The total activity of the sample two hours after removal from the reactor core is 0.7552 MBq. The sample can be given to the researcher 11 hours and 42 minutes after removal from the reactor core.
Question 1The total activity of the NaCl sample two hours after removal from the reactor core can be calculated using the equation of radioactive decay.
Activity = Initial activity × (1/2)t/h
where
Activity = activity at time t
Initial activity = activity at time zero
h = half-life
t = time
For Na-24, t = 2 hours, h = 15 hours, and initial activity = 1 MBq.
Activity of Na-24 after 2 hours= 1 × (1/2)2/15= 0.7552 MBq
Therefore, the total activity of the sample two hours after removal from the reactor core is 0.7552 MBq.
Question 2The activity of the Cl-38 after time t can be calculated using the equation below:
Activity of Cl-38 = Initial activity of Cl-38 × (1/2)t/h
where
Activity = activity at time tInitial activity = activity at time zeroh = half-life
t = time
For Na-24, t = 2 hours, h = 15 hours, and initial activity = 1 MBq.
For Cl-38, t = ?, h = 37.2 minutes (0.62 hours), and initial activity = 2.7 MBq.
The researcher wants the Cl-38 activity to be 1% of the Na-24 activity.
Activity of Cl-38 = 0.01 × activity of Na-24= 0.01 × 1= 0.01 MBq
Activity of Cl-38 = Initial activity of Cl-38 × (1/2)t/h0.01 = 2.7 × (1/2)t/0.62(1/2)t/0.62 = 0.01/2.7(1/2)t/0.62 = 0.0037037(1/2)t = log 0.0037037/ log 0.50.5t = 5.8546t = 11.7092 hours (to the nearest minute)= 11 hours 42 minutes
Therefore, the sample can be given to the researcher 11 hours and 42 minutes after removal from the reactor core.
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4. Fluorine-18 is used in PET scans. It has a nuclear half-life of 110 minutes and a biological half-life of 24.5 minutes. A radiopharmaceutical company packages and ships a 250.mg sample of F−18 for delivery to a hospital. By the time it arrives it is only 35.2mg. How much time had elapsed? Give your answer in hours
The time elapsed between packaging and delivery is approximately 182.6 minutes, which is equivalent to 3.04 hours.
To solve this question, we can use the formula for radioactive decay:
A = A0 * e^(-kt)
Where:
A = Final amount
A0 = Initial amount
k = Decay constant
t = Time elapsed
Let's determine the decay constant, k. We can do this by using the formula:
t1/2 = (ln 2) / k
Where t1/2 is the half-life of the isotope.
Given that the nuclear half-life of Fluorine-18 is 110 minutes, we can substitute this value into the equation:
110 min = (ln 2) / k
Solving for k:
k = (ln 2) / 110 ≈ 0.00631 min^-1
Now, we can use the formula for radioactive decay to find the time elapsed. We know that the sample delivered to the hospital was 35.2 mg, while the original sample was 250 mg. Therefore, the fraction that remained after delivery is:
(amount remaining / initial amount) = A / A0 = 35.2 / 250 = 0.1408
Substituting this value, along with the other values we have, into the radioactive decay formula:
0.1408 = e^(-0.00631t)
Taking the natural logarithm on both sides, we get:
ln(0.1408) = -0.00631t
Solving for t, we find:
t = -ln(0.1408) / 0.00631 ≈ 182.6 minutes
Therefore, the time elapsed between packaging and delivery is approximately 182.6 minutes, which is equivalent to 3.04 hours. Hence, the answer is 3.04 hours.
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A Cell is B.00 un in diameter' and has a cell width of 60.0 nm thrck. If densty x (mass druided by volome) of the wall is the Same as thent of pure water (1000kym
−3
). What ts the mass (in my) of the cell wall cossuming cell is splowicul and the wall is thin sphericul slell?
The mass of the cell wall, assuming the cell is spherical and the wall is a thin spherical shell, is approximately 0.91 milligrams.
