Answer: 13.5g of O2
Explanation:
2C2H5OH + 7O2 → 4CO2 + 6H2O
Molar Mass of C2H5OH = (12x2)+5+16+1= 46g/mol
Mass conc. Of C2H5OH =2x46= 92g
Molar Mass of O2 = 32g/mol
Mass conc. Of O2 = 7 x 32 = 224g
From the equation,
92g of C2H5OH required 224g O O2.
Therefore, 5.54g of C2H5OH will require = (5.54x224)/92 = 13.5g of O2
To burn 5.54 g of C2H5OH, 13.4 g of O2 is required.
Explanation:To determine the grams of oxygen gas required to burn 5.54 g of C2H5OH, we first need to convert the given mass of C2H5OH to moles using its molar mass. The molar mass of C2H5OH is 46.07 g/mol. So, 5.54 g of C2H5OH is equal to 5.54 g / 46.07 g/mol = 0.12 mol of C2H5OH.
According to the balanced equation, the mole ratio between C2H5OH and O2 is 2:7. Therefore, for every 2 moles of C2H5OH, we require 7 moles of O2.
Using this mole ratio, we can calculate the moles of O2 required: 0.12 mol C2H5OH × (7 mol O2 / 2 mol C2H5OH) = 0.42 mol O2.
To convert moles of O2 to grams, we can use its molar mass, which is 32.00 g/mol. Therefore, 0.42 mol O2 × 32.00 g/mol = 13.4 g of O2.
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A 5.21 mass % aqueous solution of urea (CO(NH2)2) has a density of 1.15 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.
Answer:
Molarity is 0.99 M
Explanation:
5.21% by mass, is a sort of concentration which shows the mass of solute in 100 g of solution.
Molarity is a sort of concentration that indicates the moles of solute in 1 L of solution (mol/L)
Let's find out the volume of solution by density.
Solution density = Solution mass / Solution volume
1.15 g/mL = 100 g / Solution volume
Solution volume = 100 g / 1.15 g/mL → 86.9 mL
We must have the volume of solution in L, so let's convert it.
86.9 mL / 1000 = 0.0869 L
Now, we have to determine the moles of solute (urea)
5.21 g . 1 mol / 60 g = 0.0868 moles
Mol/L = Molarity → 0.0868 moles / 0.0869L = 0.99 M
Answer:
[tex]\large \boxed{\text{1.00 mol/L}}[/tex]
Explanation:
Molar concentration = moles/litres
So, we need both the number of moles and the volume.
1. Volume
Assume a volume of 1 L.
That takes care of that.
2. Moles of urea
(a) Mass of solution
[tex]\text{ Mass of solution} = \text{1000 mL} \times \dfrac{\text{1.15 g solution}}{\text{1 mL}} = \text{1150 g solution}[/tex]
(b) Mass of urea
[tex]\text{Mass of urea} = \text{1150 g solution}\times \dfrac{\text{5.21 g urea}}{\text{100 g solution}} = \text{59.92 g urea}[/tex]
(c) Moles of urea
[tex]\text{Moles of urea} = \text{59.92 g urea} \times \dfrac{\text{1 mol urea}}{\text{60.06 g urea}} = \text{1.00 mol urea}[/tex]
3. Molar concentration
[tex]\text{Molar concentration} = \ \dfrac{\text{1.00 mol}}{\text{1 L}} = \textbf{1.00 mol/L}\\\text{The molar concentration of the urea is $\large \boxed{\textbf{1.00 mol/L}}$}[/tex]
What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?
Answer:
P = 1/8
Explanation:
The wave function of a particle in a one-dimensional box is given by:
[tex] \psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L}) [/tex]
Hence, the probability of finding the particle in the one-dimensional box is:
[tex] P = \int_{x_{1}}^{x_{2}} \psi^{2} dx [/tex]
[tex] P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx [/tex]
[tex] P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx [/tex]
Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:
[tex] P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})] [/tex]
[tex] P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi}) [/tex]
Solving for n=4:
[tex] P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi}) [/tex]
[tex] P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi}) [/tex]
[tex] P = \frac{1}{8} [/tex]
I hope it helps you!
The total probability of finding a particle in this one-dimensional box is [tex]\frac{1}{8}[/tex]
Given the following data:
Energy level, n = 4x = 0x = [tex]\frac{L}{8}[/tex]To determine the total probability of finding a particle in a one-dimensional box:
A particle in a one-dimensional box describes the translational motion of a particle that is trapped inside an infinitely deep well, from which it is unable to escape.
Mathematically, the wave function of a particle in a one-dimensional box is given by this formula:
[tex]\psi = \sqrt{\frac{2}{L} } sin\frac{n\pi}{L} x[/tex] ...equation 1.
Where:
[tex]\psi[/tex] is the wave function.L is the length of a box.x is the displacement.In a one-dimensional box, the probability of finding a particle is given by the formula:
[tex]P=\int\limits^{x_2}_{x_1} {\psi^2} \, dx[/tex] ...equation 2.
Substituting eqn. 1 into eqn. 2, we have:
[tex]P=\int\limits^{x_2}_{x_1} {(\sqrt{\frac{2}{L} } sin\frac{n\pi}{L} x)^2} \, dx \\\\P = \frac{2}{L} \int\limits^{x_2}_{x_1} {( sin^2(\frac{n\pi}{L} x))} \, dx\\\\P=\frac{2}{L} [\frac{L}{16} (1-4(\frac{sin\frac{n\pi}{4} }{n\pi} ))]\\\\P=\frac{1}{8} (1-4(\frac{sin\frac{n\pi}{4} }{n\pi} ))[/tex]
Substituting the value of n, we have:
[tex]P=\frac{1}{8} (1-4(\frac{sin\frac{4\pi}{4} }{4\pi} ))\\\\P=\frac{1}{8} (1-(\frac{sin\pi }{\pi} ))\\\\P=\frac{1}{8} (1-0)\\\\P=\frac{1}{8}[/tex]
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Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71 x 10¹⁵ J. What is the de Broglie wavelength of this electron (Ek = ½mv²)?
