Answer:
NH₂O
Explanation:
Given compound:
N₄H₈O₄
Unknown:
The empirical formula of the compound = ?
Solution:
The empirical formula of a compound is its simplest formula. It expresses the composition of a the compound in the simplest whole ratio of atoms of the different elements present in the compound.
For the given compound:
N₄H₈O₄
Number of moles of N = 4
H = 8
O = 4
the highest common factor is 4 and we simply divide through by this number;
N = 1
H = 2
O = 1
So, the empirical formula of compound is NH₂O
Nicotinic acid, HC6H4NO2, is a B vitamin. It is also a weak acid with Ka = 1.4 × 10-5. Calculate [H+] and the pH of a 0.041 M solution of HC6H4NO2.
Answer:
[H+] = 7.576x10⁻⁴M
pH = 3.12
Explanation:
Based on the equilibrium of the nicotinic acid in water:
HC6H4NO2(aq) + H2O(l) ⇄ C6H4NO2-(aq) + H3O+(aq)
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
As both C6H4NO2-(aq) and H3O+(aq) comes from the same equilibrium, we can approximate their concentration as X and replace:
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
1.4x10⁻⁵ = [X] [X] / [0.041M]
5.47x10⁻⁷ = X²
7.576x10⁻⁴M = X = [H+]And as pH is defined as -log [H⁺]
pH = 3.12The solubility of limestone, CaCO3, at 25˚C is 0.00067 g/100 mL. Write the chemical equation for the solubility equilibrium of this sparingly soluble salt in water. Then compute the molar solubility and the solubility product constant Ksp for CaCO3 at 25˚C.
Answer:
4.5 × 10⁻⁹
Explanation:
Step 1: Write the reaction for the solution of CaCO₃
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Step 2: Convert the solubility of CaCO₃ from g/L to mol/L
We will use the following conversion factors:
The molar mass of CaCO₃ is 100.09 g/mol.1 L= 1000 mL.There are 0.00067 g of CaCO₃ per 100 mL of solution.[tex]\frac{0.00067 gCaCO_3}{100mLSol} \times \frac{1molCaCO_3}{100.09gCaCO_3} \times \frac{1000mLSol}{1LSol} = 6.7 \times 10^{-5} M[/tex]
Step 3: Calculate the solubility product constant (Ksp)
To relate Ksp and the molar solubility (S), we need to make an ICE chart.
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
Ksp = [Ca²⁺].[CO₃²⁻] = S² = (6.7 × 10⁻⁵)² = 4.5 × 10⁻⁹
Someone plz hello me ASAP it would be appreciated on question 7 btw
Answer:
Kinetic Energy is the correct Answer.
Explanation:
At the highest point on the roller coaster (assuming it has no velocity), the object has a maximum quantity of gravitational potential energy. As the object begins moving down to the bottom, its gravitational potential energy begins to decrease and the Kinetic Energy starts to increase.
Which phase change results in an increase in entropy?
1.
12(9) - 12(s)
2.
CH2(g) - CH40
3.
Br2(1) - Br2(g)
4.
H20(1) - H20(s)
The phase change that results in an increase in entropy is the change; Br2(1) - Br2(g)
What is entropy?The term entropy refers to the degree of disorderliness in a system. We have to note that gases have a higher entropy than liquids and liquids have a higher entropy than the solids.
As such, the phase change that results in an increase in entropy is the change; Br2(1) - Br2(g)
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What is the value of E0 for the spontaneous reaction resulting from a suitable combination of these half reactions?
Standard Potentials E0
Fe2+ --> Fe3+ + e- -0.77 V
2Hg --> Hg2+ + 2e- -0.79 V
a. +0.02 V.
b. +0.36 V.
c. -1.56 V.
d. -0.02 V.
e. +1.56 V.
Answer:
+0.02 V
Explanation:
For the reaction to be spontaneous, iron must be the cathode and Mercury the anode
Given that the standard cell potential is obtained from;
E°=E° cathode - E° anode
Since;
E° cathode = -0.77V
E° anode = 0.79 V
E°cell = -0.77 -(-0.79)
E°cell = +0.02 V
pentane or 2,2,3- trimethylhexane has the higher boiling point. why?
Answer:
2,2,3- trimethylhexane because it has more carbon atoms than pentane.
