Answer:
Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.
Answer:
10.77%
Explanation:
Molar mass of Cu = mass deposited/number of moles of Cu
Molar mass of Cu = 0.4391 g/6.238x10^-3 moles
Molar mass of Cu = 70.391 g/mol
%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100
%error = 10.77%
cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Percentage remaining after 60 days =?Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Number of half-lives (n) =?
n = t / t½
n = 60 / 9.7Finally, we shall determine the percentage remaining. This can be obtained as follow:
Let the original amount be N₀
Let the amount remaining be N
Number of half-lives (n) = 60 / 9.7
N = N₀ / 2ⁿ
Divide both side by N₀
N/N₀ = 1/2ⁿ
N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾
N/N₀ = 0.0137
Multiply by 100 to express in percentage
N/N₀ = 0.0137 × 100
N/N₀ = 1.37%Therefore, the percentage remaining after 60 days is 1.37%
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Calculate [H3O+] for pH 1.86. Steps please.
Answer:
[H₃O⁺] = [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]
Explanation:
By definition pH = -log[H⁺] => [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]
Using your calculator ... I am using a TI-30XA scientific calculator.
=> start by entering the number 1.86 => then press the (+/-) function => this will insert a negative symbol => -1.86,=> next find button with "2nd" printed on face (on some calculators the button is in yellow); press this button to change to 'secondary mode',=> find the symbol (10ˣ) ... the button below this symbol is usually the 'log' button, then press it => the answer of interest will show in the display window. => ...Depending on the calculator, the answer may show as 0.0138, or 1.38x10⁻², or 1.38E-2 (=1.38 x 10⁻²). It is the user's job to insert dimensional units into answer of interest => 0.0138M, or 1.38 x 10⁻²M, or 1.38E-2M.
1.38E-2 which is 1.38 x 10⁻².
The standard enthalpies of combustion of fumaric acid and maleic acid (to form carbon dioxide and water) are - 1336.0 kJ moJ-1 and - 1359.2 kJ moJ-1, respectively. Calculate the enthalpy of the following isomerization process:
maleic acid ----> fumaric acid
Answer:
Explanation:
maleic acid ⇒ fumaric acid
ΔHreaction = ΔHproduct - ΔHreactant
ΔHproduct = -1336.0 kJ mol⁻¹
ΔHreactant = - 1359.2 kJ mol⁻¹.
ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)
= 1359.2 kJ mol⁻¹ -1336.0 kJ mol⁻¹
= 23.2 kJ mol⁻¹ .
Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.
what are the properety of covalent bond
Explanation:
1. boiling and melting point
2. electrical conductivity
3. Bond strength
4. bond length
A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.
Starting from (R)-3-methylhex-1-yne as the substrate at the center of your page, draw a reaction map showing the regiochemical and stereochemical outcome or outcomes for each of the following series of reagents. Name each of your products, including stereochemical designations for any chirality centers that are generated.
a. HgSO4, H2SO4, H2O
b. 1. 9-BBN; 2. H2O2, NaOH
c. Br2, CCl4
d. HBr
Solution :
A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.
Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.
In the context, the products including the stereochemical designations for any chirality centers starting from the (R)-3-methylhex-1-yne as the substrate are attached below.
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationWhich subshells are found in each of the following shells
electron subshell - M shell
Answer:
3
Explanation:
The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.
As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Hence, M shell contains s,p and d subshells.
Which phenomenon explained below is an example of deposition?
Select the correct answer below:
A) Hail is formed from water droplets lifted by air currents to an altitude where they turn into pellets of ice.
B) Frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
C) In the winter, the top few inches of a pond turn to ice.
D) The visible cloud arising from a boiling tea kettle is not actually steam, but droplets of liquid water that form as the
steam cools in the air.
Answer:
b
Explanation:
deposition is when water turns from gas to solid. b is the only one that fits
Deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
What is deposition?Deposition is a process that involves collection of large mass or when mean distance between molecules are reduced. It can also be explained as gathering of substances together to form a larger mass.
