Answer:
A) 577.6 years
B) 1580.9 years
Explanation:
You can know the order of a reaction given the units of the rate constant:
For a 0 order reaction are M/s=(mol/L*s)For a 1 order reaction are 1/sFor a 2 order reaction are 1/M*s=(L/mol*s)Then we know that this is a first order reaction because years is a unit of time as well as seconds. The half life of a first order reaction is given by:
[tex]t_{1/2}=\frac{ln(2)}{k}[/tex]
Here you can solve for the half life:
[tex]t_{1/2}=\frac{ln(2)}{0.0012 years^-1}=577.6 years[/tex]
Now for the rate law for a first order reaction is:
[tex][A]=[A_{o} ]*e^{-kt}[/tex]
Then, if you want to know how long does it take to reach a certain value you solve for time:
[tex]t=-ln([A]/[A]_{o}) *1/k[/tex]
The 15% of its original value is [tex]0.15*[A_{o}][/tex]
You solve for time:
[tex]t=-ln(0.15) *1/0.0012 years^-1=1580.9 years[/tex]
Hope it helps!
In the Decomposition reaction, a reactant gets split into two or more products. The half-life of the reaction is 577.6 years and the time taken to reach 15% of the original is 1580.9 years.
What is half-life?Half-life is the time taken by the radioactive isotope to decay one-half of its total amount.
From the units of the rate of the reaction, the order can be estimated as, for a first-order reaction unit is 1/s.
The half-life of the reaction can be given as,
[tex]\rm t\dfrac{1}{2} = \dfrac{\rm ln (2)}{k}[/tex]
The rate constant (k) for the decomposition reaction is 0.0012 per year.
Substituting values in the above equation:
[tex]\begin{aligned} \rm t\dfrac{1}{2} &= \dfrac{\rm ln (2)}{0.0012}\\\\&= 577.6\;\rm years\end{aligned}[/tex]
The rate law of the first-order reaction is given as,
[tex]\begin{aligned}\rm [A] = [A_{o}] \times e^{-kt}\\\\\rm t = -ln ([A][A_{o}]) \times \dfrac{1}{k}\end{aligned}[/tex]
Solving for time (t):
[tex]\begin{aligned}\rm t &= \rm -ln (0.15) \times \dfrac{1}{0.0012}\\\\& = 1580.9\;\rm years\end{aligned}[/tex]
Therefore, the half-life is 577.6 years and the time taken is 1580.9 years.
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Drag the following in order, starting with the largest particle (can be seen) on the top and ending with the smallest particle (cannot be seen) on the bottom.
atoms
four carbon atoms
nucleus of an atom
electron
Answer:
Explanation:
four carbon atoms ------ atoms ------------nucleus of an atom------------electron
Atoms are the smallest indivisible particle in any substances. But atoms are also made up of other tiny particles which are subatomic in sizes. These particles are protons, neutrons and electrons.
Protons and neutrons are called the nucleons of an atoms. They are massive particles found in the nucleus of an atom. The nucleus is a very small area but very dense.
Electrons are negatively charged subatomic particles. The bulk of the volume of the atom is occupied by electrons orbiting the nucleus.
Together, the electrons and the nucleons makes up the subatomic particles in an atom.
What volume of sample, in mL, would be needed to make 2 mL of a 5-fold dilution? (3 significant figures needed)
Answer:
0.400 mL
Explanation:
Hello, the dilution factor (in folds) is given by:
[tex]folds=\frac{total_{volume}}{sample_{volume}}[/tex]
Thus, the sample volume with three significant figures is given by:
[tex]sample_{volume}=\frac{2mL}{5"folds"}=0.400 mL[/tex]
Best regards.
A chemist prepares a solution of zinc oxalate by measuring out of zinc oxalate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's zinc oxalate solution. Round your answer to significant digits. 0.0075 umol 450 mL
The concentration of a 0.0075 umol of zinc oxalate in a 450 mL solution is 1.67 x [tex]10^{-8[/tex] mol/L or 0.0000000167 M.
Explanation:To calculate the concentration of the zinc oxalate solution, we first convert the given quantity of zinc oxalate from umol to mol.
Converting 0.0075 umol to mol gives us 0.0075 x [tex]10^{-6[/tex] mol, which is 7.5 x [tex]10^{-9[/tex] mol.
Molarity is defined as the number of moles of solute per liter of solution.
So, we also need to convert 450 mL to liters, giving us 0.45L.
The concentration (C) is then calculated by dividing the number of moles (n) by the volume (V) in liters.
So, the molarity of the zinc oxalate solution can be calculated as follows: C = n/V = 7.5 x [tex]10^{-9[/tex] mol / 0.45L which equals 1.67 x [tex]10^{-8[/tex] mol/L, or 0.0000000167 M.
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Explain why the properties of a polymer below the glass transition temperature are different from the same properties in the same polymer above the glass transition temperature.
Explanation:
At low temperatures (below the glass transition temperature) crystalline polymers are rigid like glass. This happens because all the polymer chains are perfectly arranged, all the polymer is in the crystalline form. On the other hand, when the temperature raises, upper the glass transition temperature, some polymer chains start to get loose and form some amorphous regions in between the crystalline regions of the polymer. This condition makes the polymer more flexible.
What information is provided by showing a molecule's stereochemistry?
Answer:
3-D shape of molecule and adjacent carbon atoms and their orientation.
Explanation:
Stereochemistry involves study of relative spatial positioning or arrangement of atoms which form structure of the molecules.
Stereochemistry studies focuses on the stereoisomers, which are the species which have same molecular formula but the sequence of the bonded atoms is different in the 3-D space of the atoms.