Explanation:
Hello.
In this case, since the physical properties of organic compounds are intensified as the number of carbon atoms start increasing on it, the larger the amount of carbon atom, the higher the boiling point since more energy is required to allow the liquid-phase molecules to transcend to the vapor-phase.
In such a way, since pentane has five carbon atoms and 2,2,3- trimethylhexane has nine carbon atoms, 2,2,3- trimethylhexane has the highest boiling point.
Best regards.
Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6% Write the empirical chemical formula of X.
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
Answer:
The empirical formula of X is C3H4O3.
Explanation:
What happen when a piece of sodium is exposed in air?
Answer:
I literally got this from google.
Explanation:
In ordinary air, sodium metal reacts to form a sodium hydroxide film, which can rapidly absorb carbon dioxide from the air, forming sodium bicarbonate. ... In a comparatively dry atmosphere, sodium burns quietly, giving off a dense white caustic smoke, which can cause choking and coughing.
Happy almost Halloween! :)
Two elements in the same period have the same number of _____ _____ in their electron clouds.
Answer:
outer electrons
Explanation:
Answer:
Energy levels
Explanation:
SOMEONE PLZ HELP!!!!
Answer:
4.22mL
Explanation:
V=m/d
v= 18.45g/4.37g/mL
What should be the temperature of the solvent before adding it to the sample to be recrystallized?
Answer:
Near the boiling point of the solvent
Explanation:
The process of recrystallization is hinged on the fact that the amount of solute that can be dissolved by a solvent increases with temperature. The process involves creation of a solution by dissolving a solute in a solvent at or near its boiling point. At the boiling point of the solvent, the solute has a greater solubility in the solvent; not much volume of the hot solvent is required to dissolve the solute.
Before the solution is later cooled, you can now filter out insoluble impurities from the hot solvent. The quantity of the original solute drops appreciably because impurities have been removed. At this lower temperature, the solution becomes saturated and the solute can no longer be held in solution hence it forms pure crystals of solute, which can be recovered.
Recrystallization must be carried out using the proper solvent. The solute must be relatively insoluble in the solvent at room temperature but more soluble in the solvent at elevated temperature.
what might happen to variables in a science experiment that would lead to unusable results?
Answer:
ejeb094
Explanation:
nnb3neneie9eei rje
plz help answer both will mark brainest
1. Assume this experiment (after the extractions are complete) left you with 4 pure products (aspirin, acetaminophen and caffeine, and the binder). Your lab mate distracted you after you had labeled the binder and now you can’t remember which of the remaining 3 products is which. What would be a simple test that you can perform in the laboratory to distinguish between the 3 remaining solids isolated from your experiments?
Answer:
Explanation:
Out of aspirin, acetaminophen and caffeine, aspirin is an acid because it is acetyl saliciylic acid . Hence it can be tested with litmus paper .
acetaminophen contains phenolic functional group , hence it is a weak acid . It can be tested with any test with which phenol test are done, like with neutral solution of ferric chloride .
caffeine is weak basic substance . It can also be tested with the help of testing a basic substance .
What concentration of NO−3NO3− results when 897 mL897 mL of 0.497 M NaNO30.497 M NaNO3 is mixed with 813 mL813 mL of 0.341 M Ca(NO3)2?
Answer:
Explanation:
NaNO₃ = Na⁺ + NO₃⁻¹
.497 M .497 M
moles of NO₃⁻¹ = .897 x .497 = .4458 moles
Ca( NO₃)₂ = Ca + 2 NO₃⁻¹
.341 M 2 x .341 M = .682 M
moles of NO₃⁻¹ = .813 x .682 = .5544 moles
Total moles = .4458 moles + .5544 moles
= 1.0002 moles
volume of solution = 897 + 813 = 1710 mL
= 1.710 L
concentration of nitrate ion = 1.0002 / 1.710 M
= .585 M
if you wanted to measure in irregular object's volume, which device would you use?
Answer:
a beaker
Explanation:
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic acid has a pKa of 3.75. 2. Create approximately 20 mL of buffer solution with a pH of 4.25.
Answer:
15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.