Therefore, the phenomenon explained in the given example about deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
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How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide
Answer:
MoO2
Explanation:
The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.
To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.
The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:
Moles Mo2O3 -Molar mass: 239.9g/mol-
12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo
Mass Mo -95.95g/mol-:
0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo
Mass oxygen in the oxide:
13.197 - 9.895g = 3.302g Oxygen
Moles oxygen -Molar mass: 16g/mol-:
3.302g Oxygen * (1mol / 16g) = 0.206 moles O
Now, the ratio of moles O / moles Mo is:
0.206 moles O / 0.1031 moles Mo = 2
That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:
MoO2calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume
Answer :
volume of a gas = weight * 22.4 l / gram molecular weight
volume of o2 = ?
weight given = 20.5 g
gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )
volume of oxygen = 20.5 * 22.4 / 32
volume of oxygen = 14.35 liters
Explanation:
hope this helps you
if wrong just correct me
here is the question
Answer:
1. Nitrate ions, NaNO3 - Sodium nitrate.
2. Sulphide ions, K2S - Potassium sulphide.
3. Sulphate ions, CaSO4 - Calcium sulphate.
4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.
5. Carbonate ions, CaCO3 - Calcium carbonate.
6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.
7. Phosphite ions, PH3 - Hydrogen phosphite.
8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).
9. Ethanoate ions, CH3COONa - Sodium ethanoate.
10. Methanoate ions, HCOONa - Sodium methanoate.
11. Fluoride ions, HF - Hydrogen fluoride.
12. Chloride ions, KCl - Potassium chloride.
13. Bromide ions, HBr - Hydrogen bromide.
14. Iodide ions, NaI - Sodium iodide.
15. Phosphate ions, K3PO3 - potassium phosphate.
A gas at 273K temperature has a pressure of 590 MM Hg. What will be the pressure if you change the temperature to 273K? 
Explanation:
here's the answer to your question
Determine the effect each given mutation would have on the rate of glycolysis in muscle cells.
a. loss of binding site for fructose 1 ,6-bisphophate in pyruvate kinase.
b. loss of allosteric binding site for ATP in pyruvate kinase.
c. loss of allosteric binding site for AMP in phosphofructokinase.
d. loss of regulatory binding site for ATP in phosphofructokinase.
1. Increase
2. decrease
3. No effect
Answer:
a. Decrease
b. Increase
c. Increase
d. No effect
Explanation:
Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.
a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease
b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect
c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase
d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase
A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.
B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.
C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.
D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.
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Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
Which substance has the highest boiling point?
Select the correct answer below:
A) ethyl ether
B) ethyl alcohol
C) water
D) these all have the same boiling point
Answer:
D) these all have the same boiling point
what is valency of an atom?
The number of replaceable electrons in an atom is called its valency.
Examples
Monovalent - HydrogenDivalent - OxygenValency = 8 - Number of electron in last shell [When number of electrons in last shell > 4]Valency = Number of electron in last shell [When number of electrons in last shell < 4]Thanks !
☺️☺️☺️☺️☺️☺️☺️
Answer:
the combining capacity if an atom is know as valency.
the property of an element that determines the number of other atimd with an aton if the element can combine.
Cho biết độ tan của NH4Cl trong nước ở 20oC và 70oC lần lượt là 37,2 g/100 gam nước và 60,2 gam/100 g nước. Hòa tan 166,8 gam NH4Cl vào 400 gam nước ở 70oC thu được dung dịch X. Sau đó, hạ nhiệt độ dung dịch X xuống 20oC. Tính khối lượng (gam) NH4Cl kết tinh lại trong X?
Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C
Actini(III) hydroxide Ac(OH)3 0,0022
Amonia NH3 1176 900 702 565 428 333 188
Amoni azua NH4N3 16 25,3 37,1
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hehe
1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.
