Radioactive gold-198 is used in the diagnosis of liver problems. the half-life of this isotope is 2.7 days. if you begin with a sample of 8.1 mg of the isotope, how much of this sample remains after 2.6 days?

Answers

Answer 1
Answer:

Answer:

See explanation below

Explanation:

To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:

m = m₀e^-kt  (1)

In this case, k will be the constant rate of this element. This is calculated using the following expression:

k = ln2/t₁/₂  (2)

Let's calculate the value of k first:

k = ln2/2.7 = 0.2567 d⁻¹

Now, we can use the expression (1) to calculate the remaining mass:

m = 8.1 * e^(-0.2567 * 2.6)

m = 8.1 * e^(-0.6674)

m = 8.1 * 0.51303

m = 4.16 mg remaining

Answer 2
Answer:

Final answer:

The half-life of gold-198 is the time it takes for half of it to decay. Given that the half-life is 2.7 days, and the period in consideration is 2.6 days, approximately half of the original amount of 8.1 mg, which is 4.05 mg, will remain.

Explanation:

This problem is related to the concept of half-life in radioactive decay. The half-life of a substance is the time it takes for half of it to decay. As the half-life of gold-198 is 2.7 days and we are considering a period of 2.6 days, which is almost one half-life, therefore, approximately half the substance should have decayed.

So, if you start with 8.1 mg of gold-198, at the end of one half-life (or close to it at 2.6 days), you should have approximately half of this amount remaining. Half of 8.1 mg is 4.05 mg, thus, approximately 4.05 mg remains after 2.6 days.

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A rigid container of gas has a pressure of 1.72 atm and temperature of 21 oC. If the temperature increases to 85 oC, what is the new pressure in atm?

Answers

Answer:

the new pressure is 2.09 atm

Explanation:

you have to use gay lussac's law so the formula is

p1/t1 = p2/t2

and convert C to Kelvin k=C+273.15

1.72atm/294.15 = p2/358.15

solve for p2 by multiplying 358.15 on both sides

p2=2.09 atm

Ammonia, NH3, is used as a refrigerant. At its boiling point of –33 oC, the standard enthalpy of vaporization of ammonia is 23.3 kJ/mol. How much heat is released when 50.0 g of ammonia is condensed at –33 oC?–0.466 kJ–7.94 kJ–36.6 kJ–68.4 kJ–1.17 x 103 kJ

Answers

Answer:

-68.4 kJ

Explanation:

The standard enthalpy of vaporization = 23.3 kJ/mol

which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).

To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.

This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.

Thus,  Q = -23.3 kJ/mol

Where negative sign signifies release of heat

Given: mass of 50.0 g

Molar mass of ammonia = 17.034 g/mol

Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles

Also,

1 mole of ammonia when condenses at -33 °C releases 23.3 kJ

2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ

Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.

Final answer:

The heat released when 50.0 g of ammonia condenses at its boiling point is -68.4 kJ. This is calculated by multiplying the moles of ammonia by the enthalpy of vaporization and recognizing that heat is released in condensation.

Explanation:

To solve this problem, we need to understand the concept of enthalpy of vaporization, which is the heat needed to convert 1 mole of a substance from a liquid to a gas at constant pressure and temperature. For ammonia (NH3), which boils at -33 °C, the enthalpy of vaporization is 23.3 kJ/mol. However, we want the heat released when 50.0 g (around 2.94 moles) of ammonia condenses, which is the reverse process of vaporization. Thus, the energy would be released rather than absorbed.

Now, let's calculate this value. We multiply the number of moles of ammonia by the enthalpy of vaporization:

2.94 moles x 23.3 kJ/mol = 68.4 kJ

Since this is the reverse of the process of vaporization, heat is released, so the enthalpy change is negative (-68.4 kJ). Therefore, the correct answer is -68.4 kJ.

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. An open manometer is filled with mercury. The mercury level is 12mm higher on the side open to the atmosphere. What is the total pressure of the gas, in kPa, if the atmospheric pressure is 100.8 kPa?

Answers

The total pressure of the gas, in kPa is equal to 102.4 kPa.

Given the following data:

  • Atmospheric pressure = 100.8 kPa
  • Pressure (open side) = 12 mm Hg

To determine the total pressure of the gas, in kPa:

First of all, we would convert the value of the pressure in mm Hg to kPa.

Conversion:

760 mm Hg = 101.325 kPa

12 mm Hg = X kPa

Cross-multiplying, we have:

760X = 12* 101.325\n\n760X = 1,215.9\n\nX=(1215.9)/(760)

X = 1.60 kPa

Now, we can calculate the total pressure by using this formula:

Total\;pressure = Atmospheric \;pressure + Pressure (side)\n\nTotal\;pressure = 100.8 + 1.60

Total pressure = 102.4 kPa

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Answer:

102kPa

Explanation:

760mmHg = 101.325 kPa

12mmHg = x

Upon converting from mmHg to kPa we have;

x = ( 12 mm Hg x  101.325 kPa)  / 760mmHg

x = 1.599868421 kPa

Total pressure = 100.8 kPa + 1.599868421 kPa

Total pressure = 102kPa

A titration is carried out to determine the concentration of the acid in an old bottle of aqueous HCl whose label has become unreadable.What is the HCl concentration if 32.1mL of 0.250M NaOH is required to titrate a 30.0mL sample of the acid?