Thus,
Molecule's stereochemistry tells the 3-D shape of molecule and adjacent carbon atoms and their orientation.
Stereochemistry provides information about the three-dimensional structure of a molecule, including the arrangement of atoms and their spatial orientation.
Explanation:Stereochemistry is the study of the relative arrangement of atoms in molecules and their manipulation. It provides information about the three-dimensional structure of a molecule, including the arrangement of atoms and the spatial orientation of these atoms. By showing a molecule's stereochemistry, we can understand its shape, bonding patterns, and how it interacts with other molecules.
A molecule's stereochemistry can be represented using various models, such as ball-and-stick models, wedge-and-dash representations, or space-filling models. These models help visualize the three-dimensional structure of a molecule and show the arrangement of atoms in space.
A woolly rhinocerous skeleton (Coelodonta antiquitatis) found in Poland contains 1.30% of the C-14 found in living animals. How long ago did the organism die?
Answer:
The organism died 35900 years ago
Explanation:
Half life C-14 is 5730 years.For decay of radioactive nuclides: [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex][tex]\frac{N}{N_{0}}[/tex] is the fraction of radioactive nuclide remain present after t time and [tex]t_{\frac{1}{2}}[/tex] is the half-life of radioactive nuclideHere [tex]\frac{N}{N_{0}}[/tex] is 0.013 and [tex]t_{\frac{1}{2}}[/tex] is 5730 yerasPlug-in all the values in the above equation:
[tex]0.013=(\frac{1}{2})^{\frac{t}{5730years}}[/tex]
So, t = 35900 years
Hence the organism died 35900 years ago
A chemistry student needs 50.0 g of methyl acetate for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of methyl acetate is 0.934 g.cm . Calculate the volume of methyl acetate the student should pour out. Round your answer to 3 significant digits. x s ?
To obtain 50.0 g of methyl acetate, the student should measure out 53.5 cm³, using the density of methyl acetate which is 0.934 g/cm³.
Explanation:To calculate the volume of methyl acetate the student should pour out using its density, the formula density = mass/volume can be rearranged to volume = mass/density. Given that the density of methyl acetate is 0.934 g/cm³, and the student needs 50.0 g of methyl acetate, the volume can be calculated as follows:
volume = mass/density
volume = 50.0 g / 0.934 g/cm³
volume = 53.533 g/cm³
The student should measure out 53.5 cm³ of methyl acetate to obtain 50.0 g, rounding to 3 significant digits.
Calculate ΔS°for the combustion of ammonia.
4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l)
Substance NH3(g) O2(g) N2(g) H2O(l)
S°(J/K·mol) 192 205.1 192 70
-135 J
-579 J
-387 J
579 J
Answer: The [tex]\Delta S^o[/tex] of the reaction is [tex]-579JK^{-1}[/tex]
Explanation:
Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.
The equation representing entropy change of the reaction follows:
[tex]\Delta S_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}][/tex]
For the given chemical equation:
[tex]4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)[/tex]
We are given:
[tex]\Delta S^o_{NH_3}=192Jmol^{-1}K^{-1}\\\Delta S^o_{O_2}=205.1Jmol^{-1}K^{-1}\\\Delta S^o_{N_2}=192Jmol^{-1}K^{-1}\\\Delta S^o_{H_2O}=70Jmol^{-1}K^{-1}[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(6\times \Delta S^o_{H_2O})+(2\times \Delta S^o_{N_2})]-[(4\times \Delta S^o_{NH_3})+(3\times \Delta S^o_{O_2})][/tex]
[tex]\Delta S^o_{rxn}=[(6\times 70)+(2\times 192)]-[(4\times 192)+(3\times 205.1)]=-579JK^{-1}[/tex]
Hence, the [tex]\Delta S^o[/tex] of the reaction is [tex]-579JK^{-1}[/tex]
A
pure solvent freezes at 12.0 C. A solution of 0.980 g of the solute
and 13.870 g of solvent froze at 5.1 C. The molar mass of the
solute is 178.2 g/mol. Calculate the freezing point depression
constant, Kf for the solvent.
Answer:
[tex]K_{f}[/tex] for solvent is [tex]17^{0}\textrm{C}.kg.mol^{-1}[/tex]
Explanation:
Let's assume that the solute is non-volatile as well as non-electrolyte.For a solution with non-volatile solute and non-electrolyte solute-[tex]\Delta T_{f}=K_{f}.m[/tex], where [tex]\Delta T_{f}[/tex] is depression in freezing point and m is molality of solution
Molality of solution (m) = (moles of solute/mass of solvent in kg)
= [tex]\frac{\frac{0.980}{178.2}}{0.01387}mol/kg[/tex]
= 0.396 mol/kg
[tex]\Delta T_{f}=(12.0-5.1)^{0}\textrm{C}=6.9^{0}\textrm{C}[/tex]
So, [tex]K_{f}=\frac{\Delta T_{f}}{m}=\frac{6.9}{0.396}^{0}\textrm{C}.kg.mol^{-1}=17^{0}\textrm{C}.kg.mol^{-1}[/tex]
Which response gives the products of hydrolysis
ofNH4Cl?
A. NH4+ + HCl
B. NH3 + OH- + HCl
C. NH3 + H+
D. NH4OH + HCl
E. No hydrolysis occurs.
Answer : The correct option is, (C) NH₃ + H⁺
Explanation:
Hydrolysis : It is defined as the chemical reaction in which the breakdown of compound takes place due to reaction with water.
As per question:
First ammonium chloride completely dissociates into ion.
[tex]NH_4Cl(aq)\rightarrow NH_4^+(aq)+Cl^-(aq)[/tex]
Now ammonium ion react with water to give ammonia and hydronium or hydrogen ion.
