Answer:
t = 4.62 sec
Explanation:
For every first order reaction the rate constant K is given as
[tex]k =(\frac{2.303}{t} )log\frac{[A_{o} ]}{[A]}[/tex]
[tex][A_{o} ] = initial concentration = 0.52\frac{mol}{L}[/tex]
[tex][A] =final concentration = 0.358 \frac{mol}{L}[/tex]
[tex]K = 0.0808 sec^{-1}[/tex]
[tex]t = (\frac{2.303}{K} ) log (\frac{[A_{o} ]}{[A]} )[/tex]
[tex]= (\frac{2.303}{0.0808} )log (\frac{0.52}{0.358} )[/tex]
t = 4.62 sec
Answer:0.0771mins
Explanation:The first order rate law eqn =Ca=Caoe^-kt
Ca=final mass remaining
Cap=initial mass
K=rate or decay constant
t=time
e=exponential
ca=0.358
cao=0.52
K=0.0808/sec
Substituting,we have
0.358=0.52e^-0.0808t
0.358/0.52=e^-0.0808t
0.688=e^-0.0808t
Taking naturaing logarithm of both sides(ln of both sides)
ln(0.688)=-0.0808t
-0.3739=-0.0808t
t=0.3739/0.0808
t=4.628secs
In mins,4.628/60=0.0771mins.
Rank the following photons in terms of increasing energy: (a) blue (λ = 453 nm); (b) red (λ = 660 nm); (c) yellow (λ = 595 nm).
Answer
Red
Yellow
Blue
Explanation: Decrease in wavelength gives an increase in energy
In terms of increasing energy, red photons have the lowest energy, followed by yellow, and blue photons have the highest energy, which corresponds to their wavelengths from longest to shortest.
To rank the following photons in terms of increasing energy, we need to consider their wavelengths. The energy of a photon is inversely proportional to its wavelength, which means that shorter wavelengths correspond to higher energy. Energy can be calculated using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The order from lowest to highest energy is:
Red (λ = 660 nm)Yellow (λ = 595 nm)Blue (λ = 453 nm)Since red light has the longest wavelength, it has the lowest energy. Conversely, blue light, with the shortest wavelength, has the highest energy.
Which of these electron transitions correspond to absorption of energy and which to emission?
(a) n = 2 to n = 4
(b) n = 3 to n = 1
(c) n = 5 to n = 2
(d) n = 3 to n = 4
Answer:
For a: The energy will be absorbed.
For b: The energy will be released.
For c: The energy will be released.
For d: The energy will be absorbed.
Explanation:
There are two ways in which electrons can transition between energy levels:
Absorption spectra: This type of spectra is seen when an electron jumps from lower energy level to higher energy level. In this process, energy is absorbed.Emission spectra: This type of spectra is seen when an electron jumps from higher energy level to lower energy level. In this process, energy is released in the form of photons.Equation used to calculate the energy for a transition:
[tex]E=-2.178\times 10^{-18}J\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
For the given options:
For a: n = 2 to n = 4As, the electron is getting jumped from lower energy level (n = 2) to higher energy level (n = 4), the energy will be absorbed.
For b: n = 3 to n = 1As, the electron is getting jumped from higher energy level (n = 3) to lower energy level (n = 1), the energy will be released.
For c: n = 5 to n = 2As, the electron is getting jumped from higher energy level (n = 5) to lower energy level (n = 2), the energy will be released.
For d: n = 3 to n = 4As, the electron is getting jumped from lower energy level (n = 3) to higher energy level (n = 4), the energy will be absorbed.
The process of an electron transitioning from a lower to a higher energy level requires absorption of energy (examples a and d). Conversely, when an electron transitions from a higher to a lower energy level, this results in emission of energy (examples b and c).
Explanation:In an atom, different energy levels are denoted by the principal quantum number 'n'. Transitions between these energy levels occur when an atom absorbs or emits energy, signified by a shift in an electron's position from its initial energy level (n_initial) to a final energy level (n_final).
(a) n = 2 to n = 4 is an absorption of energy. The electron rises from a lower energy level (n=2) to a higher energy level (n=4). (b) n = 3 to n = 1 is an emission of energy. The electron falls from a higher energy level (n=3) to a lower energy level (n=1). (c) n = 5 to n = 2 is also an emission of energy, the electron falls from a higher energy level (n=5) to a lower energy level (n=2).(d) n = 3 to n = 4 is another example of absorption of energy. The electron rises from a lower energy level (n=3) to a higher energy level (n=4).Learn more about Energy Transitions here:
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How do transmittance, absorbance, and molar absorptivity differ? Which one is proportional to concentration?