The structures are shown in the image attached.
A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.
Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.
I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2) and tetrahydrofuran (image 3).
All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.
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11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
how many moles of oxygen atoms are present in 0.4 moles of oxygen gas
Answer:
Each molecule of O2 is made up of 2 oxygen atoms. So 1 mole of O2 molecules is made up of 2 moles of oxygen atoms. Therefore 1 mole of oxygen gas contains 2 moles of oxygen atoms. And 0.4 moles of oxygen gas contains 0.8 moles of oxygen atoms.
There are 0.8 moles of oxygen atoms in 0.4 moles of oxygen gas.
Oxygen gas (O₂) consists of two oxygen atoms bonded together. Therefore, to determine the number of moles of oxygen atoms present in a given amount of oxygen gas, we can simply multiply the number of moles of oxygen gas by the number of oxygen atoms per molecule, which is 2.
Given that we have 0.4 moles of oxygen gas, we can calculate the number of moles of oxygen atoms as follows:
Number of moles of oxygen atoms = Number of moles of oxygen gas × Number of oxygen atoms per molecule
= 0.4 moles × 2
= 0.8 moles
Therefore, there are 0.8 moles of oxygen atoms present in 0.4 moles of oxygen gas.
This calculation is based on the stoichiometry of oxygen gas, which indicates that each molecule of O₂ contains two oxygen atoms. By considering the mole ratio between oxygen gas and oxygen atoms, we can determine the number of moles of oxygen atoms in a given quantity of oxygen gas.
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state function and non state function
Answer:
State functions represent quantities or properties of a thermodynamic system, while non-state functions represent a process during which the state functions change. For example, the state function PV is proportional to the internal energy of an ideal gas, but the work W is the amount of energy transferred as the system performs work.
Explanation:
How do I do this? What are the answers to the 5 questions shown?
Answer:
1. C₃H₆O₃
2. C₆H₁₂
3. C₆H₂₄O₆
4. C₆H₆
5. N₂O₄
Explanation:
1. Determination of the molecular formula.
Empirical formula => CH₂O
Mass of compound = 90 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂O]ₙ = 90
[12 + (2×1) + 16]n = 90
[12 + 2 + 16]n = 90
30n = 90
Divide both side by 30
n = 90/30
n = 3
Molecular formula = [CH₂O]ₙ
Molecular formula = [CH₂O]₃
Molecular formula = C₃H₆O₃
2. Determination of the molecular formula.
Empirical formula => CH₂
Mass of compound = 84 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂]ₙ = 84
[12 + (2×1)]n = 84
[12 + 2]n = 84
14n = 84
Divide both side by 14
n = 84/14
n = 6
Molecular formula = [CH₂]ₙ
Molecular formula = [CH₂]₆
Molecular formula = C₆H₁₂
3. Determination of the molecular formula.
Empirical formula => CH₄O
Mass of compound = 192 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₄O]ₙ = 192
[12 + (4×1) + 16]n = 192
[12 + 4 + 16]n = 192
32n = 192
Divide both side by 32
n = 192/32
n = 6
Molecular formula = [CH₄O]ₙ
Molecular formula = [CH₄O]₆
Molecular formula = C₆H₂₄O₆
4. Determination of the molecular formula.
Empirical formula => CH
Mass of compound = 78 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH]ₙ = 78
[12 + 1]n = 78
13n = 78
Divide both side by 13
n = 78/13
n = 6
Molecular formula = [CH]ₙ
Molecular formula = [CH]₆
Molecular formula = C₆H₆
5. Determination of the molecular formula.
Empirical formula => NO₂
Mass of compound = 92 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[NO₂]ₙ = 92
[14 + (2×16)]n = 92
[14 + 32]n = 92
46n = 92
Divide both side by 46
n = 92/46
n = 2
Molecular formula = [NO₂]ₙ
Molecular formula = [NO₂]₂
Molecular formula = N₂O₄
A skydiver slows down from 65 m/s to 5 m/s by opening the parachute. If this
takes 0.75 seconds, what is the skydiver's acceleration?