Question in incomplete, complete question is:
Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of [tex]4.71\times 10^{-15}J[/tex] . What is the de Broglie wavelength of this electron (Ek = ½mv²)?
Answer:
[tex] 6.762\times 10^{-12} m[/tex] is the de Broglie wavelength of this electron.
Explanation:
To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:
[tex]\lambda=\frac{h}{\sqrt{2mE_k}}[/tex]
where,
= De-Broglie's wavelength = ?
h = Planck's constant = [tex]6.624\times 10^{-34}Js[/tex]
m = mass of beta particle = [tex] 9.1094\times 10^{-31} kg[/tex]
[tex]E_k[/tex] = kinetic energy of the particle = [tex]4.71\times 10^{-15}J[/tex]
Putting values in above equation, we get:
[tex]\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}[/tex]
[tex]\lambda = 6.762\times 10^{-12} m[/tex]
[tex] 6.762\times 10^{-12} m[/tex] is the de Broglie wavelength of this electron.
Suppose a helium-3 nuclide transforms into a helium-4 nuclide by absorbing a proton and emitting a positron. Complete the nuclear chemical equation below so that it describes this nuclear reaction.
Answer: The nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.
Explanation:
Nuclear reaction are defined as the reactions in which nucleus of an atom is involved.
Positron emission is defined as the emission process in which positron particle is emitted. In this process, a proton gets converted to neutron and an electron neutrino particle.
[tex]_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e[/tex]
The chemical equation for the reaction of He-3 with a proton follows:
[tex]_2^3\textrm{He}+_1^1\textrm{H}\rightarrow _2^4\textrm{He}+_{+1}^0e[/tex]
Hence, the nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.
The nuclear equation where a helium-3 nuclide transforms into a helium-4 nuclide by absorbing a proton and emitting a positron is written as ³He + ¹H → ⁴He + e⁺. This ensures that mass and charge are conserved in the reaction.
Explanation:To complete the nuclear chemical equation where a helium-3 nuclide (³He) transforms into a helium-4 nuclide (⁴He) by absorbing a proton (¹H) and emitting a positron (e⁺), we must ensure that both mass and charge are conserved in the reaction. In this case, the equation can be represented as:
³He + ¹H → ⁴He + e⁺
The mass number on the left side of the equation is 3 (from helium-3) plus 1 (from the proton), totaling 4, which matches the mass number of helium-4 on the right side of the equation. The atomic number (number of protons) is also conserved through this reaction: 2 (from helium-3) + 1 (from the proton) equals 2 (from helium-4) + 1 (from the emitted positron), with positrons having a positive charge but no atomic number associated.
A solution is made by dissolving 5.61 g of a new polymer in enough water to make 260 mL of solution. At 25.0 oC, the osmotic pressure of the solution is 0.174 atm. What is the molar mass of the polymer in g/mol?
Answer:
3030.2 g/mol is the molar mass for our polymer
Explanation:
Formula for Osmotic pressure → π = M . R .T
where π is pressure (atm)
M is molarity mol/L
R, the Universal Gases Constant (0.082 L.atm/mol.K)
T, Absolute T° (T°C + 273)
25°C + 273 = 298 K
Let's replace the values
0.174 atm = M . 0.082l.atm/mol.K . 298K
0.174 atm / (0.082l.atm/mol.K . 298K) = M
7.12×10⁻³ mol/L
As molarity is mol/L, and we have the volume of solution (in mL we must convert to L) we can find out the moles of our polymer that corresponds to the mass we used.
260mL . 1L/1000mL = 0.260L
7.12×10⁻³ mol/L = mol / 0.260L
7.12×10⁻³ mol . 0.260 = mol → 1.85×10⁻³
These moles refers to te 5.61 g of solute, to if we want to determine the molar mass, we should do:
g/mol → 5.61 g / 1.85×10⁻³ mol = 3030.2 g/mol
Final answer:
The molar mass of the polymer is approximately 28.93 g/mol.
Explanation:
To calculate the molar mass of the polymer, we can use the osmotic pressure formula:
π = MRT
Where π is the osmotic pressure, M is the molar mass, R is the ideal gas constant, and T is the temperature in Kelvin.
Given that the osmotic pressure is 0.174 atm and the temperature is 25.0 °C (which needs to be converted to Kelvin), we can rearrange the formula to solve for M:
M = π / (RT)
Now we can plug in the values:
M = 0.174 atm / (0.0821 L·atm/mol·K * 298 K)
M = 0.00706 mol / 0.024438 L
M = 28.93 g/mol
Therefore, the molar mass of the polymer is approximately 28.93 g/mol.
One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configuration for . Would this atom be expected to be paramagnetic, and how many unpaired electrons are present? It has unpaired electron(s). Consider the ground-state electron configuration for . Would this atom be expected to be paramagnetic, and how many unpaired electrons are present?
Answer:
Li, N, Ni, Te
Explanation:
First let us consider the number of unpaired electrons in each of the given atoms.
Li-1
N-3
Ni-2
Te-2
For an atom to be paramagnetic, it must possess at least one unpaired electron in its valence shell. Ba and Hg are not paramagnetic because all electrons in their outermost shells are spin paired.
The more the number of unpaired electrons in the outermost shell of an atom, the greater its paramagnetic behavior.
Whether an atom is paramagnetic depends on the presence of unpaired electrons in its ground state electron configuration. The number of unpaired electrons determines the strength of its paramagnetism. For example, a nitrogen atom has three unpaired electrons leading it to exhibit paramagnetism.
Explanation:The question has asked about the paramagnetism and unpaired electrons present in an atom according to the quantum mechanical model. However, the specific atom is not given in the question. In general, if an atom has unpaired electrons in its ground state electron configuration, it will exhibit paramagnetic behavior. The number of unpaired electrons will determine the strength of this effect.
For example, let's consider an atom of nitrogen. According to quantum mechanics, a nitrogen atom has 7 electrons. In its ground state, it has two electrons in 1s orbital, two in 2s, and three in 2p orbitals. Only the three 2p electrons are unpaired. Therefore, a nitrogen atom exhibits paramagnetism due to these three unpaired electrons.