Explanation:
To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid
4.25 = 3.75 + log [A⁻] / [HA]
0.5 = log [A⁻] / [HA]
3.162 = [A⁻] / [HA] (1)
As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:
0.10M * 20x10⁻³L =
2x10⁻³moles = [A⁻] + [HA] (2)
Replacing (2) in (1):
3.162 = 2x10⁻³moles - [HA] / [HA]
3.162 [HA] = 2x10⁻³moles - [HA]
4.162[HA] = 2x10⁻³moles
[HA] = 4.805x10⁻⁴ moles
[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles
That means, to create the buffer you must add:
[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =
15.2mL of the 0.10M sodium formate solution[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =
4.8mL of the 0.10M formic acid solutionThe Adálie penguins feed on krill that live on the underside of ice sheets. Which of the following would help Adálie penguins survive during the melting of Antarctic sea ice?
The correct answer is Migrating farther into the sea
Explanation:
The Adálie penguins rely on ice sheets to obtain food as the krill they feed on is only found underside of the ice. This implies as the Antarctic sea ice melts Adálie penguins need to look for new places to find krill as if there are no ice sheets there are no krill for these penguins. Due to this, it is likely these penguins are forced to migrate or move farther into the sea to survive because it is likely in these zones there are ice sheets and therefore places where they can find krill. Despite this, this process might lead to a significant decline in the population of penguins due to the competence for food and the process of migration itself. Thus, the one that would help Adáline penguins to survive during the melting of Antarctic sea ice is migrating farther into the sea.
Have a nice day! :)
The Adálie penguins feed on krill that live on the underside of ice sheets. Adálie penguins survive during the melting of Antarctic sea ice help migrating farther into the sea. Therefore, option A is correct.
What is migration ?The term migration is defined as the movement of either people or animals from one area to another.
The Adálie penguins rely on ice sheets for food because the krill they eat is only found on the underside of the ice. As a result, it is likely that these penguins are forced to migrate or move further into the sea to survive.
Regardless, this process may result in a significant decline in penguin population due to food competence and the migration process itself.Thus, the one that will assist Adáline penguins in surviving during the melting of Antarctic sea ice is migrating deeper into the sea.
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Your question is incomplete,most probably your question was
The Adálie penguins feed on krill that live on the underside of ice sheets. Which of the following would help Adálie penguins survive during the melting of Antarctic sea ice? Migrating farther into the sea Breeding during summer months Growing strong toe nails to grip ice Spending more time raising their young
Consider the diagram below.
What does C represent?
A) enthalpy of reaction
B) activation energy
C) activated complex
D) energy of the reactants
Answer:
A) enthalpy of reaction
Explanation:
The region C signifies the enthalpy of reaction.
This diagram is the energy profile of an endothermic reaction. In such reaction, heat is absorbed from the surrounding. At the end of the reaction, the heat of product is lesser than that of the reactants.
Enthalpy changes are heat changes accompanying a physical and chemical change. An enthalpy is the difference between the sum of the heat contents of products and sum of the heat contents of reactants.it is indeed A) enthalpy of reaction
3. A certain Chemical Industry company has a quality control job opening. The job is open for any major with basic knowledge of chemistry. You decided to apply. In the interview the HR personnel gives you a sealed folder from a certain lot to test your laboratory experience, as well as your quantitative and volumetric analysis skills. The chemical contained in the sample is benzoic acid (C-H602) and it is known to be a monoprotic acid. In order to get the job, you need to determine if the sample's purity is acceptable based on their standards. Inside the folder you found a vial with a solid sample labeled BA-I, a periodic table, and the following data: 1.250 g of the sample required 20.15 mL of 0.500 M concentration of NaOH to reach the end point. The lot can be denied if the purity is below 99.5 % purity.
1) What is the purity in the sample?
2) Is it the purity acceptable?
3) Would you repeat the titration experiment?
Answer:
1) 97.6%
2) No the purity is not acceptable because the standard is 99.5% purity.
3) Yes I will repeat the titration experiment to confirm my result.