To calculate the mass of the cell wall, we first need to determine the volume of the wall.
The given diameter of the cell is 0.00 μm, which means the radius (r) of the cell is half of that, so r = 0.00/2 = 0.00 μm = 0.00 nm.Now, we need to find the volume of the cell wall, which can be approximated as a thin spherical shell. The volume of a thin spherical shell can be calculated using the formula:
V = 4/3 * π * (r_outer^3 - r_inner^3)
Since the cell is spherical, the inner radius of the shell is the same as the radius of the cell (r), and the outer radius of the shell is the sum of the radius of the cell (r) and the thickness of the wall (60.0 nm). Thus, the outer radius (r_outer) of the shell is:
r_outer = r + thickness = 0.00 + 60.0 = 60.0 nm
Substituting the values into the formula, we have:
V = 4/3 * π * (60.0^3 - 0.00^3)
= 4/3 * π * (216,000 nm^3)
= 288,000 π nm^3
Next, we need to calculate the mass of the cell wall using the density of pure water. The density (ρ) is given as 1000 kg/m^3, which is equivalent to 1000 kg/1,000,000,000 nm^3 since 1 m = 1,000,000,000 nm. Thus, the mass (m) of the cell wall is:
m = ρ * V
= 1000 kg/1,000,000,000 nm^3 * 288,000 π nm^3
= 0.000288 π kg
Now, we can calculate the mass of the cell wall by substituting the value of π (pi) as 3.14159:
m = 0.000288 * 3.14159 kg
= 0.000905 kg
≈ 0.91 mg
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Which of the following characteristics of the Moon is the best evidence that the Moon has been extensively heated? Mass of the moon Lack of iron core Oxygen isotopes like the Earth Lack of volatiles
The best evidence that the moon has been extensively heated is the lack of volatiles.
The following are the characteristics of the Moon
:Mass of the moon.Lack of iron core.
Oxygen isotopes like the Earth.
Lack of volatiles.
Volatiles are materials with low boiling points that exist in solid or liquid form at the Earth's surface. Water and carbon dioxide are two examples of volatile materials. The lack of volatiles on the moon is a strong indication that the moon was subjected to high temperatures. It also indicates that volatiles have been expelled from the moon's surface due to the loss of gas molecules that occurred as a result of the heat. As a result, the absence of volatiles is the best evidence that the Moon has been extensively heated.
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A pot is filled to the brim with 2329 milliliters of water at a temperature of 20ºC.
How much volume of water will spill over if the temperature of the water is increased to 82ºC? Express your answer in milliliters.
The coefficient fo thermal expansion of water is 210E-6 per Celsius.
Therefore, approximately 0.3007748 milliliters of water will spill over when the temperature is increased from 20ºC to 82ºC.
To determine the volume of water that will spill over when the temperature is increased, we need to consider the expansion of water due to the temperature change.
The formula to calculate the change in volume due to thermal expansion is given by:
ΔV = V0 * α * ΔT,
where:
ΔV is the change in volume,
V0 is the initial volume,
α is the coefficient of thermal expansion of water (210E-6 per Celsius),
ΔT is the change in temperature.
Given:
Initial volume, V0 = 2329 milliliters
Initial temperature = 20ºC
Final temperature = 82ºC
To find the change in volume, we can substitute the values into the formula:
ΔV = V0 * α * ΔT
ΔV = 2329 ml * (210E-6 per ºC) * (82ºC - 20ºC)
Now, let's calculate the change in volume:
ΔV = 2329 ml * (210E-6 per ºC) * (82ºC - 20ºC)
ΔV = 2329 ml * 210E-6 * 62ºC
Performing the calculation:
ΔV = 0.3007748 ml
Therefore, approximately 0.3007748 milliliters of water will spill over when the temperature is increased from 20ºC to 82ºC
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frequently used in shampoos. the detergent sodium dodecyl sulfate (sds) denatures proteins. suggest how sds destroys protein structure.