Likewise, you can determine the magnetic properties of any atom by examining its electron configuration and identifying the number of unpaired electrons.
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Wastewater from a cement factory contains 0.280 g of Ca2 ion and 0.0550 g of Mg2 ion per 100.0 L of solution. The solution density is 1.001 g/mL. Calculate the Ca2 and Mg2 concentrations in ppm (by mass).
Answer:
The correct answer is: 2.8 ppm Ca²⁺ and 0.55 ppm Mg²⁺
Explanation:
ppm (by mass) is equal to: mass solute (mg)/mass solution(kg)
First, we use the solution density to calculate the mass of 100.0 L solution in kg:
mass solution= 100.0 L x 1.001 g/ml x 1000 ml/ 1L x 1kg/1000 g= 100.1 kg
Then, we divide the mass of each ion (in mg) into the mass solution to obtain the ppm:
mg Ca²⁺= 0.280 g x 1000 mg/1g= 280 mg
mg Mg²⁺= 0.0550 g x 1000 mg/1g= 55 mg
ppm Ca²⁺= 280 mg Ca²⁺/ 100.1 kg solution= 2.797 mg Ca²⁺/kg solution= 2.8 ppm
ppm Mg²⁺= 55 mg Mg²⁺/ 100.1 kg solution= 0.549 mg Mg²⁺/kg solution= 0.55 ppm
A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give your answer to 3 decimal places.
Answer: The molality of potassium hydroxide solution is 0.608 m
Explanation:
We are given:
3.301 mass % of potassium hydroxide solution.
This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution
Mass of solvent = Mass of solution - Mass of solute (KOH)
Mass of solvent = (100 - 3.301) g = 96.699 g
To calculate the molality of solution, we use the equation:
[tex]\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}[/tex]
Where,
[tex]m_{solute}[/tex] = Given mass of solute (KOH) = 3.301 g
[tex]M_{solute}[/tex] = Molar mass of solute (KOH) = 56.1 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent = 96.699 g
Putting values in above equation, we get:
[tex]\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m[/tex]
Hence, the molality of potassium hydroxide solution is 0.608 m
Complete and balance the molecular equation for the reaction of aqueous sodium carbonate, Na 2 CO 3 Na2CO3 , and aqueous nickel(II) chloride, NiCl 2 NiCl2 . Include physical states. molecular equation:
Na 2 CO 3 ( aq ) + NiCl 2 ( aq ) ⟶ 2 NaCl ( aq ) + NiCO 3 ( s ) Na2CO3(aq)+NiCl2(aq)⟶2NaCl(aq)+NiCO3(s) Enter the balanced net ionic equation for this reaction. Include physical states. net ionic equation:
Answer: The net ionic equation is written below.
Explanation:
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of sodium carbonate and nickel (II) chloride is given as:
[tex]Na_2CO_3(aq.)+NiCl_2(aq.)\rightarrow 2NaCl(aq.)+NiCO_3(s)[/tex]
Ionic form of the above equation follows:
[tex]2Na^{+}(aq.)+CO_3^{2-}(aq.)+Ni^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow NiCO_3(s)+2Na^+(aq.)+2Cl^-(aq.)[/tex]
As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
[tex]Ni^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow NiCO_3(s)[/tex]
Hence, the net ionic equation is written above.
The balanced molecular equation for the reaction of Na2CO3 and NiCl2 is Na2CO3(aq) + NiCl2(aq) ⟶ 2NaCl(aq) + NiCO3(s). The balanced net ionic equation, excluding the spectator ions, is CO32-(aq) + Ni2+(aq) ⟶ NiCO3(s).
Explanation:The reaction between aqueous sodium carbonate (Na2CO3) and aqueous nickel(II) chloride (NiCl2) results in the formation of sodium chloride (NaCl) and nickel carbonate (NiCO3). Represented as molecular equation, it looks like this: Na2CO3(aq) + NiCl2(aq) ⟶ 2NaCl(aq) + NiCO3(s).
For the net ionic reaction, we exclude the spectator ions, which in this case are Na+ and Cl-. These ions remain unreacted in the solution. Therefore, the net ionic equation will be: CO32-(aq) + Ni2+(aq) ⟶ NiCO3(s).
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A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS₂ (Kb = 2.43 °C/m).
If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide?
Answer : The formula of a sulfur molecule in carbon disulfide is, [tex]S_8[/tex]
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
or,
[tex]\Delta T_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]\Delta T_b[/tex] = boiling point of elevation = [tex]0.107^oC[/tex]
[tex]k_b[/tex] = boiling point constant of carbon disulfide = [tex]2.43^oC/m[/tex]
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte
[tex]w_2[/tex] = mass of solute (sulfur sample) = 0.210 g
[tex]w_1[/tex] = mass of solvent (carbon disulfide) = 17.8 g
[tex]M_2[/tex] = molar mass of solute (sulfur sample) = ?
Now put all the given values in the above formula, we get:
[tex]0.107^oC=1\times (2.43^oC/m)\times \frac{(0.210g)\times 1000}{M_2\times (17.8g)}[/tex]
[tex]M_2=267.93g/mol[/tex]
The molar mass of sulfur sample is, 267.93 g/mol
Now we have to determine the formula of a sulfur molecule in carbon disulfide.
Let the compound of sulfur be, [tex]S_n[/tex]
Molar mass of [tex]S_n[/tex] = n × Molar mass of S
267.93 g/mol = n × 32 g/mol
n = 8
Thus, the formula of a sulfur molecule in carbon disulfide is, [tex]S_8[/tex]
The Tris/Borate/EDTA buffer (TBE) is commonly made as a 5x solution. What volumes of 5x TBE and water are required to make 500 mls of a 0.5x solution which is often used in electrophoresis?
Answer:
50 ml (5x TBE) + 540 ml (water)
Explanation:
To prepare 0.5x TBE solution from 5x TBE solution we need to use the following dilution formula:
C1 x V1 = C2 x V2, where:
- C1, V1 = Concentration/amount (start), and Volume (start)
- C2, V2 = Concentration/amount (final), and Volume (final)
* So when we applied this formula it will be:
5 x V1 = 0.5 x 500
V1= 50ml
- To prepare 0.5x we will take 50ml from 5x and completed with 450ml water and the final volume will going to be 500ml.