The balanced hydrolysis reaction will be:
[tex]NH_4^++H_2O\rightarrow NH_3+H_3O^+[/tex]
Hence, the correct option is, (C) NH₃ + H⁺
Final answer:
The correct response to the products of hydrolysis of NH4Cl is 'NH3 + H+', because, during hydrolysis, NH4Cl separates into NH4+ and Cl- ions, with NH4+ reacting with water to form ammonia (NH3) and hydronium ions (H3O+). So the correct option is C.
Explanation:
The question asks which response gives the products of hydrolysis of NH4Cl. During hydrolysis, water is involved in breaking down a compound. In the case of NH4Cl, when it is dissolved in water, it separates into NH4+ ions and Cl− ions. The Chloride ion (Cl−) does not hydrolyze as it's the conjugate base of a strong acid (HCl) and has no significant basicity. On the other hand, the Ammonium ion (NH4+) is the conjugate acid of a weak base (NH3), and it will hydrolyze in water. The NH4+ accepts a hydroxide ion (OH−) from water, forming NH3 and H3O+ (hydronium ion). Therefore, the hydrolysis of NH4Cl will result in ammonia (NH3) and hydronium ions (H3O+).
Option C is the correct response: NH3 + H+. When NH4Cl hydrolyzes, it forms ammonia (NH3) and hydronium ions (H3O+), not hydrochloric acid (HCl). The correct formula of the products reflects the ammonia and hydronium ions formed.
The trash cans distributed by the city of Mobile are approximately 4 feet tall and have a square cross section with a side of approximately 30 inches. Assuming that the trash can is rectangular, approximate its capacity: 4. a. in gallons b. in metric tons of water at 4°C
Explanation:
Length of trash cans = l = 4 feet
Breadth of trash cans = b = 4 feet
Height of trash cans = h = 30 inches = 2.5 feet
1 inches = 0.0833333 feet
Capacity of trash can = Volume of the rectangular trash can = V
V = l × b × h
[tex]V=4 feet\times 4 feet \times 2.5 feet= 40 feet^3[/tex]
a) [tex]1 feet^3=7.48052 gallons[/tex]
[tex]V=40 feet^3=40\times 7.48052 gallons=299.221 gallons[/tex]
b) Density of water at 4°C = 1 kg /L
1 metric tonne of water = 264.17 gallons of water
[tex]V=299.221 gallons=\frac{299.221}{264.17}[/tex] metric tonne of water
[tex]V=1.1327[/tex] metric tonne of water
Write a chemical equation for the dissolution of the AgCl precipitate upon the addition of NH,(aq).
Explanation:
White precipitate of silver chloride get dissolves in excess ammonia to formation of complex between silver ions, chloride ions and ammonia molecules.
The chemical reaction is given as:
[tex]AgCl(s)+2NH_3(aq)\rightarrow Ag[(NH_3)_2]^+Cl^-(aq)[/tex]
When 1 mole of silver chloride is added to 2 mole of an aqueous ammonia it form coordination complex of diaaminesilver(I) chloride.
Final answer:
The chemical equation for the dissolution of AgCl precipitate with the addition of ammonia is [tex]AgCl(s) + 2NH_3(aq) < = > Ag(NH_3)_2+(aq) + Cl-(aq).[/tex] Ammonia interacts with [tex]Ag^+[/tex] to form a complex ion, increasing the solubility of AgCl significantly.
Explanation:
The dissolution of AgCl precipitate upon the addition of NH3(aq) is described by the following chemical equation:
[tex]AgCl(s) + 2NH_3(aq) \ < = > Ag(NH_3)_2+(aq) + Cl−(aq)[/tex]
When ammonia (NH3) is added to a solution containing AgCl, it reacts with the [tex]Ag^+[/tex] ions to form the complex ion [tex]Ag(NH_3)_2^+[/tex]. This acts to decrease the concentration of [tex]Ag^+[/tex] ions in solution. According to Le Chatelier's principle, the solubility of AgCl will increase to re-establish equilibrium, which leads to the dissolution of the AgCl precipitate.
The net effect of adding ammonia is a significant increase in the solubility of AgCl, as indicated by a change in the equilibrium constant from 1.8 x [tex]10^-10[/tex] in pure water to 3.0 x [tex]10^3[/tex] in the presence of dissolved ammonia.
Five different substances are given to you to be dissolved in water. Which substances are most likely to undergo dissolution in water? Check all that apply. View Available Hint(s) Check all that apply. lithium iodide, LiI heptane, C7H16 octane, C8H18 sodium fluoride, NaF potassium iodide, KI
Answer:
Lithium iodide, sodium fluoride and potassium iodide are more likely to undergo dissolution in water.
Explanation:
Water is a polar substance, which means that it will be a good solvent for other polar substances and salts.
LiI, NaF and KI are all salts that easily dissociate in water to produce ions. The ions will be surrounded by water molecules (solvatation) due to electrostatic interactions.
Heptane and octane are both non-polar substances that have no significant charges to interact with water molecules. Therefore, this substances will not dissolve in water.
Final answer:
Lithium iodide, sodium fluoride, and potassium iodide are ionic compounds that are most likely to dissolve in water, a polar solvent. Nonpolar substances like heptane and octane are less likely to dissolve in water.
Explanation:
When considering the solubility of substances in water, we must account for the polarity of both the solute and the solvent. Water is a polar solvent, which means it can easily dissolve other polar compounds and ionic compounds due to its ability to form ion-dipole interactions. Considering the substances provided, lithium iodide (LiI), sodium fluoride (NaF), and potassium iodide (KI) are ionic compounds and are most likely to undergo dissolution in water.