Sodium dodecyl sulfate (SDS), a commonly used detergent in shampoos, can disrupt protein structure through a process called denaturation.
Denaturation refers to the alteration of a protein's native structure, leading to loss of its functional properties. Here's how SDS can contribute to protein denaturation:
Hydrophobic interactions: SDS molecules have a hydrophobic tail that can interact with the hydrophobic regions of proteins. Proteins have hydrophobic amino acid residues buried within their interior, contributing to their structural stability.
When SDS interacts with these hydrophobic regions, it can disrupt the hydrophobic interactions holding the protein's structure together, causing unfolding and denaturation.
Disruption of hydrogen bonds: Proteins rely on hydrogen bonds for their secondary and tertiary structures. SDS can disrupt these hydrogen bonds by competitively binding to the polar regions of the protein, weakening the interactions.
As a result, the protein's folded structure can unravel.
Electrostatic interactions: SDS is an anionic detergent, meaning it carries a negative charge. It can interact with positively charged regions of proteins through electrostatic interactions. This interaction can disrupt the protein's stability by altering the balance of charges and affecting its overall structure.
Denaturation by stripping water molecules: SDS has a strong affinity for water molecules and can solubilize hydrophobic substances.
In the presence of SDS, water molecules that surround the protein can be displaced, leading to protein unfolding and denaturation.
Overall, the interactions of SDS with proteins can disrupt their hydrophobic interactions, hydrogen bonds, electrostatic interactions, and water interactions, resulting in the denaturation and loss of protein structure and function.
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for the compound copper i sulfate give the correct formula
The correct formula for the compound copper I sulfate is Cu2SO4. Here's the explanation:Chemical formulas are used to express the chemical composition of compounds in a simple manner.
In these formulas, chemical symbols are employed to signify elements while subscripts are used to specify the number of atoms or ions of each element in the molecule.For example, the formula for water is H2O, which means that each water molecule contains two hydrogen atoms and one oxygen atom.
Copper I sulfate, like all other ionic compounds, has a formula that reflects its chemical composition. Copper I sulfate is formed by the combination of copper I ions (Cu+) and sulfate ions (SO42-).
The copper I ion has a charge of +1, while the sulfate ion has a charge of -2. As a result, in order to achieve electrical neutrality, two copper I ions must combine with one sulfate ion.To express the chemical composition of copper I sulfate in formula notation, we can use subscripts to indicate the number of atoms or ions of each element present in the molecule.Therefore, the formula for copper I sulfate is Cu2SO4.
The two copper I ions and one sulfate ion in the molecule are indicated by the subscripts 2 and 1, respectively.
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Diastolic Blood Pressure of Females For the diastolic blood pressure measurements of females listed in Data Set 1 "Body Data" in Appendix B, the highest measurement is 98 mmHg. The 147 diastolic blood pressure measurements of females have a mean of
x
ˉ
=70.2 mmHg and a standard deviation of s=11.2 mmHg. a. What is the difference between the highest diastolic blood pressure and the mean of the diastolic blood pressures for females? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the highest diastolic blood pressure to a z score. d. Using the criteria summarized in Figure 3-6 on page 123, is the highest blood pressure significantly low, significantly high, or neither?
The mean of the diastolic blood pressures for females is 98 - 70.2 = 27.8 mmHg. The highest diastolic blood pressure is significantly high.
a. The difference between the highest diastolic blood pressure and the mean of the diastolic blood pressures for females is 98 - 70.2 = 27.8 mmHg.
b. The difference found in part (a) is 27.8 mmHg, and the number of standard deviations is 27.8/11.2 ≈ 2.48.c. To convert the highest diastolic blood pressure to a z score, we use the formula:z = (x - µ)/σ,where x is the value, µ is the mean, and σ is the standard deviation.z = (98 - 70.2)/11.2 ≈ 2.48d. Using the criteria summarized in Figure 3-6 on page 123, a z score of 2.48 is significantly high. Therefore, the highest diastolic blood pressure is significantly high.