When glucose goes from an open-chain form [Image 1] to a cyclic form [Image 2] in solution, it loses one type of functional group and gains another type of functional group. Which choice below corresponds to the functional groups unique to the open-chain and cyclic forms, respectively?
ketone; secondary alcohol
aldehyde; epoxide
primary alcohol; epoxide
aldehyde; ether
ketone; ether
The aldehyde group changes to ether group.
Explanation:In the straight chain glucose molecule, there's the 6 carbon atoms attached with each other where there are several hydrogen and hydroxyl radicals attached. The C1 of glucose contains the aldehyde group and the C5 is the most reactive atom of glucose. In a solution, the lone pair of the oxygen atom of hydroxyl group of C5 attacks the electrophilic centre of C1, forming a carbon oxygen ether bond. This bond makes the double bonded oxygen weaker and one of the bond breaks forming oxygen radicle, which then attracts the hydrogen radicle emitted by the ether oxygen to form a hydroxyl group. This is how the ring structure forms in a solution.
Thus, the aldehyde group of glucose changes to ether group.
The plot shows the absorbance spectra for solutions of caffeine, benzoic acid, and Mountain Dew® soda, each in 0.010 M HCl . A plot contains three spectra with Absorbance on the y axis and wavelength in nanometers on the x axis. Absorbance measurements range from 0.0 to 1.5, with increments every 0.2. Wavelength measurements range from 200 to 300, with icrements every 100 nanometers. A note states that All solutions contain 0.010 molar H C L. The spectrum for benzoic acid at a concentration of 8.74 milligrams per liter, begins at an absorbance of about 1.3 and 200 nanometers. Absorbance sharply decreases to 0.2 at 210 nanometers, then rises and peaks at an absorbance close to 0.8 and 228 nanometers. The spectrum for a 1 to 50 dilution of Mountain Dew soda begins at an absorbance of 1.5 and 200 nanometers, then sharply decreases in absorbance to 0.6 at 215 nanometers, then rises and peaks at absorbance close to 0.75 at 228 nanometers, decreases, then peaks again at absorbance 0.27 at 275 nanometers. The spectrum for a caffeine at a concentration of 10.88 milligrams per liter begins at an absorbance of 1.45 at 200 nanometers, then rises slightly, with a peak at absorbance 1.5 at 205 nanometers, then decreases sharply in absorbance until 0.2 at 245 nanometers. The plot then peaks again at an absorbance close to 0.6 at 270 nanometers. What is the approximate absorbance of benzoic acid at 228 nm?
Complete Question
The diagram of the complete question is shown on the first uploaded image.
Answer:
a) The approximate absorbance of benzoic acid at 228 nm? : A = 0.8
b) The molar absorptivity(∈)
of benzoic acid at 228 nm? : ∈ = [tex]1.12 * 10^{4} M^{-1} cm^{-1}[/tex]
Explanation:
Looking at the absorbance spectra, we can see that the approximate absorbance of benzoic acid at 228 is 0.8.
mass concentration of benzoic acid = [tex]8.74 \frac{mg}{L} =8.74 * 10^{-3} \frac{g}{L}[/tex]
the molar concentration of benzoic acid = (mass concentration of benzoic acid) / (molar mass of benzoic acid )
molar concentration of benzoic acid = [tex](8.74 *10^{-3} \frac{g}{L} ) /(122.12\frac{g}{mol} )[/tex]
molar concentration of benzoic acid = [tex]7.157 * 10^{-5}M[/tex]
molar absorptivity (∈) of benzoic acid = (absorbance) /[ (molar concentration of benzoic acid ) × (path length) ]
∈ = (0.8) / [ ([tex]7.157 *10^{-5}M[/tex]) ×(1.00 cm)]
= [tex]1.12 * 10^{4} M^{-1} cm^{-1}[/tex]
The approximate absorbance of benzoic acid at 228 nanometers as described in the provided spectra plot is close to 0.8. The absorbance reveals the amount of light absorbed by the solution, which can further be used to determine the concentration of a compound.
Explanation:Based on the provided description of the absorbance spectra, the approximate absorbance of benzoic acid at 228 nanometers is close to 0.8. In many analytical procedures, the values of absorbance help determine the concentration of a compound.
It's important to understand that absorbance is a measure of the quantity of light absorbed by a solution, in this case, the solution of benzoic acid. The higher the absorbance, the more light is absorbed and less is transmitted through the solution. Each compound has a unique absorbance spectrum, serving like a 'fingerprint' which helps in identifying substances.
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If 3.52 L of nitrogen gas and 2.75 L of hydrogen gas were allowed to react, how many litres of ammonia gas could form? Assume all gases are at the same temperature and pressure.
Answer:
V NH3(g) = 1.833 L
Explanation:
balanced reaction:
N2(g) + 3H2(g) → 2NH3(g)assuming STP:
∴ V N2(g) = 3.52 L
∴ V H2(g) = 2.75 L
ideal gas:
PV = RTn∴ moles N2(g) = PV/RT
⇒ mol N2(g) = (1 atm)(3.52 L)/(0.082 atm.L/K.mol)(298 K)
⇒ mol N2(g) = 0.144 mol
∴ moles H2(g) = PV/RT
⇒ mol H2(g) = (1)(2.75)/(0.082)(298) = 0.113 mol (limit reagent)
∴ moles NH3(g) = (0.113 moles H2(g))(2 moles NH3 / 3 mol H2) = 0.075 mol
∴ V NH3(g) = RTn/P
⇒ V NH3(g) = ((0.082 atm.L/K.mol)(298 K)(0.075 mol))/(1 atm)
⇒ V NH3(g) = 1.833 L
Final answer:
From 3.52 L of nitrogen and 2.75 L of hydrogen, 1.83 L of ammonia can be produced, considering hydrogen as the limiting reactant based on the stoichiometry of the balanced chemical equation.