On the other hand, heptane ([tex]C_{7} H_{16}[/tex]) and octane ([tex]C_{8} H_{18}[/tex]) are hydrocarbons, which are nonpolar substances. Nonpolar substances do not dissolve well in polar solvents like water, thus making these substances less likely to dissolve in water. Instead, they would be more soluble in nonpolar solvents such as hexane or other hydrocarbons.
Magnesium has 12 protons. What charge do you expect it to have when it ionizes? Why?
Answer: This element has +2 charge on it.
Explanation:
An ion is formed when an atom looses or gains electrons.
When an atom looses electrons, it leads to the formation of positive ion known as cation.When an atom gains electrons, it leads to the formation of negative ion known as anion.Magnesium is the 12th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^2[/tex]
This element will loose 2 electrons to attain stable electronic configuration and leads to the formation of a cation having formula [tex]Mg^{2+}[/tex]
Hence, this element has +2 charge on it.
What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x10^1-7 J (A) 2.18x10- (B) 7.84x10- (C) 8.62x10-13 (D) 5.34x10-5
The kinetic energy acquired by the electron in a hydrogen atom when it absorbs a light radiation can be calculated using E = hf, where E is the energy of the radiation and f is the frequency of the radiation.
Explanation:When an electron in a hydrogen atom absorbs a light radiation, it gains kinetic energy. To calculate the energy gained, we can use the equation E = hf, where E is the energy of the radiation, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the radiation. In this case, the energy of the radiation is given as 1.08 x 10^-7 J.
Since the electron absorbs the radiation, we know that the energy gained will be equal to the energy of the radiation. Therefore, the kinetic energy acquired by the electron in the hydrogen atom is 1.08 x 10^-7 J.
At what temperature are the liquid and the vapour of Bromine in equilibrium (e.g. boiling point)?
Explanation:
Boiling point is defined as the point at which liquid state and vapor state of a substance are existing in equilibrium.
Equilibrium is defined as the state in which rate of forward and rate of backward reaction are equal to each other.
For example, [tex]Br(l) \rightleftharpoons Br(g)[/tex]
So, when we boil bromine which is present in liquid state then at the boiling point its vapors will exist in equilibrium. And unless all the liquid state of bromine will not convert into vapors its temperature will not change.
Therefore, we can conclude that at boiling point the liquid and the vapur of Bromine are in equilibrium.
Many important biochemicals are organic acids, such as pyruvic acid ( p K a = 2.50 ) and lactic acid ( p K a = 3.86 ) . The conjugate bases are pyruvate and lactate, respectively. For each acid, determine which form—the acid or the conjugate base—predominates at pH 7.4. A graphic shows a check mark inscribed on a colored circle. 6
Answer:
Pyruvic acid: conjugate base
Lactic acid: conjugate base
Explanation:
The ratio of conjugate base to conjugate acid can be found using the Henderson-Hasselbalch equation when the pH and pKa are known.
pH = pKa + log([A⁻]/[HA])
The equation can be rearranged to solve for the ratio:
pH - pKa = log([A⁻]/[HA])
[A⁻]/[HA] = 10^(pH-pKa)
Now we can calculate the ratio for the pyruvic acid:
[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 2.50) = 79433
[A⁻] = 79433[HA]
There is a much higher concentration of the conjugate base.
Similarly for lactic acid:
[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 3.86) = 3467
[A⁻] = 3467[HA]
For lactic acid the conjugate base also dominates at pH 7.4
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C. Determine the molar heat of vaporization of substance X using the derived form of the Clausius-Clapeyron equation given below. (Include the sign of the value in your answer.) ____ kJ/mol
The molar heat of vaporization of substance X can be determined using the Clausius-Clapeyron equation. The molar heat of vaporization of substance X is -61.78 kJ/mol.
Explanation:The molar heat of vaporization of substance X can be determined using the Clausius-Clapeyron equation. The equation is given by:
ln(P₂/P₁) = -(ΔHvap/R)((1/T₂) - (1/T₁))
We can solve for ΔHvap by substituting the values given: P₁ = 100 mm Hg, T₁ = 1080 °C (or 1353 K), P₂ = 600 mm Hg, and T₂ = 1220 °C (or 1493 K).
ln(600/100) = -(ΔHvap/8.314)((1/1493) - (1/1353))
Solving for ΔHvap gives us a value of -61.78 kJ/mol. Therefore, the molar heat of vaporization of substance X is -61.78 kJ/mol.
In a gas grill, 29 lbs propane C3H8 are
burned with just enough air for complete combustion at a party. How
many lbs of combustion products are formed? Round your answer to
the nearest whole number.
Answer : The mass of combustion products formed are 134 lbs.
Explanation :
The balanced chemical reaction will be:
[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]
Given :
Mass of [tex]C_3H_8[/tex] = 29 lbs = 13154.2 g
conversion used : 1 lbs = 453.592 g
Molar mass of [tex]C_3H_8[/tex] = 44 g/mole
First we have to calculate the moles of [tex]C_3H_8[/tex].
[tex]\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{13154.2g}{44g/mole}=298.9moles[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex] and [tex]H_2O[/tex].
From the balanced chemical reaction we conclude that,
As, 1 mole of [tex]C_3H_8[/tex] react to give 3 moles of [tex]CO_2[/tex]
So, 298.9 mole of [tex]C_3H_8[/tex] react to give [tex]298.9\times 3=896.7[/tex] moles of [tex]CO_2[/tex]
and,
As, 1 mole of [tex]C_3H_8[/tex] react to give 4 moles of [tex]H_2O[/tex]
So, 298.9 mole of [tex]C_3H_8[/tex] react to give [tex]298.9\times 4=1195.6[/tex] moles of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]CO_2[/tex] and [tex]H_2O[/tex].