Explanation:
Equation of the reaction;
C7H6O2(aq) + NaOH(aq) ---------> C7H5ONa(aq) + H2O(aq)
From the information provided;
Number of moles of NaOH reacted = concentration × volume = 20.15/1000 × 0.500 = 0.01 moles
From the reaction equation;
1 mole of C7H6O2 reacts with 1 mole of NaOH
Hence 0.01 moles of C7H6O2 will react with 0.01 moles of NaOH
Mass of C7H6O2 reacted = number of moles of C7H6O2 × molar mass of C7H6O2
Molar mass of C7H6O2 = 122.12 g/mol
Mass of C7H6O2 reacted = 0.01 moles × 122.12 g/mol = 1.22 g
Percentage by mass of pure C7H6O2 in the impure sample = 1.22/1.250 × 100 = 97.6 %
Use the Rydberg Equation to calculate the energy in Joules of the transition between n = 7 and n = 3 for the hydrogen atom. Find the frequency in Hz of this transition if the wavelength is 1000nm.
Answer:
The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].
The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].
Explanation:
The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.
Let [tex]\lambda_\text{vac}[/tex] denote the wavelength of the photon released when measured in vacuum.Let [tex]R_\text{H}[/tex] denote the Rydberg constant for hydrogen. [tex]R_\text{H} \approx 1.09678 \times 10^{7}\; \rm m^{-1}[/tex].Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the principal quantum number of the initial and final main energy level of that electron. (Both [tex]n_1\![/tex] and [tex]n_2\![/tex] should be positive integers; [tex]n_1 > n_2[/tex].)The Rydberg Equation gives the following relation:
[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].
Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:
[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].
In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:
[tex]\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}[/tex].
Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].
Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:
[tex]\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}[/tex].
Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].
Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:
[tex]\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}[/tex].
The energy of the transition between n = 7 and n = 3 is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.
Using the Rydberg Equation for energy;
ΔE = -RH(1/n^2final - 1/n^2initial)
Given that;
nfinal = 3
ninitial = 7
RH = 2.18 × 10^-18 J
ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)
ΔE = - 2.18 × 10^-18(0.11 - 0.02)
ΔE = - 1.96 × 10^-19 J
For the second part;
Since the wavelength is 1000nm, we have;
λ = 1000nm
c = 3 × 10^8 m/s
f = ?
c = λf
f = c/λ
f = 3 × 10^8 m/s/1000 × 10^-9 m
f = 3 × 10^8 m/s/ 1 × 10^-6 m
f = 3 × 10^14 Hz
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Which condition produces an anion?
A. more neutrons than electrons
B. more protons than electrons
C. more electrons than neutrons
D. more electrons than protons
Bromine is found above iodine in Group 17 of the periodic table. If an ion formed by bromine has a charge of 1-, what is the charge on an ion
formed by iodine?
A. -7
B. -2
C. -1
D. +1
Answer:
Explanation:
Iodine is the most flexible of the atoms in Group 17 of the periodic table. It can have more than 1 charge, especially when combined with oxygen. The answer that you are expected to give, I think is C. Iodine's dominant charge is - 1
Please help! I'm confused on a few of these, 100 points!
2. Determine the possible traits of the calves of : Da red (RR) bull is mated with a red (RR) cow 1 a red * (RR) bullis mated with a white (WW) Cow 2 Da roan * (RW) is mated with a red(RR)Cow 3 3. Illustrate your answers using a Punnett square. 4. Write your answers on the paper.
During a solar eclipse, which of the following is true?
HELP
Answer:
The moon blocks the Sun's light from hitting the surface of the earth
Explanation:
Which is one way that minerals crystallize from materials dissolved in water?
from the air
from solutions that evaporate
from hot water solutions when water boils
from the soil
Answer:
the second answer its science behind it
Answer:
b
Explanation:
Peroxyacylnitrate (PAN) is one of the components
of smog. It is a compound of C, H, N, and O.
Determine the percent composition of oxygen and
the empirical formula from the following percent
composition by mass: 19.8 percent C,
2.50 percent
H, 11.6 percent N. What is its molecular formula
given that its molar mass is about 120 g?
C – 19,9%, H – 2,2%, N – 11,6%, O – x%
[tex]M=120\frac{g}{mol}[/tex]
1 percentage
The entire molecule is 100% and all the components of the compound add up to 100%.
100% - 19,9% - 2,5% - 11,6% = 66,1%
The compound contains 66,1% oxygen.