Explanation:
The question involves a stoichiometric calculation based on the reaction between hydrogen and nitrogen gases to form ammonia. Given the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g), we can determine how many litres of ammonia gas could form from 3.52 L of nitrogen gas and 2.75 L of hydrogen gas, assuming all gases are at the same temperature and pressure. Since the reaction consumes nitrogen and hydrogen in a 1:3 ratio to produce ammonia in a 2 moles product per 1 mole of nitrogen ratio, we first identify the limiting reactant. Here, hydrogen gas (H2) is the limiting reactant because we need 3 volumes of hydrogen for every volume of nitrogen, but we have less than that (2.75 L instead of 3.52*3 L). The amount of ammonia produced is therefore determined by the amount of hydrogen available. Since 3 volumes of H2 produce 2 volumes of NH3, 2.75 L of H2 would produce (2.75 L * (2/3)) = 1.83 L of NH3.
30 mL of 0.25 M acetic acid are titrated with 0.05 M KOH. What is the pH after addition of 75 mL KOH? Group of answer choices
Answer:
The PH of the mixture is 4.74
Explanation:
The number of millimoles of acetic acid is calculated using the formula:
No of millimoles= Molarity * Volume( in ml)
= 0.25M * 30ml = 7.5 moles
Number of millimoles of KOH is calculated using:
Number of millimoles = Molarity * Volume ( in ml)
=0.05M * 75ml
= 3.75 moles
The PH of the solution is derived using:
pH = pKa + log [salt] / acid
= [tex] -log [ 1.8 * 10^5 ] + log [ 3.75 mmoles/ 3.75 mmoles] [/tex]
=4.74
How many microliters of original sample are required to produce a final dilution of 10-2 in a total volume of 0.2 mL? 1 microliter is 10-6 L or 10-3 mL.
Answer:
The required volume of the original sample required is 2 micro liter
Explanation:
assuming the original sample concentration is 1N
after final dilution of 10-2 solution concentration becomes 0.01 N
normality of original sample = 1 N
normality of final solution = 0.01 N
volume of original sample= ?
volume of final solution = 0.2 mL
Considering thef formula below :
N1V1 = N2V2
V1 = (N2V2)/N1
= (0.01*0.2)/1
= 0.002 mL
1 milli liter = 1000 micro liter
0.002 mL = 2 micro liter
The original sample required is 2 micro liter
To produce a final dilution of 10^-2 in a total volume of 0.2mL, 2 microliters of original sample are required.
Explanation:To find out how many microliters of original sample are needed to generate a final dilution of 10^-2 in a total volume of 0.2 mL, you can use the formula: V1 = V2 × D2 / D1. Here, V1 is the volume of the original sample needed, V2 is the final volume required (which is 0.2 mL), D2 is the final dilution (which is 10^-2), and D1 is the original dilution (which is 1 for undiluted samples).
So, plugging these values into the formula gives: V1 = 0.2 mL × 10^-2 / 1 = 0.002 mL. Convert this volume from milliliters to microliters (1 mL = 1000 μL), so V1 = 0.002 mL * 1000 = 2 μL.
So, 2 microliters of original sample are required to produce a final dilution of 10^-2 in a total volume of 0.2 mL.
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What happens to the volume of a gas when you double the number of moles of gas while keeping the temperature and pressure constant?
Answer: The volume of the gas also gets double when number of moles are doubled.
Explanation:
The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.
The equation used to calculate number of moles is given by:
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
where,
[tex]V_1\text{ and }n_1[/tex] are the initial volume and number of moles
[tex]V_2\text{ and }n_2[/tex] are the final volume and number of moles
We are given:
[tex]n_2=2n_1[/tex]
Putting values in above equation, we get:
[tex]\frac{V_1}{n_1}=\frac{V_2}{2n_1}\\\\V_2=\frac{2n_2\times V_1}{n_1}=2V_2[/tex]
Hence, the volume of the gas also gets double when number of moles are doubled.
Final answer:
Doubling the number of moles of a gas while keeping temperature and pressure constant will double the volume of the gas, in accordance with Avogadro's law.
Explanation:
When we double the number of moles of gas while keeping the temperature and pressure constant, according to Avogadro's law, the volume of the gas also doubles. This is because Avogadro's law states that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas. Therefore, if the amount of gas is doubled, the volume also doubles, assuming the gas behaves ideally.
What is the mass of cyclohexane solvent, in kg, if 9.76 mL are used and the cyclohexane has a density of 0.779 g/mL?
Answer:
0.0076kg
Explanation:
To get the mass, we use the relation among density, mass and volume.
Mass = density * volume
Here mass? , density = 0.779g/ml , volume = 9.76ml
Mass = 9.76 * 0.779 = 7.60g
Answer is wanted in kg so we divide by 1000. This is 7.60/1000 = 0.0076kg
In addition to continuous radiation, fluorescent lamps emit sharp lines in the visible region from a mercury discharge within the tube. Much of this light has a wavelength of 436 nm. What is the energy (in J) of one photon of this light?
Answer:
[tex]4.56 x 10^{-19}J[/tex]
Explanation:
Electromagnetic radiations consist of quanta of energy called photons which have energy, E which is equal to:
E = hν.....................................(1)
where h is the Planck's constant which is [tex]6.626 x10^{-34}Js[/tex] and ν is the frequency of light radiation.
But ν = c/λ ....................................(2)
Putting equation (2) into (1), we have
E = hc/λ..........................................(3)
c is the speed of light (c =[tex]3 x 10^{8}m/s[/tex]) while λ is the wavelength of light.
Wavelength λ = 436nm = [tex]436 x 10^{-9}m[/tex]
Therefore the energy E of one photon of this light, using equation (3) is
[tex]E=\frac{6.626 x 10^{-34} x3 x10^{8} }{436 x 10^{-9} } = 4.56 x 10^{-19} J[/tex]
The energy of one photon of light with a wavelength of 436 nm is 4.546 × 10^-19 J.
Explanation:To calculate the energy of one photon, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 × 10-34 J·s), c is the speed of light (3.00 × 108 m/s), and λ is the wavelength.
Plugging in the given wavelength of 436 nm (which is equal to 4.36 × 10-7 m) into the equation, we have:
E = (6.626 × 10-34 J·s)(3.00 × 108 m/s) / (4.36 × 10-7 m) = 4.546 × 10-19 J
Therefore, the energy of one photon of light with a wavelength of 436 nm is 4.546 × 10-19 J.