Molar mass of [tex]CO_2[/tex] = 44 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass }CO_2[/tex]
[tex]\text{Mass of }CO_2=896.7mole\times 44g/mole=39454.8g=86.98lbs[/tex]
and,
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass }H_2O[/tex]
[tex]\text{Mass of }H_2O=1195.6mole\times 18g/mole=21520.8g=47.44lbs[/tex]
The total mass of products = Mass of [tex]CO_2[/tex] + Mass of [tex]H_2O[/tex]
The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs
Therefore, the mass of combustion products formed are 134 lbs.
What is the pH of a solution that is 0.10 M formic acid and 0.0065 M formate (the conjugate base)? Ka of formic acid = 1.77 x 10-4
Answer:
pH = 2.56
Explanation:
The Henderson-Hasselbalch equation relates the pH to the Ka and ratio of the conjugate acid-base pair as follows:
pH = pKa + log([A⁻]/[HA]) = -log(Ka) + log([A⁻]/[HA])
Substituting in the value gives:
pH = -log(1.77 x 10⁻⁴) + log((0.0065M) / (0.10M))
pH = 2.56
The pH of a 0.10 M formic acid and 0.0065 M formate solution can be calculated using the Henderson-Hasselbalch equation and a given Ka for formic acid of 1.77 x 10-4. The resulting pH is approximately 3.04.
Explanation:In order to find the pH of a solution consisting of a weak acid (formic acid, HCOOH) and its conjugate base (formate, HCOO-), you'd use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the molarity of the conjugate base (here, formate, 0.0065 M) and [HA] is the molarity of the weak acid (here, formic acid, 0.10 M). The pKa is found by taking the negative logarithm of the Ka value, so pKa = -log(Ka) = -log(1.77 x 10-4) = 3.75.
Using these values in the Henderson-Hasselbalch equation: pH = 3.75 + log(0.0065/0.10) this results in a pH of approximately 3.04
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A solution is prepared by condensing 4.00 L of a gas,
measuredat 270C and 748 mmHg pressure into 58.0g
ofbenzene. Calculate the freezing point of this solution?
Answer:
-2.3 ºC
Explanation:
Kf (benzene) = 5.12 ° C kg mol – 1
1st - We calculate the moles of condensed gas using the ideal gas equation:
n = PV / (RT)
P = 748/760 = 0.984 atm
T = 270 + 273.15 = 543.15 K
V = 4 L
R = 0.082 atm.L / mol.K
n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol
Then, you calculate the molality of the solution:
m = n / kg solvent
m = 0.088 mol / 0.058 kg = 1.52mol / kg
Then you calculate the decrease in freezing point (DT)
DT = m * Kf
DT = 1.52 * 5.12 = 7.8 ° C
Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:
T = 5.5 - 7.8 = -2.3 ºC
2 of 20 Which intermolecular force or bond is primarily responsible for the solubility of CH3OH in water? lonic bonding Hydrogen bonding Covalent bonding Dipole-dipole force lon-dipole force Navigator F10 F11 F12 PSC
Answer:
Hydrogen bonding
Explanation:
As a rule of thumb, "likes dissolve like", meaning polar solutes dissolve in polar solvents and nonpolar solutes in nonpolar solvents. In this case, water is polar (dipolar moment = 1.85 Debye) dissolves methanol which is also polar (dipolar moment = 1.69 Debye). Besides being dipoles, both molecules have atoms of Hydrogen with a covalent bond to more electronegative atoms of Oxygen. When this happens, stronger dipole-dipole interactions appear known as Hydrogen bonding. There is an electrostatic attraction between H (positive charge density) and O (negative charge density).
The concentration of chlorobenzene (C&HsCl) in water is 100 mol/m3. density is 1.00 g/cm3 The solution (a) What is the weight fraction of chlorobenzene? (b) What is the chlorobenzene concentration in PPM? (c) What is the mole fraction of chlorobenzene? (d) What is the molarity of chlorobenzene? (e) What is the molality of chlorobenzene? The concentration of chlorobenzene (C&HsCl) in air is 0.100 mol/m3 at 25 °C and 1 atm. The molecular weight of air may be taken to be 28.84 gmol. (a) What is the weight fraction of chlorobenzene? (c) What is the mole fraction of chlorobenzene? (b) What is the chlorobenzene concentration in PPM?
Answer:
Part 1
(a) 0.0113
(b) 11300 ppm
(c) 1.82 *10⁻³
(d) 0.100 M
(e) 0.101 m
Part 2
(a) 9.45 *10⁻³
(b) mole fraction = 2.45 *10⁻³
(c) 11.3 ppm
Explanation:
Chlorobenzene formula is C₆H₅Cl
Part 1: We are given a concentration of chlorobenzene in water of 100 mol/m³, and a density of the solution of 1.00 g/cm³.
(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / mass solution
We know there are 100 moles of C₆H₅Cl per m³ of solution.
To get the mass of C₆H₅Cl we'll convert the moles to mass by using the molar mass:
Molar mass C₆H₅Cl = 6*12.011 + 5*1.00794 + 35.4527 = 112.558 g/mol
mass C₆H₅Cl = moles C₆H₅Cl * molar mass C₆H₅Cl
mass C₆H₅Cl = 100 moles * 112.558 g/mol = 11255.8 g
11255.8 g of C₆H₅Cl are in 1 m³ of solution.
