Answer:
Heats the earth evenly;can acclerate the water cycle;is the perfect distance from earth.Explanation:
Radiation is the most important process of heat propagation, because it is through it that the heat of the sun reaches the earth. Without this process there would be no life on earth. While conduction and convection occur only in material media, radiation also occurs in a vacuum, so it is possible for the sun to heat the earth evenly.
The heat radiated by the sun heats the water of rivers, lakes, seas and oceans occurring the phenomenon of evaporation during the water cycle. At this point, the liquid state of water changes to its gaseous state as it moves from the earth's surface to the atmosphere. For this reason we can say that the sun can accelerate the water cycle.
Although the Earth's orbit around the Sun is an ellipse, not a circle, the distance from Earth to the Sun varies by only 3%, with the Earth being closest to the Sun from January 4-7 each year, depending on leap year. But it's easy to remember that the northern hemisphere of the earth is also closer to the sun in January and it's winter there, while it's summer in the southern hemisphere. So we cannot say that the earth is closer to the sun in summer, but between January 4th and 7th.
Living conditions on earth are a puzzle to astronomy. A little closer to the sun - and we'd be a greenhouse and toxic high-pressure gas plant like Venus. A little farther - and the gases would escape, the water would freeze, there would be at most microbes, as on Mars. The earth is the perfect distance to house water, oxygen, carbon and complex life forms.
Surface Waves are seismic waves that propagate to the Earth's surface and result from interference between P waves and S waves (called volumetric waves). It is to the surface waves that are due to the great destruction caused by the earthquakes. The sun does not interfere with the creation of these waves, which is a factor caused by the movement of the earth and other geological terms.
Carl crumbled a piece of paper before throwing it in the trash can. Which is true about the paper after it has been crumbled
Final answer:
Crumpling a piece of paper doesn't change its mass but alters its shape, leading to reduced air resistance.
Explanation:
When Carl crumbled a piece of paper before throwing it in the trash can, the physical shape of the paper changed, but not its mass. This action relates to a physics concept concerning gravity and air resistance.
An important observation is that when different objects are dropped from the same height, like a crumpled piece of paper and a textbook, they will hit the ground at approximately the same time if we ignore air resistance.
This is because all objects in a vacuum fall at the same rate regardless of mass, due to the uniform acceleration caused by gravity. In real-life conditions, air resistance affects how objects fall.
Crumpling the paper reduces its surface area, thereby reducing air resistance and allowing it to fall faster compared to an uncrumpled sheet of paper.
If you crumple one piece of paper into a small ball and then crumple two pieces of paper, making a bigger ball but with double the mass, they will still hit the ground at the same time if dropped from the same height because the acceleration due to gravity is constant.
Given the starting solution had a concentration of 1.25 m, how many moles of co[h2o]6cl2 were available in the amount of starting solution you used
To calculate the number of moles of co[h2o]6cl2 in the starting solution, we can use the molarity (concentration) of the solution and the volume of the solution used. If 1L of the solution is used, the number of moles would be equal to the molarity of the solution.
Explanation:The question is about finding out the number of moles of co[h2o]6cl2 present in a starting solution with a given molarity (concentration). Molarity, in a nutshell, is defined as the number of moles of solute per liter of solution. Given that the molarity of the solution is 1.25M, we can directly relate this to the number of moles using the definition of molarity. For example, if we used 1L of the solution, the number of moles would be 1.25moles. If less or more than 1L of solution was used, the number of moles would be less or more, respectively, which is directly proportional to the volume of the solution used.
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The equilibrium constant for the reaction sr(s) + mg2+(aq) ⇌ sr2+(aq) + mg(s) is 3.69 × 1017 at 25°c. calculate e o for a cell made up of sr / sr2+ and mg / mg2+ half-cells.
To calculate the standard electrode potential (E°) for a cell composed of Sr/Sr²⁺ and Mg/Mg²⁺ half-cells, use the Nernst equation with the given equilibrium constant value. The resulting E° is approximately 1.038 V.
To find the standard electrode potential (E°) for the cell, we can use the Nernst equation and the relationship between the equilibrium constant (Keq) and the standard electrode potential:
Nernst equation:
E° = (RT / nF) * ln(Keq)
Where:
R = 8.314 J/(mol·K)T = 298 K (since the temperature is 25°C)n = 2 (number of electrons transferred, as each Sr replaces an Mg)F = 96485 C/molPlugging in the values, we get:
E° = (8.314 * 298 / (2 * 96485)) * ln(3.69 × 1017)
Simplify the constants:
E° = (0.0257 V) * ln(3.69 × 1017)
Evaluating the natural logarithm:
E° = 0.0257 * 40.38
E° ≈ 1.038 V
Therefore, the standard electrode potential (E°) for the cell is approximately 1.038 V.