Today's demand curve for gasoline could shift in response to a change ina.today's price of gasoline. b.the expected future price of gasoline. c.the number of sellers of gasoline. d.All of the above are correct.
Answer:
d.All of the above are correct.
Explanation:
The curve of demand moves left or right continuously. Income, patterns and preferences, related products prices as well as the population size and composition are the key factors causing demand change.
The demand curve for gasoline can shift in response to changes in price, expected future price, and the number of sellers. Option b is the correct option.
Explanation:The correct answer is d. All of the above are correct. The demand curve for gasoline reflects the relationship between the price and quantity demanded, so a change in the price of gasoline can shift the curve. The expected future price of gasoline can also influence current demand, as consumers may adjust their purchasing behavior based on their expectations. Additionally, the number of sellers of gasoline can impact market supply, which in turn affects the equilibrium price and quantity.
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Calculate the moles of O atoms in 0.658 g of Mg(NO3)2.
Answer:
There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.
Explanation:
Mass of magnesium nitrate = m = 0.658 g
Molar mass of magnesium nitrate = M = 148 g/mol
Moles of magnesium nitrate = n
[tex]n=\frac{m}{M}=\frac{0.658 g}{148 g/mol}=0.004446 mol[/tex]
1 mole of magnesium nitrate has 6 moles of oxygen atoms. Then 0.004446 moles magnesium nitrate will have :
[tex]6\times 0.004446 mol=0.026676 mol\approx 0.0267[/tex]
There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.
Beta (β) sheets are a type of secondary structure in proteins. A segment of a single chain in an antiparallel β sheet has a length of 80.5 Å . How many residues are in this segment?
Answer:
Explanation:
The structural repeating unit of beta sheet is 7 anstrom/2 aminoacids. So,
[tex]\frac{80.5 Angstrom}{3.5} = 23 aminoacids[/tex]
If a segment of a single chain in an antiparallel beta sheet has a length of 80.5 angstrom, then there will be 23 residues in this segment.
To find the number of residues in an 80.5 Å long segment of an antiparallel beta-pleated sheet, divide the total length by the length of one residue (3.5 Å per residue), which gives approximately 23 residues.
Explanation:The student asked how many residues are in a segment of a single chain in an antiparallel beta-pleated sheet with a length of 80.5 Å. To determine the number of amino acid residues in the segment, we can use the typical amino acid residue length in a ß-pleated sheet, which is approximately 3.5 Å per residue in an extended conformation. This measurement considers the distance hydrogen bonds can span between the carbonyl oxygen and amino hydrogen along the peptide chain in the secondary structure.
By dividing the total length of the chain (80.5 Å) by the length of one residue (3.5 Å), you can calculate the number of residues:
Number of residues = Total length ÷ Length per residue
Number of residues = 80.5 Å ÷ 3.5 Å/residue
Number of residues ≈ 23 residues
This calculation does not account for slight variations that may occur in different proteins or specific contexts, but it provides a general estimate for the number of amino acids in a segment of a beta sheet.
A 50.0 mL sample of waste process water from a food processing plant weighing 50.1 g was placed in a properly prepared crucible that had a tare weight of 17.1234 g. After drying in a 103oC oven it weighed 19.9821 g. After the crucible was placed in a 550oC oven it weighted 18.7777 g. What are the TS and VS (in mg/L)?
Answer:
TS (total solid, mg/L) = 57,174 mg/L
VS (volatile solid, mg/L) = 24,088 mg/L
Explanation:
First step to solve this problem is to know data and questions:
Data:
Sample volume = 50 mL
Sample weight = 50.1 g
P1 (weight empty crucible) = 17.1234 g
P2 (weight after drying a 103º in oven) = 19.9821 g
P3 (weight after incinatrion 550º) = 18.7777 g
Questions:
TS = ?
VS = ?
Formula:
We need to use formulas to calculate total solid (TS) and volatile solid (VS), these are:
[tex]TS =\frac{(P2-P1)*\frac{1,000 mg}{1g} }{Sample volumen (L)}[/tex]
[tex]VS = \frac{(P2-P3)*\frac{1,000 mg}{1g} }{Sample volume (L)}[/tex]
We have to transform mL to L so we will divide mL by 1,000 the sample volume:
[tex]Sample Volumen (L) = 50 mL * \frac{1L}{1,000 mL} = 0.05 L[/tex]
In the formula the value of 1,000 results by the convertion factor to transform grams to miligrams (we have to multiplie by 1,000)
Now we need to replace data on previous formulas and we will get TS and VS expressed in mg/L:
[tex]TS = \frac{(19.9821 g-17.1234g)*\frac{1,000 mg}{1g} }{0.05L} = 57,174mg/L[/tex]
[tex]VS = \frac{(19.9821g-18.7777g)*\frac{1,000mg}{1g} }{0.05L}=24,088 mg/L[/tex]
We divide TS and VS formula by sample volume because the exercise is asking us to express the results in mg/L.
Calculate the molality of a solution containing 100.7 g of glycine (NH2CH2COOH) dissolved in 3.466 kg of H2O.
Answer : The molality of a solution is, 0.387 mol/kg
Explanation : Given,
Mass of solute (glycine) = 100.7 g
Mass of solvent (water) = 3.466 kg
Molar mass of glycine = 75.1 g/mole
Formula used :
[tex]Molality=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent}}[/tex]
Now put all the given values in this formula, we get:
[tex]Molality=\frac{100.7g}{75.1g/mol\times 3.466kg}[/tex]
[tex]Molality=0.387mol/kg[/tex]
Thus, the molality of a solution is, 0.387 mol/kg
Answer:
The molality of the solution is 0.387 molal
Explanation:
Step 1: Data given
Mass of glycine = 100.7 grams
Mass of H2O = 3.466 kg
Molar mass of glycine = 75.07 g/mol
Step 2: Calculate moles glycine
Moles glycine = mass glycine / molar mass glycine
Moles glycine = 100.7 grams / 75.07 g/mol
Moles glycine = 1.341 moles
Step 3: Calculate molality of the solution
Molality = moles glycine / mass H2O
Molality = 1.341 moles / 3.466 kg
Molality = 0.387 molal
The molality of the solution is 0.387 molal
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3 and the density of alcohol is 790 kg/m3.