Next we'll convert 1 m³ of solution to mass by using the density
mass solution = volume solution * density of solution
[tex]mass solution = 1m^{3} *\frac{(100cm)^{3} }{ 1m^{3}} * \frac{1.00 g}{cm^{3} } = 1.00 *10^{6} g[/tex]
weight fraction C₆H₅Cl = 11256 g / 1.00 *10⁶ g = 0.0113
(b) ppm stands for "parts per million" and it is usually expressed as mg per Liter of solution
We already calculated that there are 11256 g or more exactly 11300 g of C₆H₅Cl in 1 m³ of solution, so lets convert to mg/L:
[tex]\frac{11300 g}{1 m^{3} } * \frac{1000 mg}{1 g} * \frac{1 m^{3} }{1000 L} = 11300 mg/L[/tex]
So the solution is 11300 ppm
(c) mole fraction = moles of C₆H₅Cl / total moles in solution
total moles = moles C₆H₅Cl + moles water
moles water = mass water / molar mass water
mass water = mass solution - mass C₆H₅Cl
moles of C₆H₅Cl = 100 moles
mass water = 1.00 *10⁶ g of solution - 11256 g = 988744 g of water
moles water = 988744 g / 18.0153 g/mol = 54884 moles water
total moles = 100 + 54884 = 54984 moles
mole fraction = 100 moles of C₆H₅Cl / 54984 moles = 1.82 *10⁻³
(d) Molarity = moles C₆H₅Cl / Liters of solution
We know the solution is 100 mol / m³ so we just have to convert the m³ to L:
[tex]\frac{100 mol}{m^{3} } * \frac{1 m^{3}}{1000 L} = 0.100 mol / L = 0.100 M[/tex]
(e) Molality = moles C₆H₅Cl / kg water
We know that there are 100 moles per 988744 g of water, so we need to convert the grams of water to kilograms.
[tex]Molality = \frac{100 moles}{988744 g} *\frac{1000 g}{1 kg} = 0.101 m[/tex]
____________________________________
Part 2: Concentration of C₆H₅Cl in air is 0.100 mol/m³, at 25 °C and 1 atm.
Molar mass air = 28.84 g/mol
(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / total mass
mass C₆H₅Cl = 0.100 mol * 112.558 g/mol = 11.26 g
total mass = mass C₆H₅Cl + mass air
mass air = moles air * molar mass air
moles air = total moles - moles C₆H₅Cl
We can calculate the total moles by using the ideal gas law:
P V = n R T
where P is pressure in atm, V is volume in L, n is the number of moles, R is the gas constant and T is temperature in Kelvin.
n = P V / R T
P = 1 atm
V = 1 m³ = 1000 L
R = 0.08206 L atm K⁻¹ mol⁻¹
T = 25 + 273.15 = 298 K
n = (1 atm * 1000 L) / (0.08206 L atm K⁻¹ mol⁻¹ * 298 K) = 40.89 moles
moles air = 40.89 - 0.100 = 40.79 moles air
mass air = 40.79 mol * 28.84 g/mol = 1176.4 g
total mass = 1176.4 g + 11.26 g = 1188 g
weight fraction = 11.26 g / 1188 g = 9.45 *10⁻³
(b) mole fraction = moles C₆H₅Cl / total moles
mole fraction = 0.100 / 40.89 = 2.45 *10⁻³
(c) ppm = mg C₆H₅Cl / Liters
We already know there are 11.26 g C₆H₅Cl in 1 m³, which is the same as 1000 L, so:
[tex]\frac{11.26 g}{1000 L} *\frac{1000 mg}{1 g} = 11.3 mg/L[/tex]
The concentration is 11.3 ppm
Final answer:
a) The weight fraction of chlorobenzene is 11.26. b) The chlorobenzene concentration in ppm is 100,000 ppm. c) The mole fraction of chlorobenzene is 0.1. d) The molarity of chlorobenzene is 0.1 M. e) The molality of chlorobenzene is 0.1 m.
Explanation:
a) The weight fraction of chlorobenzene can be calculated by dividing the mass of chlorobenzene by the total mass of the solution. Since the density of the solution is 1.00 g/cm3, the mass of chlorobenzene can be calculated as 100 mol/m3 * 112.6 g/mol / 1000 cm3 = 11.26 g/cm3. Therefore, the weight fraction of chlorobenzene is 11.26 g/cm3 / (1.00 g/cm3) = 11.26.
b) To convert the concentration of chlorobenzene from mol/m3 to ppm (parts per million), we need to multiply by 10^6. Therefore, the chlorobenzene concentration in ppm is 100 mol/m3 * 10^6 ppm/mol = 100,000 ppm.
c) The mole fraction of chlorobenzene can be calculated by dividing the number of moles of chlorobenzene by the total number of moles in the solution. Since the concentration of chlorobenzene is given in mol/m3, the number of moles of chlorobenzene can be calculated as 100 mol/m3 * (1 L / 1000 cm3) = 0.1 mol. The total number of moles in the solution is 1 L * (1 mol/m3) = 1 mol. Therefore, the mole fraction of chlorobenzene is 0.1 mol / 1 mol = 0.1.
d) The molarity of chlorobenzene can be calculated by dividing the number of moles of chlorobenzene by the volume of the solution in liters. Since the concentration of chlorobenzene is given in mol/m3, the number of moles of chlorobenzene can be calculated as 100 mol/m3 * (1 L / 1000 cm3) = 0.1 mol. The volume of the solution is 1 L. Therefore, the molarity of chlorobenzene is 0.1 mol / 1 L = 0.1 M.
e) The molality of chlorobenzene can be calculated by dividing the number of moles of chlorobenzene by the mass of the solvent in kilograms. Since the density of the solution is 1.00 g/cm3, the mass of the solvent (water) can be calculated as 1 L * (1000 cm3 / 1 L) * (1.00 g/cm3) = 1000 g. The number of moles of chlorobenzene can be calculated as 100 mol/m3 * (1 L / 1000 cm3) = 0.1 mol. Therefore, the molality of chlorobenzene is 0.1 mol / 1 kg = 0.1 m.