A sound is first produced by making something BLANK . The sound then travels through a BLANK to reach the ears, which are the parts of the body that allow for sounds to be heard.
A sound is first produced by making something vibrates and then the sound travels through a medium and reach our ears.
Which are the parts of the body that allow for sounds to be heard?The ear is responsible for hearing sounds as well as for balance in the human body. The ear has three parts i.e. the outer, middle and inner ears. We hear the sound when the body starts vibration.
So we can conclude that a sound is first produced by making something vibrates and then the sound travels through a medium and reach our ears.
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if you have 100 grams of S8, how many moles of S8 is that
How many carbon-carbon sigma bonds are present in each of the following molecules: benzene, cyclobutane?
Final answer:
Benzene has six carbon-carbon sigma bonds because each carbon atom forms two sigma bonds with its adjacent carbon atoms. Cyclobutane has four carbon-carbon sigma bonds since it is a four-membered ring with each carbon bonded to two other carbons.
Explanation:
The question asks about the number of carbon-carbon sigma bonds in benzene and cyclobutane. Let's examine each molecule separately.
Benzene (C₆H₆)
In benzene, the carbons form a hexagonal ring and each carbon atom is bonded to two other carbon atoms. The structure of benzene is often depicted with alternating single and double bonds. However, due to resonance, the electrons are delocalized and each carbon-carbon bond is equivalent, resembling a structure with one-and-a-half bonds. Despite the presence of alternating double bonds, each carbon is involved in two sigma bonds with adjacent carbons and one sigma bond with a hydrogen atom, making a total of six carbon-carbon sigma bonds in benzene.
Cyclobutane (C₄H₈)
Cyclobutane is a four-membered ring where each carbon atom is bonded to two other carbons forming single bonds. Thus, for cyclobutane, there are four carbon-carbon sigma bonds, one between each pair of adjacent carbon atoms.
what is the volume of 12.0 g of cl2 gas at stp
Answer:
[tex]V=3.79L[/tex]
Explanation:
Hello,
In this case, we are talking about an ideal gas problem in which 12.0 g of chlorine gas is at standard both pressure (1atm) and temperature (0°C), therefore, to compute the corresponding volume, one applies the ideal gas law, then converts from grams to moles considering the diatomic chlorine and subsequently solves for volume as shown below:
[tex]PV=nRT\\V=\frac{nRT}{P}=\frac{12.0gCl_2*\frac{1molCl_2}{70.9gCl_2} *0.082 \frac{atm*L}{mol*K} * 273.15K}{1atm} \\V=3.79L[/tex]
Best regards.
What crossed-aldol product results when butanal is heated in the presence of excess benzaldehyde and sodium hydroxide? draw the molecule on the canvas by choosing buttons from the tools (for bonds), atoms, and advanced template toolbars. the single bond is active by defa?
When butanal is heated with excess benzaldehyde and sodium hydroxide, the crossed-aldol product formed is 4-phenyl-2-butanol, which is obtained through the aldol condensation reaction.
Explanation:Crossed-aldol product formation occurs when two different carbonyl compounds, such as aldehydes or ketones, react in an aldol condensation. The products are a mixture of different aldol adducts, resulting from the nucleophilic addition of one carbonyl compound to the other. This reaction allows for the creation of complex molecules with varied functionalities.
When butanal is heated in the presence of excess benzaldehyde and sodium hydroxide, the crossed-aldol product formed is 4-phenyl-2-butanol. The product is obtained by the aldol condensation reaction between butanal and benzaldehyde. In this reaction, a molecule of water is eliminated, and the resulting product contains both the aldehyde and the alcohol functional groups.
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Given that delta hvap is 58.2 kj/mol and the boiling point is 83.4 c 1atm if one mole of this substance is vaporized at 1atm calculate delta ssurr
Why are Noble gases inert?
A. They cannot form bonds because they have 0 electrons in their valence shell.
B. They are actually so reactive that they bond instantly with any substance they touch. This means they cannot be found in their pure form in nature.
C. They have a full shell of valence electrons.
D. They bond only with atoms of their own element.
Among the alkali earth metals, the tendency to react with other substances
A) increases from bottom to top within the group.
B) is shown by the ways they react with water.