Here is the full question
Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.
Express your answer in two significant figures and include the appropriate units (in cm)
Answer:
ΔH ≅ 3.73 cm
Explanation:
The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:
P = ρ × g × h
where;
ρ is the density of the fluid
g is the gravitational constant
h is the height from the surface
From the question above;
For glycerine; we have:
density of glycerine = 1260 kg/m³
gravitational constant = 9.8 m/s²
height = ???
∴
[tex]P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g[/tex] ----- equation (1)
On the other hand for alcohol:
density of alcohol is given as = 790 kg/m³
gravitational constant = 9.8 m/s²
height = 10 cm
∴
[tex]P_{(a)= 790kg/m^3*9.8m/s^2*10[/tex] ----------- equation (2)
if we equate equation 1 and 2 together; we have
[tex]P_{(g)= P_{(a)[/tex]
[tex]1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm[/tex]
Making [tex]h_g[/tex] the subject of the formula, we have :
[tex]h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}[/tex]
[tex]h_g[/tex] = 6.269 cm
The difference in the height denoted by ΔH can therefore be calculated as:
ΔH [tex]= H_a-H_g[/tex]
ΔH [tex]= 10cm - 6.269cm[/tex]
ΔH = 3.731 cm
ΔH ≅ 3.73 cm (to two significant figures)
Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooctane is 0.792 g/mLg/mL, what mass of CO2CO2 (in kilograms) is produced each year by the annual U. S. gasoline consumption of 4.6×1010L4.6×1010L?
Answer:
1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane
Explanation:
Let's state the combustion reaction:
C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O
Let's calculate the mass of isooctane that reacts.
Density = Mass / Volume
Density . Volume = Mass
First of all, let's convert the volume in L to mL, so we can use density.
4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL
0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g
This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction
3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol
Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide
Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8) = 2.55×10¹² moles of CO₂
Now, we can determine the mass of produced CO₂ by multipling:
moles . molar mass
2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂
If we convert to kg 1.12×10¹⁴ g / 1000 = 1.12×10¹¹ kg
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an atomic mass of 122.9042 u . Antimony has an average atomic mass of 121.7601 u . What is the percent natural abundance of each isotope?
Answer:
Percentage abundance of 121 Sb is = 57.2 %
Percentage abundance of 123 Sb is = 42.8 %
Explanation:
The formula for the calculation of the average atomic mass is:
[tex]Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})[/tex]
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, 121 Sb :
% = x %
Mass = 120.9038 u
For second isotope, 123 Sb:
% = 100 - x
Mass = 122.9042 u
Given, Average Mass = 121.7601 u
Thus,
[tex]121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042[/tex]
[tex]120.9038x+122.9042\left(100-x\right)=12176.01[/tex]
Solving for x, we get that:
x = 57.2 %
Thus, percentage abundance of 121 Sb is = 57.2 %
percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %
Considering the definition of atomic mass, isotopes and atomic mass of an element,
Definition of atomic mass AFirst of all, the atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.
Definition of isotopesThe same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.
Atomic mass of an elementOn the other hand, the atomic mass of an element is the weighted average mass of its natural isotopes. In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.
Percent natural abundance of each isotope of antimonyIn this case, 121 Sb and 123 Sb are the naturally isotopes of antimony. 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an atomic mass of 122.9042 u.
Being the average atomic mass of antomony 121.7601 u, the average mass of antimony can be calculated as:
average mass of antimony= percent natural abundance of 121 Sb× atomic mass of 121 Sb + percent natural abundance of 123 Sb× atomic mass of 123 Sb
Being percent natural abundance of 123 Sb= 1 - percent natural abundance of 121 Sb, and substituting the corresponding values, you get:
121.7601 u= percent natural abundance of 121 Sb× 120.9038 u + (1 - percent natural abundance of 121 Sb)× 122.9042 u
Solving:
121.7601 u= percent natural abundance of 121 Sb× 120.9038 u + 1× 122.9042 u - percent natural abundance of 121 Sb× 122.9042 u
121.7601 u= percent natural abundance of 121 Sb× (-2.0004 u) + 122.9042 u
121.7601 u - 122.9042 u= percent natural abundance of 121 Sb× (-2.0004 u)
-1.1441= percent natural abundance of 121 Sb× (-2.0004 u)
-1.1441÷ (-2.0004 u) = percent natural abundance of 121 Sb
0.5719 ×100 % = percent natural abundance of 121 Sb
57.19% = percent natural abundance of 121 Sb
So:
percent natural abundance of 123 Sb= 1 - percent natural abundance of 121 Sb
percent natural abundance of 123 Sb= 1 - 0.5719
percent natural abundance of 123 Sb= 0.4281
percent natural abundance of 123 Sb= 0.4281× 100
percent natural abundance of 123 Sb= 42.81%
Finally, the percent natural abundance of 121 Sb is 57.19% and the percent natural abundance of 123 Sb is 42.81%
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Functional groups confer specific chemical properties to the molecules of which they are a part. In this activity, you will identify which compounds exhibit certain chemical properties as well as examples of those six different compounds.
Answer:
This question is incomplete, here is the complete question:
Functional groups confer specific chemical properties to the molecules of which they are a part. In this activity, you will identify which compounds exhibit certain chemical properties as well as examples of those six different compounds.
(a) alcohol: is highly polar and may act as a weak acid (OH)
(b) carboxylic acid: acts as an acid (COOH)
(c) aldehyde: may be a structural isomer of a ketone (C=O)
(d) thiol: forms disulfide bonds (S-H)
(e) amine: acts as a base (NH2)
(f) organic phosphate: contributes negative charge (PO4)
Explanation:
Functional groups are groups of atoms attached to the carbon skeleton of an organic compounds and confers some particular properties. For example:
(a) alcohol: is highly polar and may act as a weak acid (OH)
. In this functional group, oxygen is very electronegative and tend to attract electrons. Water is attracted to this group and because of this, compounds that have OH group will disolve in water. Organic compounds with this functional group are alcohols and carbohydrates for example.