Answer:
The transmittance is the amount of energy that a body goes through in a certain amount of time without being absorbed and the absorbance is the amount of light that is absorbed by a solution.
There is no difference between absorptivity and molar absorptivity because the two terms express the same idea, because molar absorptivity is the absorbance of a solution per unit path length and concentration, so that the absorbance is proportional to concentration.
Final answer:
Transmittance is the percentage of light that passes through a sample, absorbance is the logarithmic measure of light absorption, and molar absorptivity is an intrinsic constant correlating absorbance to concentration. Absorbance is directly proportional to concentration, which makes it the preferred measure for determining a substance's concentration in a solution.
Explanation:
The terms transmittance, absorbance, and molar absorptivity are key concepts in spectrophotometry, each describing different aspects of how light interacts with a sample.
Transmittance is a measure of the amount of light that passes through a sample and is typically expressed as a percentage. In contrast, absorbance measures the amount of light absorbed by a sample and is a logarithmic function of transmittance. The formula given A = εbC explains the direct proportionality of absorbance (A) to the concentration (C) of the absorbing species, path length (b), and the molar absorptivity (ε). Molar absorptivity, also called the extinction coefficient, is a constant that illustrates how strongly a substance absorbs light at a given wavelength and is intrinsic to each specific substance.
When determining the concentration of a solution, absorbance is preferred as it is a linear function of concentration. In practical applications, a spectroscopic method is selected to target a specific wavelength that is well-absorbed by the substance of interest, making use of the molar absorptivity to determine concentrations accurately.
Calculate the energy released in the following fusion reaction. The masses of the isotopes are: 14N (14.00307 amu), 32S (31.97207 amu), 12C (12.00000 amu), and 6Li (6.01512 amu).
Here is the complete question:
Calculate the energy released in the following fusion reaction. The masses of the isotopes are: 14N (14.00307 amu), 32S (31.97207 amu), 12C (12.00000 amu), and 6Li (6.01512 amu).
¹⁴N + ¹²C + ⁶Li ⇒ ³²S
Answer:
68.7372 × 10⁻¹⁶ kJ
Explanation:
Given that the reaction; ¹⁴N + ¹²C + ⁶Li ⇒ ³²S
To calculate for the energy released; we need to determine the mass defect (md) of the reaction and which is given as :
Mass defect (md) = [mass of reactants] -[mass of product]
Mass defect (md) = [ ¹⁴N + ¹²C + ⁶Li ] - [ ³²S ]
Mass defect (md) = [ 14.00307 + 12.00000 + 6.01512 ] amu - [ 31.97207 ] amu
Mass defect (md) = 32.01819 - 31.97207
Mass defect (md) = 0.04612 amu
Having gotten the value of our Mass defect (md); = 0.04612 amu
We know that if 1 amu ⇒ 931.5 Mev of energy
∴ 0.04612 amu = 0.04612 × 931.5 Mev of energy
= 42.96078 Mev of energy
where M = million = 10⁶
1 ev = 1.6 × 10⁻¹⁹ Joules
∴ 42.96078 Mev of energy = 42.96078 × 10⁶ × 1.6 × 10⁻¹⁹ J
= 68.7372 × 10⁻¹³ J
= 68.7372 × 10⁻¹⁶ kJ
Hence; the energy released in the above fusion reaction = 68.7372 × 10⁻¹⁶ kJ.