C) varies in an unpredictable way within the group.
D) does not vary among the members of the group.
In this lab, you will use the flame test to identify the metal ion in two boxes of unidentified fireworks. Write an investigative question that you can answer by doing this experiment.
The first one is chlorine ion, the second one is metal ion.
The solubility product for chromium(iii) fluoride is ksp = 6.6 × 10–11. what is the molar solubility of chromium(iii) fluoride?
To determine the molar solubility of chromium(III) fluoride, we need to use the solubility product constant (Ksp) of 6.6 × 10⁻¹¹. The molar solubility of chromium(III) fluoride is 8.184 × 10⁻⁴M.
Explanation:To determine the molar solubility of chromium(III) fluoride, we need to use the solubility product constant (Ksp) of 6.6 × 10⁻¹¹ for chromium(III) fluoride. The Ksp expression is Ksp = [Cr³⁺][F⁻]³.
Let's assume the molar solubility of chromium(III) fluoride is x M. Since the stoichiometry of the balanced equation for its dissolution is 1:3 (one chromium ion per three fluoride ions), the equilibrium expression becomes Ksp = x(3x)³.
Now, we can set up the expression and calculate the molar solubility by solving the equation: Ksp = 108x⁴ = 6.6 × 10⁻¹¹. Solving for x gives x = 8.184 × 10⁻⁴ M. Therefore, the molar solubility of chromium(III) fluoride is 8.184 × 10⁻⁴ M.
Consider 2NH3(g)+3CuO(s)→N2(g)+3Cu(s)+3H2O(g) 2NH 3 (g)+3CuO(s)→N 2 (g)+3Cu(s)+3H 2 O(g). What volume (in mL) of NH3 NH 3 is required to completely react 45.2 g of CuO at STP? Enter your solution as a numerical value with no units.
The concentration of the solid phase _____ the quantity present.
depends on
changes with the amount of
is equal to
always remains the same regardless of
Answer: Option (b) is the correct answer.
Explanation:
A solution that contains more number of solute particles than the solvent particles is known as a concentrated solution.
For example, concentrated solution of sulfuric acid.
On the other hand, a solution that contains less number of solute particles than the solvent particles is known as a dilute solution.
For example, a dilute solution of HCl.
Hence, we can conclude that concentration of a substance changes with change in the amount of solute.
Therefore, the concentration of the solid phase changes with the amount of the quantity present.
A mixture of 100 g of k2cr207 and 200 g of water is stirred at 60 °c until no more of the salt dissolves. the resulting solution is poured off, leaving the undissolved solid behind. the solution is now cooled to 20°c. what mass of k2cr207 crystallizes from the solution during the cooling?
No K₂Cr₂O₇ will crystallize out of the solution during cooling.
What is crystallization?Crystallization is a separation technique used to purify a solid substance by selectively dissolving it in a suitable solvent at a high temperature and then cooling the solution to obtain pure crystals of the solute.
Given:
The solubility of K₂Cr₂O₇ in water at 60°C is 127 g/100 mL, and at 20°C is 13.9 g/100 mL.
Dissolved 100 g of K₂Cr₂O₇ in 200 g of water, which is 200 mL of water.
At 60°C, the solution can dissolve 127 g/100 mL × 2 L
= 254 g of K₂Cr₂O₇.
100 g of K₂Cr₂O₇, the solution is saturated and no more of the salt can dissolve.
When the solution is cooled to 20°C, the solubility of K₂Cr₂O₇ is only 13.9 g/100 mL.
The amount of water in the solution at 20°C is 200 mL. The maximum amount of K₂Cr₂O₇ that can remain in solution at this temperature is:
13.9 g/100 mL × 2 L = 278 g
Dissolved 100 g of K₂Cr₂O₇ in the solution, the amount that will crystallize out is:
100 g - 278 g = -178 g
This result is negative, indicating that all the K₂Cr₂O₇ will remain in solution at 20°C.
Therefore, no K₂Cr₂O₇ will crystallize out of the solution during cooling.
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How much energy does an electric hair dryer use if it draws 8.3 amps of current when using 120 volts for 5.0 minutes?
Answer: The energy drawn by electric hair dryer is 2,98,800 J
Explanation:
To calculate the power of the electric hair dryer, we use the equation:
[tex]P=I\times V[/tex]
where,
P = power of electric hair dryer
I = Current of electric hair dryer = 8.3 A
V = voltage of electric hair dryer = 120 V
Putting values in above equation, we get:
[tex]P=8.3\times 120=996W[/tex]
To calculate the energy of electric hair dryer, we use the equation:
[tex]E=P\times t[/tex]
where,
E = energy of electric hair dryer = ?
