgBr precipitate can be isolated.

Finally, the remaining solution will only contain potassium ions (K+), which do not require further separation steps as potassium salts are highly soluble in water. By following this procedure, effective separation of the copper (II), silver, and potassium ions can be achieved.

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