(b) carboxylic acid: acts as an acid (COOH)
. The two electronegative oxygens of this group keep electrons attached to them and do not share them with tha Hydrogen atom. Because of this, the hydrogen atom can dissociates from the molecule, giving an acid.
(c) aldehyde: may be a structural isomer of a ketone (C=O)
. As in the case of -OH group, oxygen of this group is very electronegative. When the group is presents at the end of the carbon skeleton we cal it aldehyde, while when it is in another part of carbon structure we call it ketone.
(d) thiol: forms disulfide bonds (S-H)
. When two thiol groups (S-H) are oxidazed, losting their hydrogen atoms, they form disulfide bonds that gives stability to the molecules that contains them. An example of this is found in proteins when two aminoacids with thiol groups (cysteine, for example) interact and form disulfide bonds that stabilizes the protein.
(e) amine: acts as a base (NH2)
. The nitrogen atom of this groups is slightly electronegative so it tends to pick up hydrogen ions from the surroundings given rise to a base. This group i typicall of aminoacids (glycine, for example)
(f) organic phosphate: contributes negative charge (PO4). In this group, phosphorus easily changes its oxydation state, thus, giving rise to a compound with electronegative oxygens. An important biological molecule that contains this group is ATP
Functional groups are important components of organic molecules, conferring specific chemical properties to the molecules. Common functional groups include hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl. The presence of these groups can greatly affect the chemical properties and function of the molecule.
Explanation:Functional groups are vital components of organic molecules, conferring specific chemical properties upon the molecules in which they occur. These groups are attached to a 'carbon backbone' of chains or rings of carbon atoms, with possible substitutions of elements like nitrogen or oxygen. For example, the C-Cl portion of the chloroethane molecule is a functional group imparting specific chemical reactivity. The nature of the functional groups in an organic molecule are significant determinants of its chemical properties.
Functional groups take part in particular chemical reactions. Some of the important functional groups include hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl. These functional groups may attach to the carbon backbone at various points along the chain or ring structure. The presence of these groups majorly contributes to the varying chemical properties of different organic molecules and their functions in living organisms.
For example, carboxyl groups (-COOH) confer acidic properties to a molecule, while amine groups (NH₂) make a molecule more basic. Molecules with identical molecular formulas may have different structures due to variations in their functional groups, morphing into different types of isomers.
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A thermos contains 80.0g of water at 23.4 degrees C. Suppose 0.200 moles of KCl are dissolved in the water. What will be the final temperature of the solution? Assume that there is no energy transfer between the solution and the thermos, and that the specific heat is 4.184J/g*degrees C. Also, the delta H of solvation for KCl at 25 degrees C is 17.1 kJ/mol.
Answer:
32.04°C will be the final temperature of the solution.
Explanation:
Moles of potassium chloride = 0.200 mol
MAs sof KCl= 0.200 mol × 74.5 g/mol= 14.9 g
Enthalpy of solvation of potassium nitrate =
[tex]\Delta H_{solv}=17.1 kJ/mol[/tex]
Energy released when 0.200 moles of KCl is dissolved in water = Q
[tex]Q=17.1kJ/mol\times 0.200 mol=3.42 kJ=3420 J[/tex]
(1 kJ = 1000 J)
Heat released on dissolving 0.200 moles of KCl is equal to heat absorbed by water = Q
Mass of solution , m= 80.0 g +14.9 g = 94.9 g
Specific heat of water = c = 4.184 J/g°C
Initial temperature of the water = [tex]T_1=23.4^oC[/tex]
Final temperature of the water = [tex]T_2=?[/tex]
[tex]Q=m\times c\times (T_2-T_1)[/tex]
[tex]3420 J=94.9g\times 4.184 J/g^oC\times (T_2-23.4^oC)[/tex]
[tex]T_2=32.04^oC[/tex]
32.04°C will be the final temperature of the solution.
You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label making it impossible to read. You know the concentration is below 1.0 M as you can make out "0.xxx" at the beginning of the label. In order to determine the concentration, you decide to precipitate out the lead in the solution as PbSO4.
If you added 1.0 mL of the unknown Pb(NO3)2 to a test tube, what is the amount of H2SO4 in mL you will need to add to be sure the H2SO4 is the excess reagent?
Answer:
Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL
Explanation:
Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃
For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.
Knowing that Concentration in mol/L = (number of moles)/(volume in L)
Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole
According to the reaction,
1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄
0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄
Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄
So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.
Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)
Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.
QED!!!
Assuming concentrated H₂SO₄ is used for the precipitation, the minimum volume of concentrated H₂SO₄ required to be the excess reagent is 0.0556 mL.
What volume of H₂SO₄ is required for H₂SO₄ to be in excessThe equation of the reaction between H₂SO₄ and Pb(NO₃)₂ is given below:
Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃Assuming that the concentration of Pb(NO₃)₂ is 1.0M.
Number of moles of Pb(NO₃)₂ in 1.0 mL solution is calculated as follows:
Number of moles = molarity * volume in LVolume of Pb(NO₃)₂ = 1.0 mL = 0.001 L
Number of moles of Pb(NO₃)₂ = 1.0 * 0.001
Number of moles of Pb(NO₃)₂ = 0.001 moles
From the equation of reaction:
1 mole of Pb(NO₃)₂ reacts with 1 mole of H₂SO₄
0.001 mole of Pb(NO₃)₂ will react with 0.001 moles of H₂SO₄
Therefore number of moles H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄
Assuming the H₂SO₄ required is concentrated H₂SO₄:
Concentration of concentrated H₂SO₄ = 18.0M
The volume of concentrated H₂SO₄ required is calculated as follows:
Volume = moles/molarityVolume of H₂SO₄ = 0.001/18
Volume of H₂SO₄ = 0.0000556 L
Volume of concentrated H₂SO₄ required = 0.0556 mL.
Therefore, the minimum volume of concentrated H₂SO₄ required to be the excess reagent is 0.0556 mL.
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