P = Power of electric hair dryer = 996 W
t = time taken = 5.0 min = 300 s (Conversion factor: 1 min = 60 sec)
Putting values in above equation, we get:
[tex]E=996\times 300=2,98,800J[/tex]
Hence, the energy drawn by electric hair dryer is 2,98,800 J
n comparison to molecules that interact by London dispersion forces only, the melting point of similar-sized molecules forming hydrogen bonds would most likely be about the same. unpredictable. lower. higher.
Answer:
d
Explanation:
Sucrose is very soluble in water. at 25◦c, 211.4 grams of sucrose will dissolve in 100 g of water. given that the density of the saturated sucrose solution is 1.34 g/ml, what is the molarity of the solution
The molarity of the sucrose solution is calculated by finding the number of moles of sucrose and the volume of the solution in liters. By using the provided data, we came up with a molarity of 2.67 M.
Explanation:To calculate the molarity of the solution, we need to know the number of moles of sucrose and the volume of the solution in liters. Given that the molecular weight of sucrose (C12H22O11) is approximately 342 g/mol, the number of moles of sucrose in 211.4 g can be calculated as (211.4 g) / (342 g/mol) = 0.618 moles.
To get the volume of the solution, first calculate the total mass of the solution which is mass of water + mass of sucrose = 100 g + 211.4 g = 311.4 g. Since the density of the solution is provided as 1.34 g/ml, convert this to kg/L to get the volume. So, 311.4 g / 1.34 g/ml = 232.39 ml or 0.232 L.
Finally, we calculate the molarity which is moles/volume in liters. So, molarity = 0.618 moles / 0.232 L = 2.67 M.
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Answer:
2.66 M
from ut quest
In the lewis structure of ch3oh, how many bonding pairs of electrons are there?
the number of atoms bonded to the central atom is 4
Explanation:
What is the ground-state electron configuration for the element cobalt (z = 27)?
The ground-state electron configuration for cobalt (Z = 27) is [Ar] 3d7 4s2.
Explanation:The ground-state electron configuration for the element cobalt (Co), which has an atomic number (Z) of 27, is as follows:
Co: [Ar] 3d74s2
This configuration indicates that cobalt has 2 electrons in the outermost 4s subshell and 7 electrons in the 3d subshell after the [Ar] noble gas core.
The ground-state electron configuration for the element cobalt (Z = 27) is 1s²2s²2p⁶3s²3p⁶4s²3d⁷. Co has 27 protons, 27 electrons, and 33 neutrons. The electron configuration represents the distribution of electrons in the energy levels and orbitals of an atom.
A stock solution of naoh is 0.800 m. what volume of this solution (in ml) is needed to prepare 2.00 l of 0.342 m naoh?
Question 7 what is the volume of 28.0 g of nitrogen gas at stp? 44.8 l 33.6 l 11.2 l 22.4 l none of the above
Find the new concentration of a solution if 25.0 mL of water is added to 125.0 mL of 0.150 M NaCl solution. What is the final volume? mL
Answer:
150.0 is the final volume
next question is a
and the next is 0.125 M
Explanation:
got it right in edge in 2020
To calculate the new concentration of NaCl solution after dilution, the final volume of the solution is determined to be 150.0 mL by adding the initial volume and the volume of water added. Using the molarity equation, the new concentration is found to be 0.125 M.
Explanation:To find the new concentration of the NaCl solution after dilution, we use the concept of molarity and the principle of conservation of mass. The amount of solute (NaCl in this case) stays the same, but the volume increases when more solvent (water) is added.
The initial molarity (Mi) of the NaCl solution is 0.150 M and the initial volume (Vi) is 125.0 mL. The volume of water added is 25.0 mL.
The final volume (Vf) is the sum of the initial volume and the volume of water added:
Vf = Vi + volume of water added
Vf = 125.0 mL + 25.0 mL
Vf = 150.0 mL
The new concentration Cf can be calculated using the formula:
Ci × Vi = Cf × Vf
(0.150 M) × (125.0 mL) = Cf × (150.0 mL)
Cf = (0.150 M × 125.0 mL) / 150.0 mL
Cf = 0.125 M
The final volume of the solution is 150.0 mL.
A 3.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific test site. The half-life of the radioactive element is 28 years. How much of the original sample will remain in the year 2030? Choose the closest.