A. 45 m/s2 up
B. 80 m/s2 up
C. 45 m/s2 down
D. 80 m/s2 down
Answer:
D. -80m/s^2
Explanation:
V = u + at
5 = 65 + a (0.75)
0.75a = -60
a = -60/0.75
a = -80m/s^2
Therefore, is decelerating at 80m/s^2
Answer:
[tex]\boxed {\boxed {\sf D. \ 80 \ m/s^2 \ down}}[/tex]
Explanation:
We are asked to find the acceleration of a skydiver. Acceleration is the change in velocity over the change in time, so the formula for calculating acceleration is:
[tex]a= \frac{v_f-v_i}{t}[/tex]
The skydiver was initially traveling 65 meters per second, then he slowed down to a final velocity of 5 meters per second. He slowed down in 0.75 seconds.
[tex]\bullet \ v_f = 5 \ m/s \\\bullet \ v_i= 65 \ m/s \\\bullet \ t= 0.75 \ s[/tex]
Substitute the values into the formula.
[tex]a= \frac{ 5 \ m/s - 65 \ m/s}{0.75 \ s}[/tex]
Solve the numerator.
[tex]a= \frac{-60 \ m/s}{0.75 \ s}[/tex]
Divide.
[tex]a= -80 \ m/s^2[/tex]
The acceleration of the skydiver is -80 meters per second squared or 80 meters per second squared down. The skydiver is slowing down or decelerating, so the acceleration is negative or down.
What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
At a constant temperature, a sample of gas occupies 1.5 L at a pressure of 2.8 ATM. What will be the pressure of this sample, in atmospheres, if the new volume is 0.92 L?
Using boyles law
[tex]\boxed{\sf v\propto \dfrac{1}{p}}[/tex]
[tex]\\ \sf\longmapsto P_1V_1=P_2V_2[/tex]
[tex]\\ \sf\longmapsto P_2=\dfrac{P_1V_1}{V_2}[/tex]
[tex]\\ \sf\longmapsto P_2=\dfrac{2.8\times 1.5}{0.92}[/tex]
[tex]\\ \sf\longmapsto P_2=\dfrac{4.2}{0.92}[/tex]
[tex]\\ \sf\longmapsto P_2=4.56atm[/tex]
[tex]\\ \sf\longmapsto P_2\approx 4.6atm[/tex]
Answer:
[tex]\boxed {\boxed {\sf 4.6 \ atm}}[/tex]
Explanation:
We are asked to find the new pressure given a change in volume. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:
[tex]P_1V_1= P_2V_2[/tex]
Initially, the gas occupies 1.5 liters at a pressure of 2.8 atmospheres.
[tex]1.5 \ L * 2.8 \ atm = P_2V_2[/tex]
The volume is changed to 0.92 liters, but the pressure is unknown.
[tex]1.5 \ L * 2.8 \ atm = P_2* 0.92 \ L[/tex]
We are solving for the final pressure, so we must isolate the variable P₂. It is being multiplied by 0.92 liters. The inverse operation of multiplication is division, so we divide both sides by 0.92 L.
[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L} = \frac{P_2* 0.92 \ L}{0.92 \ L}[/tex]
[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L}= P_2[/tex]
The units of liters cancel each other out.
[tex]\frac {1.5 * 2.8 \ atm}{0.92 }=P_2[/tex]
[tex]\frac {4.2}{0.92} \ atm= P_2[/tex]
[tex]4.565217391 \ atm = P_2[/tex]
The original measurements of pressure and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 5 up to a 6.
[tex]4.6 \ atm \approx P_2[/tex]
The pressure is approximately 4.6 atmospheres.
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?
Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