Plants breathe in carbon dioxide to make sugar through photosynthesis. How much sugar can they create if 200 grams of carbon dioxide are used? 6 CO2 + 6 H2O -> C6H12O6 + 6 O2 A. 136.48 grams B. 180.16 grams C. 30.03 grams D. 0.76 grams
Answer: The correct answer is Option A.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For carbon dioxide:Given mass of carbon dioxide = 200 g
Molar mass of carbon dioxide = 44 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of carbon dioxide}=\frac{200g}{44g/mol}=4.54mol[/tex]
The given chemical equation of photosynthesis reaction follows:
[tex]6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2[/tex]
By Stoichiometry of the reaction:
If 6 moles of carbon dioxide produces 1 mole of glucose.
Then, 4.54 moles of carbon dioxide will produce = [tex]\frac{1}{6}\times 4.54=0.756mol[/tex]
Now, calculating the mass of glucose from equation 1, we get:
Molar mass of glucose = 180.2 g/mol
Moles of glucose = 0.756 moles
Putting values in equation 1, we get:
[tex]0.756mol=\frac{\text{Mass of glucose}}{180.2g/mol}\\\\\text{Mass of glucose}=136.48g[/tex]
Hence, the correct answer is Option A.
Discuss the advantages of using building information modeling (BIM).
Answer:
The advantage of using building information modeling (BIM) are as follows:
1.Model based cost estimation
2. Preconstruction project visualization
3.Safe construction site
4. Improve scheduling
5.Improve coordination and clash detection
6.Reduced mitigated risk and cost
7.Improve prefabrication
8.Better collaboration and communication
9. Strong facility management
10.Improve sequencing
The peregrine falcon has been measured as Traveling up to 350 km/hr in a dive. if this falcon can fly to the moon at this speed, how many seconds would it take?
Answer:
4 × 10⁶ sec
Explanation:
The distance between the earth and the moon = 384,400 km
The speed of the peregrine falcon = 350 km/hr
Considering,
Distance = Speed × Time
So,
Time = Distance / Speed = 384,400 km / 350 km/hr = 1098.28571 hrs
Also,
1 hr = 3600 sec
So,
1098.28571 hrs = 3600 × 1098.28571 s ≅ 4 × 10⁶ sec
Thus time taken by peregrine falcon if falcon fly to the moon = 4 × 10⁶ sec
It would take the peregrine falcon approximately 3,953,044 seconds to fly to the moon at a speed of 350 km/hr.
Explanation:If the peregrine falcon can travel up to 350 km/hr in a dive, let's calculate how long it would take for the falcon to fly to the moon at this speed. The average distance from Earth to the moon is approximately 384,400 km.
To find the time it would take, we can use the formula:
Time = Distance/Speed
So, Time = 384,400 km / 350 km/hr = 1098.29 hours.
Converting hours to seconds, we have:
Time = 1098.29 hours x 60 minutes x 60 seconds = 3,953,044 seconds.
Therefore, it would take the peregrine falcon approximately 3,953,044 seconds to fly to the moon at a speed of 350 km/hr.
What are the three ways resources are classified?
Answer:
On the basis of ownership, distribution, durabilities.
Explanation:
In 1933, the concept of resources was given Zimmerman as a resource signifies possibility and may b a means to an end. According to him resources satisfy human wants and needs, having two important functions of utility and ability. Based on durability, they are either Fund or exhaustible resources. These get destroyed after use. Flow or inexhaustible resource remains unchanged even after continuous use like reiver, sea, and land. Based on ownership, they are International means used by the global population. National resource means used by one nation in one country. And individual searches money or property resources etc. And based on availability they are Ubiquitous and localized .like sunshine and the air is available everywhere.The benzene boiling temperature (C6H6) is 80.1ºC dissolving 36 g pentane, C5H12 at 500 g benzene increases the boiling point of the solution to 82.73ºC
A. Consider the benzene boiling point constant. Show calculations.
B. In dissolving 1.2 g of unknown solute in 50 g of benzene, a solution with a boiling point of 80.36ºC is obtained, which is the molar mass of the solute (assume that i = 1) (show calculations)
Answer:
A)Boiling point constant of benzene = 2.63°C/m
B) 242.77 g/mol is the molar mass of the solute.
Explanation:
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] =Elevation in boiling point
[tex]K_b[/tex] = boiling point constant od solvent= 3.63 °C/m
1 - van't Hoff factor
m = molality
A) Mas of solvent = 500 g = 0.500 kg
T = 80.1°C ,[tex]T_b[/tex] =82.73°C
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b[/tex]= 82.73°C - 80.1°C = 2.63°C
[tex]2.63^oC=K_b\times \frac{36 g}{72 g/mol\times 0.500 kg}[/tex]
[tex]K_b=\frac{2.63^oC\times 72 g/mol\times 0.500 kg}{36 g}=2.63 ^oC/m[/tex]
Boiling point constant of benzene = 2.63°C/m
B) Mass of solute = 1.2 g
Molar mas of solute = M
Mass of solvent = 50 g= 0.050 kg
i = 1
T = 80.1°C ,[tex]T_b[/tex] =80.36°C
[tex]\Delta T_b=T_b-T[/tex]=80.36°C - 80.1°C = 0.26°C
[tex]0.26^oC=1\times K_b\times \frac{1.2 g}{M\times 0.050 kg}[/tex]
[tex]M=1\times 2.63^oC/m\times \frac{1.2 g}{0.26^oC\times 0.050 kg}[/tex]
M = 242.77 g/mol
242.77 g/mol is the molar mass of the solute.