DO NOT GUESS, ONLY COMMENT IF YOU KNOW
Answer : The correct answer for amount of radioisotope remain in 2030 is 0.619 g .
Radioactive Decay is emission of radiations ( in form of alpha , beta particle etc ) by unstable atom .
Radioactive decay is FIRST ORDER reaction . So , the equation of first order can be used to find decay constant , amount of radioisotopes or half life .
The equation for radioactive decay is given as :
[tex] ln (\frac{N}{N_0}) = -k * t [/tex]
Where : N = amount of radioisotope at time t
N₀ = amount of radioisotope initially present
k = decay constant t = time
Half life :
It is time when amount of radioisotope decrease to 50 % of its original amount . Half life [tex] (T_\frac{1}{2} ) [/tex] and decay constant can be related :
[tex] T_\frac{1}{2} = \frac{ln 2 }{k} = \frac{0.693}{k} [/tex]
Following are the steps can be used to determine amount of radioisotope (N) :
1) To find decay constant :
Given : [tex] (T_\frac{1}{2} ) [/tex] = 28 yrs
Decay constant can be calculated using half life by plugging value in half life formula :
[tex] 28 yrs = \frac{0.693}{k} [/tex]
On multiplying both side by k
[tex] 28 yrs * k= \frac{0.693}{k} *k [/tex]
On dividing both side by 28 yrs
[tex] \frac{28 yrs * k}{28 yrs} = \frac{0.693}{28 yrs} [/tex]
k = 0.02475 yrs⁻¹
2) To find amount of radioisotope (N):
Given : Amount of radioisotope originally present = 3.5 g
Time = 2030 - 1960 = 70 yrs
decay constant = 0.02475 yrs⁻¹
Amount of radioisotope (N) = ?
Plugging these values in the formula as:
[tex] ln (\frac{N}{3.5 } ) = - 0.02475 yrs^-^1 * 70 yrs [/tex]
[tex] ln (\frac{N}{3.5 } ) = - 1.7325 [/tex]
[tex] ln\frac{N}{No} [/tex] can be converted using the formula ( [tex] ln (\frac{a}{b} ) = ln a - ln b [/tex] )
ln N - ln (3.5 ) = - 1.7325
(ln 3.5 = 1.253 )
ln N -1.253 = -1.7325
Adding both side 1.253
ln N -1.253 + 1.253 = -1.7325 + 1.253
ln N = -0.4795
Taking anti ln of -0.4795
N = 0.619 g
Hence amount of radioisotope remained in 2030 is 0.619 g
Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is 1.86°c/m. remember to include the value of i.
Answer : The freezing point of a solution is, [tex]0.32^oC[/tex]
Explanation :
First we have to calculate the Van't Hoff factor (i) for [tex]NaNO_3[/tex].
The dissociation of [tex]NaNO_3[/tex] will be,
[tex]NaNO_3\rightarrow Na^++NO_3^-[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]Na^++NO_3^-[/tex] = 1 + 1 = 2
Now we have to calculate the freezing point of a solution.
Formula used for lowering in freezing point :
[tex]\Delta T_f=i\times k_f\times m[/tex]
or,
[tex]T_f^o-T_f=i\times k_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]T_f[/tex] = temperature of solution = ?
[tex]T^o_f[/tex] = temperature of pure water = [tex]0^oC[/tex]
[tex]k_f[/tex] = freezing point constant = [tex]1.86^oC/m[/tex]
m = molality = 0.08500 m
i = Van't Hoff factor = 2
Now put all the given values in this formula, we get the freezing point of a solution.
[tex]0^oC-T_f=2\times (1.86^oC/m)\times 0.08500m[/tex]
[tex]T_f=0.32^oC[/tex]
Therefore, the freezing point of a solution is, [tex]0.32^oC[/tex]
what is the molarity of 10.0 g of kcl in 0.500 l of solution
Define atom. list the subatomic particles, describe their relative masses,charges, and positions in the atom.
which of the following is a balanced chemical equation for the reaction of magnesium with nitrogen gas to form magnesium nitride
Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio of the mass of 14 6c to the total mass of carbon in the atmosphere is r, and the decay constant of 14 6c is λ. assume that, at any time, 14 6c is a negligible fraction of the total mass of carbon and that the measured activity of the sample is purely due to 14 6c. also assume that the ratio of mass of 14 6c to total carbon mass in the atmosphere (the source of the carbon in the sample) is the same at present and on the day when the number of 14 6c atoms in the sample was set. express your answer in terms of the mass ma of a 14 6c atom, mc, a, r, and λ.