Answers

Answer:

The concentration of HCl is - 0.2675 M

Explanation:

At equivalence point,

Moles of HCl = Moles of NaOH

Considering:-

Molarity_(HCl)* Volume_(HCl)=Molarity_(NaOH)* Volume_(NaOH)

Given  that:

Molarity_(NaOH)=0.250\ M

Volume_(NaOH)=32.1\ mL

Volume_(HCl)=30.0\ mL

Molarity_(HCl)=?\ M

So,  

Molarity_(HCl)* Volume_(HCl)=Molarity_(NaOH)* Volume_(NaOH)

Molarity_(HCl)* 30.0=0.250* 32.1

Molarity_(HCl)=(0.250* 32.1)/(30.0)=0.2675\ M

The concentration of HCl is - 0.2675 M

5. The partition coefficient of Compound A is 7.5 in dichloromethane (a.k.a. methylene chloride) with respect to water. a. If 5 grams of Compound A were dissolved in 100 mL of water, how much of Compound A would be extracted with four 25-mL portions of dichloromethane

Answers

Answer:

4.93g are extracted

Explanation:

Partition coefficient (P) is defined as the ratio of solute dissolved in the organic solvent and the solute dissolved in the aqueous phase.

That is:

P = 7.5 = Concentration in dichloromethane / Concentration in water.

Knowing this, in the first extraction with 25mL of dichloromethane you will extract:

7.5 = (X/25mL) / (5g - X) / 100mL

Where X is the amount of compound A that is extracted.

7.5 = 100X / (125 - 25X)

937.5 - 187.5X = 100X

937.5 = 287.5X

3.26g of A are extracted in the first extraction.

In water will remain 5g - 3.26g = 1.74g

In the second extraction you will extract:

7.5 = (X/25mL) / (1.74g - X) / 100mL

7.5 = 100X / (43.5 - 25X)

326.25 - 187.5X = 100X

326.25 = 287.5X

1.13g are extracted in the second extraction.

And remain: 1.74g - 1.13g = 0.61g

In the third extraction you will extract:

7.5 = (X/25mL) / (0.61g - X) / 100mL

7.5 = 100X / (15.25 - 25X)

114.375 - 187.5X = 100X

114.375 = 287.5X

0.40g are extracted in the third extraction.

And remain: 0.61g - 0.40g = 0.21g

In the second extraction you will extract:

7.5 = (X/25mL) / (0.21g - X) / 100mL

7.5 = 100X / (5.25 - 25X)

39.375 - 187.5X = 100X

39.375 = 287.5X

0.14g are extracted in the fourth extraction.

Thus, after the three extractions you will extract: 0.14g + 0.40g + 1.13g + 3.26g = 4.93g are extracted

Final answer:

The process involves using the partitioncoefficient to determine how much of Compound A will prefer the dichloromethane solvent over the water. Following a calculation process through four rounds of extraction, it is concluded that approximately 4.999g of Compound A will be extracted using four 25mL portions of dichloromethane.

Explanation:

The partition coefficient of a compound is a measure of how much it prefers one solvent over another. Given that the partition coefficient of Compound A is 7.5 in dichloromethane with respect to water, we can predict how much of this compound could be extracted using four separate 25 mL portions of dichloromethane.

Here's the step-by-step calculation process:

  1. We start with 5 grams of Compound A in 100 mL of water. Given the partition coefficient, in the initial phase, 5/(7.5+1)=0.625g remains in water and 7.5/8.5*5=4.375g goes into the dichloromethane.
  2. After one extraction with 25ml of dichloromethane, the amount left in the water will be 0.625g*1/(7.5+1)=0.069g.
  3. After the second extraction: 0.069g*1/(7.5+1) = 0.008g.
  4. After the third extraction: 0.008g*1/(7.5+1) = 0.0009g.
  5. After the fourth extraction: 0.0009g*1/(7.5+1) = 0.0001g.

In total, around 4.999g of compound A will be extracted using four 25mL portions of dichloromethane.

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Please match word and definition < electrons and electric current move easily 1. electric current < Prevents electrons from moving 2. insulator flow of an electrical charge in a circuit 3. electric circuit 4. conductor provides a path for electrons to flow​

Answers

Due to property of conduction, conductor allows electrons and electric current to move easily,insulators prevent electric current from moving, electric current is the  flow of an electrical charge in a circuit and electric circuit provides a path for electrons to flow​.

What is conduction?

Conduction is defined as a process as a means of which heat is transferred from the hotter end of the body to it's cooler end.Heat flows spontaneously from a body which is hot to a body which is cold.

In the process of conduction,heat flow is within the body and through itself.In solids the conduction of heat is due to the vibrations and collisions of molecules while in liquids and gases it is due to the random motion of the molecules .

When conduction takes place, heat is usually transferred from one molecule to another as they are in direct contact with each other.There are 2 types of conduction:1) steady state conduction 2) transient conduction.According to the type of energy conduction is of three types:

1) heat conduction

2) electrical conduction

3)sound conduction

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Answer:

1st box:4

2nd box:2

3rd box:3

4th box:1

Explanation:

I am not 100% sure on this, but i am pretty confident that this is the answer! BUT If this is somehow wrong, please tell me so i can comment a better answer! :)