2 molar masses
[tex]M_{C}=12,01\frac{g}{mol}[/tex]
[tex]M_{H}=1,008\frac{g}{mol}[/tex]
[tex]M_{O}=15,999\frac{g}{mol}[/tex]
[tex]M_{N}=14,007\frac{g}{mol}[/tex]
3 masses
The compound has a molar mass of 120g/mol. So one molecule weighs 120 g. To find out how much the percentage of a component weighs, you have to calculate it using the molar mass.
carbon
19,8% of 120g
[tex]m=120g*0,198\\m=23,76g[/tex]
One molecule contains 23,76g of carbon.
hydrogen
2,5% of 120g
[tex]m=120g*0,025\\m=3g[/tex]
One molecule contains 3g of hydrogen.
oxygen
66,1% of 120g
[tex]m=120g*0,661\\m=79,32g[/tex]
One molecule contains 79,32g of oxygen.
nitrogen
11,6% of 120g
[tex]m=120g*0,0,116\\m=13,92g[/tex]
One molecule contains 13,92g of nitrogen.
4 amount of substance
carbon
[tex]n=\frac{23,76g}{12,01\frac{g}{mol} }\\n=1,98mol[/tex]
The compound contains about 2 moles of carbon.
hydrogen
[tex]n=\frac{3g}{1\frac{g}{mol} }\\n=3mol[/tex]
The compound contains about 3 moles of hydrogen.
oxygen
[tex]n=\frac{79,32g}{15,999\frac{g}{mol} }\\n=4,96mol[/tex]
The compound contains about 5 moles of oxygen.
nitrogen
[tex]n=\frac{13,92g}{14,007\frac{g}{mol} }\\n=0,99mol[/tex]
The compound contains about 1 moles of nitrogen.
5. molecular formula
The formula results from the ratio of the amounts of substance.
[tex]n_{C} :n_{H} :n_{O} :n_{N} =2:3:5:1\\C_{2}H_{3}NO_{5}[/tex]
The molecular formula of the given compound is C₂H₃NO₅, and percent composition of oxygen in it is 66.1%.
How do we calculate mass from % composition?Mass of any composition of any compound will be calculated by using the below formula as:
Mass of component = (% composition)×(mass of compound) / 100
Given mass of compound = 120g/mol
Total composition of compound (100%) = Percent composition of all components
% composition of oxygen = 100 - (19.8 + 2.50 + 11.6) = 66.1%
Moles will be calculated as:
n = W/M, where
W = given mass
M = molar mass
For carbon atom:Mass of Carbon component = (0.198)(120g) = 23.76g
Moles of Carbon atom = 23.76g / 12.01g/mol = 1.98mol = 2 moles
For nitrogen atom:Mass of Nitrogen component = (0.116)(120g) = 13.92g
Moles of Nitrogen atom = 13.92g / 14.007g/mol = 0.99mol = 1 moles
For oxygen atom:Mass of Oxygen component = (0.661)(120g) = 79.32g
Moles of Oxygen atom = 79.32g / 15.99g/mol = 4.96mol = 5 moles
For hydrogen atom:Mass of Hydrogen component = (0.025)(120g) = 3g
Moles of Hydrogen atom = 3g / 1g/mol = 3 moles
So, the molecular formula of the compound on the basis of moles of given entities is C₂H₃NO₅.
Hence required molecular formula is C₂H₃NO₅.
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PLS PLS PLS PLS PLS PLS PLS help ASAP!!!!! Scientists call all of the compounds that contain carbon and are found in living things Organic because ________.
WILL DO BRAINLIEST!
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?
Answer:
3.13%.
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 0.15
Half life (t½) = 2 mins
Number of half-life (n) = 5
Fraction of sample remaining =.?
Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:
Amount remaining (N) = 1/2ⁿ × original amount (N₀)
NOTE: n is the number of half-life.
N = 1/2ⁿ × N₀
N = 1/2⁵ × 0.15
N = 1/32 × 0.15
N = 0.15/32
N = 4.69×10¯³
Therefore, 4.69×10¯³ is remaining after 5 half-life.
Finally, we shall the fraction of the sample remaining after 5 half-life as follow:
Original amount (N₀) = 0.15
Amount remaining (N) = 4.69×10¯³
Fraction remaining = N/N₀ × 100
Fraction remaining = 4.69×10¯³/0.15 × 100
Fraction remaining = 3.13%