the age of the sample (t) can be expressed as:** t = (1 / λ) * ln(a / r)
**1. Relate Activity and Initial Abundance:**
The activity (a) of the sample is proportional to the current number of ¹⁴⁶C atoms in the sample (N) relative to the initial number of ¹⁴⁶C atoms (N₀) at the time the sample was isolated from the atmosphere:
a ∝ N / N₀
**2. Relate Number of Atoms to Mass:**
The number of ¹⁴⁶C atoms (N) is related to the total mass of carbon (mc) and the mass of a single ¹⁴⁶C atom (ma) through Avogadro's constant (Na):
N = (mc / ma) * Na
**3. Consider Decay and Substitute:**
Over time, ¹⁴⁶C atoms decay with a decay constant (λ). We can express the change in the number of ¹⁴⁶C atoms using the radioactive decay equation:
dN/dt = -λN
Since we're looking for the age (t), we can integrate this equation to find the relationship between N and t:
N(t) = N₀ * e^(-λt) // (Equation 1)
Substitute N₀ with the initial number of ¹⁴⁶C atoms proportional to the initial ¹⁴⁶C/¹²C ratio in the atmosphere (r * mc / ma * Na):
N(t) = (r * mc / ma * Na) * e^(-λt) // (Equation 2)
**4. Relate Activity and Current ¹⁴⁶C/¹²C Ratio:**
The current activity (a) is also proportional to the current ¹⁴⁶C/¹²C ratio in the sample:
a ∝ (mc / ma * Na) * [N(t) / mc]
Substitute N(t) from Equation 2:
a ∝ (mc / ma * Na) * [ (r * mc / ma * Na) * e^(-λt) ] / mc
**5. Solve for Age (t):**
Cancel common factors and simplify:
a ∝ r * e^(-λt)
Taking the natural logarithm of both sides:
ln(a/r) = -λt
The final expression for the age of the sample [tex]t[/tex] in terms of the given variables is [tex]t = -\frac{1}{\lambda} \ln \left( \frac{a}{\lambda \cdot \frac{m_c \cdot r}{m_a}} \right)[/tex]
Step 1: Understanding Activity
The activity [tex]A[/tex] of a radioactive sample is related to the number of radioactive nuclei present [tex]N[/tex] by the equation:
[tex]A = \lambda N[/tex]
Here, [tex]\lambda[/tex] is the decay constant and represents the probability of decay per unit time.
Step 2: Relating Activity to Carbon-14
Given that the measured activity [tex]a[/tex] of the sample is purely due to [tex]^{14}C[/tex] and can be expressed as:
[tex]a = \lambda N_t[/tex]
Where [tex]N_t[/tex] is the current number of [tex]^{14}C[/tex] atoms in the sample.
Step 3: Finding Original Amount of Carbon-14
If we denote the original number of [tex]^{14}C[/tex] atoms in the sample as [tex]N_0[/tex], this can be related to the total mass of carbon in the sample and the ratio [tex]r[/tex] as follows:
[tex]N_0 = \frac{m_c \cdot r}{m_a}[/tex]
Where [tex]m_a[/tex] is the mass of a single [tex]^{14}C[/tex] atom.
Step 4: Calculating Age
The relationship between current [tex]N_t[/tex], original [tex]N_0[/tex], and time [tex]t[/tex] can be expressed using the formula for exponential decay:
[tex]N_t = N_0 e^{-\lambda t}[/tex]
Thus, rearranging this expression, we find:
[tex]t = -\frac{1}{\lambda} \ln \left( \frac{N_t}{N_0} \right)[/tex]
Step 5: Substituting Values
Substituting in our earlier expressions:
[tex]t = -\frac{1}{\lambda} \ln \left( \frac{N_t}{\frac{m_c \cdot r}{m_a}} \right)[/tex]
This simplifies further to:
[tex]t = -\frac{1}{\lambda} \left( \ln(N_t) - \ln\left(\frac{m_c \cdot r}{m_a}\right) \right)[/tex]
Thus, we can express [tex]t[/tex] as:
[tex]t = -\frac{1}{\lambda} \ln(N_t) + \frac{1}{\lambda} \ln\left(\frac{m_c \cdot r}{m_a}\right)[/tex]
[tex]t = -\frac{1}{\lambda} \ln \left( \frac{a}{\lambda \cdot \frac{m_c \cdot r}{m_a}} \right)[/tex]