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A rigid container holds 0.20 g of hydrogen gas. How much heat is needed to change the temperature of the gas from 50 K to 100 K ? Part B How much heat is needed to change the temperature of the gas from 250 K to 300 K ? Express your answer with the appropriate units. - Part C How much heat is needed to change the temperature of the gas from 2250 K to 2300 K ? Express your answer with the appropriate units.
A) The amount of heat needed is approximately 103.925 joules.
B) The amount of heat needed to change the temperature of the gas from 250 K to 300 K is also approximately 103.925 joules
C) The amount of heat needed to change the temperature of the gas from 2250 K to 2300 K is also approximately 103.925 joules.
To calculate the amount of heat needed to change the temperature of a gas, we can use the equation:
Q = n * C * ΔT
where:
Q is the heat energy (in joules),
n is the number of moles of the gas,
C is the molar heat capacity of the gas (in joules per mole per kelvin), and
ΔT is the change in temperature (in kelvin).
Part A:
Given that the container holds 0.20 g of hydrogen gas, we need to determine the number of moles of hydrogen.
The molar mass of hydrogen is approximately 2 g/mol. So, the number of moles (n) can be calculated as follows:
n = mass / molar mass
n = 0.20 g / 2 g/mol
n = 0.10 mol
Now we need to determine the molar heat capacity of hydrogen gas.
For diatomic gases like hydrogen, the molar heat capacity at constant volume (Cv) is approximately 5/2 R,
where R is the gas constant.
In this case, we'll assume the ideal gas constant R = 8.314 J/(mol·K).
Therefore,
Cv = (5/2) * R
Cv = (5/2) * 8.314 J/(mol·K)
Cv = 20.785 J/(mol·K)
Now we can calculate the amount of heat needed to change the temperature of the gas from 50 K to 100 K using the equation:
Q = n * Cv * ΔT
Q = 0.10 mol * 20.785 J/(mol·K) * (100 K - 50 K)
Q = 0.10 mol * 20.785 J/(mol·K) * 50 K
Q = 103.925 J
Therefore, the amount of heat needed is approximately 103.925 joules.
Part B:
The calculations for Part B are similar, but with different values for ΔT.
ΔT = 300 K - 250 K = 50 K
Q = n * Cv * ΔT
Q = 0.10 mol * 20.785 J/(mol·K) * 50 K
Q = 103.925 J
The amount of heat needed to change the temperature of the gas from 250 K to 300 K is also approximately 103.925 joules.
Part C:
Using the same approach as above, we can calculate the amount of heat needed for Part C.
ΔT = 2300 K - 2250 K = 50 K
Q = n * Cv * ΔT
Q = 0.10 mol * 20.785 J/(mol·K) * 50 K
Q = 103.925 J
The amount of heat needed to change the temperature of the gas from 2250 K to 2300 K is also approximately 103.925 joules.
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What atmospheric pressure would be required for carbon dioxide to boil at a temperature of 20
∘
C ? See the vapor pressure curve for carbon dioxide in the drawing.
5.5×10
6
Pa
7.5×10
6
Pa
3.5×10
6
Pa
The atmospheric pressure that would be required for carbon dioxide to boil at a temperature of 20 degrees Celsius is [tex]5.5*10^6 Pa.[/tex]
This is because the point at which the vapor pressure curve for carbon dioxide intersects with the 20 degrees Celsius temperature line on the graph is approximately [tex]5.5*10^6 Pa.[/tex]
At this pressure, the boiling point of carbon dioxide is equal to the temperature of the surroundings, which is 20 degrees Celsius.The vapor pressure curve for a substance is a graphical representation of its vapor pressure as a function of temperature.
It shows the pressure at which a liquid will boil at different temperatures. In the case of carbon dioxide, the vapor pressure curve shows that at atmospheric pressure (1 atm), carbon dioxide will only boil at a temperature of -78.5 degrees Celsius.
However, at higher pressures, the boiling point of carbon dioxide increases.
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