Final answer:
The boiling point constant (Kb) of benzene was found to be 2.63°C/m using the data from part A. Using this constant and the boiling point elevation data in part B, the molar mass of the unknown solute was calculated to be 240 g/mol.
Explanation:
The subject of this question is Chemistry, focusing specifically on the boiling point elevation of solutions and the calculation of molar mass from boiling point data. The problem pertains to colligative properties of solutions, which are properties that depend on the number of particles in a solution but not their identity.
Part A: Calculating Boiling Point Elevation
To find the boiling point constant (Kb) of benzene, we use the formula for boiling point elevation, ΔTb = Kb × m, where ΔTb is the change in boiling point and m is the molality of the solution. Given that the boiling point of benzene increases from 80.1°C to 82.73°C, the elevation in boiling point is 2.63°C. Molality (m) is calculated by the number of moles of pentane divided by the mass of benzene in kilograms. The molar mass of pentane (C5H12) is 72.15 g/mol, and so 36 g corresponds to 0.5 moles. The mass of benzene is 500 g, which is 0.5 kg. Thus, m = 0.5 moles / 0.5 kg = 1 mol/kg. Using these values, we can calculate Kb: 2.63°C = Kb × 1 mol/kg, so Kb = 2.63°C/m.
Part B: Determining Molar Mass of an Unknown Solute
For part B, to find the molar mass of the solute, we first calculate the change in boiling point of benzene, which is 80.36°C - 80.1°C = 0.26°C. With the boiling point constant of benzene from Part A (Kb = 2.63°C/m), and assuming i (the van 't Hoff factor) is 1 for a non-electrolyte, we establish the molality of the solution as m = ΔTb / Kb = 0.26°C / 2.63°C/m = 0.1 mol/kg. Knowing the molality and the mass of solvent (benzene), we can now calculate the molar mass (M) of the unknown solute using the formula M = mass of solute / (molality × mass of solvent in kg). With the mass of solute as 1.2 g and the mass of solvent as 0.05 kg, M = 1.2 g / (0.1 mol/kg × 0.05 kg)= 240 g/mol.
A substance in a specific state of matter was transferred from a cylindrical shaped container to a cube shaped container. The substance took different shapes in each container. Which of the following could be another characteristic of the substance. A. It is partially compressible B. It’s particles are arranged in a fixed pattern. C. It has very strong intermolecular forces between particles D. The particles move up and down without changing their position
Answer:
D. The particles move up and down without changing their position
Explanation:
These are gases. They are typically known for their randomness and no fixed arrangement of their atoms.
Gases generally assume the volumes of the containers they fill. They spread easily and readily to fill the volume where they occupy. Also, gases are readily compressible as they lack intermolecular attraction between their molecules.
Answer: Option (A) is the correct answer.
Explanation:
In a solid substance, the atoms are held together by strong intermolecular forces of attraction. Hence, the atoms are not able t move freely but they are able to vibrate at their mean position.
As a result, solids have a definite shape and volume and also, they are incompressible in nature.
In liquids, the atoms are held together by less strong intermolecular forces of attraction as compared to solids. Therefore, atoms of a liquid are able to slide past each other. So, they are partially compressible in nature.
Also, liquids acquire the shape of container in which they are placed.
But in gases, the molecules are held together by weak intermolecular forces of attraction. Hence, atoms are able to move far away from each other as they have high kinetic energy.
Gases are highly compressible in nature.
Therefore, in the given situation as substance took different shapes in each container.
Thus, we can conclude that it is partially compressible could be another characteristic of the substance.
Nickel has an FCC structure and an atomic radius of 0.124 nm. (12 points) a. What is the coordination number? b. Calculate the APE C. Calculate the lattice parameter, "a" d. Calculate the theoretical density
Answer:
a)CN=12
b)APF=74 %
c)a=0.35 nm
d)ρ=9090.9 [tex]Kg/m^3[/tex]
Explanation:
Given that
Nickel have FCC structure
We know that in FCC structure ,in FCC 8 atoms at corner with 1/8 th part in one unit cell and 6 atoms at faces with 1/2 part in one unit cell .
Z=8 x 1/8 + 1/2 x 6 =4
Z=4
Coordination number (CN)
The number of atoms which touch the second atoms is known as coordination number.In other word the number of nearest atoms.
CN=12
Coordination number of FCC structure is 12.These 12 atoms are 4 atoms at the at corner ,4 atoms at 4 faces and 4 atoms of next unit cell.
APF
[tex]APF=\dfrac{Z\times \dfrac{4}{3}\pi R^3}{a^3}[/tex]
We know that for FCC
[tex]4R=\sqrt{2}\ a[/tex]
Now by putting the values
[tex]APF=\dfrac{4\times \dfrac{4}{3}\pi R^3}{\left(\dfrac{4R}{\sqrt2}\right)^3}[/tex]
APF=0.74
APF=74 %
[tex]4R=\sqrt{2}\ a[/tex]
[tex]4\times 0.124=\sqrt{2}\ a[/tex]
a=0.35 nm
Density
[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
We know that M for Ni
M=58.69 g/mol
a=0.35 nm
[tex]N_A=6.023\times 10^{23}\ atom/mol[/tex]
[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
[tex]\rho=\dfrac{4\times 58.69}{6.023\times 10^{23}\times( 0.35\times 10^{-9})^3}\ g/m^3[/tex]
ρ=9090.9 [tex]Kg/m^3[/tex]