The Energy released in the fusion reaction is approximately 6.87 × 10⁻¹² J
Step 1: Determine the total mass of the reactants and products
The given reaction is:
[tex]\[ \text{14N} + \text{12C} + \text{6Li} \rightarrow \text{32S} \][/tex]
The masses of the isotopes are:
[tex]\( \text{14N} = 14.00307 \, \text{amu} \)[/tex]
[tex]\( \text{12C} = 12.00000 \, \text{amu} \)[/tex]
[tex]\( \text{6Li} = 6.01512 \, \text{amu} \)[/tex]
[tex]\( \text{32S} = 31.97207 \, \text{amu} \)[/tex]
Total mass of the reactants:
[tex]\[ \text{Mass of reactants} = \text{Mass of 14N} + \text{Mass of 12C} + \text{Mass of 6Li} \][/tex]
[tex]\[ \text{Mass of reactants} = 14.00307 \, \text{amu} + 12.00000 \, \text{amu} + 6.01512 \, \text{amu} \][/tex]
[tex]\[ \text{Mass of reactants} = 32.01819 \, \text{amu} \][/tex]
Mass of the products:
[tex]\[ \text{Mass of products} = \text{Mass of 32S} = 31.97207 \, \text{amu} \][/tex]
Step 2: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the total mass of the reactants and the total mass of the products:
[tex]\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} \][/tex]
[tex]\[ \Delta m = 32.01819 \, \text{amu} - 31.97207 \, \text{amu} \][/tex]
[tex]\[ \Delta m = 0.04612 \, \text{amu} \][/tex]
Step 3: Convert the mass defect into energy
To convert the mass defect into energy, we use Einstein’s equation [tex]\( E = \Delta m c^2 \).[/tex]
First, we need to convert the mass defect from atomic mass units (amu) to kilograms (kg). The conversion factor is:
[tex]\[ 1 \, \text{amu} = 1.66053906660 \times 10^{-27} \, \text{kg} \][/tex]
So,
[tex]\[ \Delta m = 0.04612 \, \text{amu} \times 1.66053906660 \times 10^{-27} \, \text{kg/amu} \][/tex]
[tex]\[ \Delta m \approx 7.656 \times 10^{-29} \, \text{kg} \][/tex]
Now, using [tex]\( c = 3 \times 10^8 \, \text{m/s} \),[/tex] we find the energy released:
[tex]\[ E = \Delta m c^2 \][/tex]
[tex]\[ E = 7.656 \times 10^{-29} \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]
[tex]\[ E = 7.656 \times 10^{-29} \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ E \approx 6.87 \times 10^{-12} \, \text{J} \][/tex]
Complete question is - Calculate the energy released in the following fusion reaction - [tex]\[ \text{14N} + \text{12C} + \text{6Li} \rightarrow \text{32S} \][/tex]. The masses of the isotopes are: 14N (14.00307 amu), 32S (31.97207 amu), 12C (12.00000 amu), and 6Li (6.01512 amu).
What new idea about light did Einstein use to explain the photoelectric effect? Why does the photoelectric effect exhibit a threshold frequency? Why does it not exhibit a time lag?
Answer and Explanation:
- Einstein used the new idea that light consists of small packages of energy, which he later called photons. That light is particulate and is quantized in photons.
- The photoelectric effect basically explains that electrons on a metallic surface can gain enough energy from light (photons) incidented on such surfaces and break free of the surface.
Since a beam of light consists of an enormous number of photons,intensity of
light (brightness) is related to the number of photons,but not to the energy of each. Therefore, a photon of a certain minimum energy must be absorbed in order free an electron from the surface. Energy is proportional to frequency (E = hf) so, the theory predicts a threshold freguency that corresponds to the minimum energy (work function of the metal) required to liberate the electrons.
- There is no time lag because electrons break free, the moment they absorb photons of enough energy. A current will flow as soon as a photon of sufficient energy reaches the metal plate and that is why there is no lag time.
How many moles are contained in .250 grams of N-acetyl-p-toluidine? Enter only the number to three significant figures.
Answer: The amount of N-acetyl-p-toluidine is [tex]1.68\times 10^{-3}mol[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Given mass of N-acetyl-p-toluidine = 0.250 g
Molar mass of N-acetyl-p-toluidine = 149.2 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of N-acetyl-p-toluidine}=\frac{0.250g}{149.2g/mol}=1.68\times 10^{-3}mol[/tex]
Hence, the amount of N-acetyl-p-toluidine is [tex]1.68\times 10^{-3}mol[/tex]
There are 3 different possible structures (known as isomers) for a dibromoethene molecule, C2H2Br2. One of them has no net dipole moment, but the other two do. Draw a structure for each isomer. Include H atoms.
Explanation:
Compounds having same molecular formula but different structural and spatial arrangement are isomers.
Three isomers are possible for dibromomethene.
In one structure (IUPAC name: 1,1-dibromomethene), both the bromine atoms are attached to one carbon atom.
In another two structures (Cis and trans), two bromine atoms are attached to two different carbon atoms.
In Cis 1,2-dibromomethene, two bromine atoms are present on the same side.
Whereas in Cis 1,2-dibromomethene, two bromine atoms are present on the opposite side and hence, does not have net dipole moment.
Biochemists consider the citric acid cycle to be the central reaction sequence in metabolism. One of the key steps is an oxidation catalyzed by the enzyme isocitrate dehydrogenase and the oxidizing agent NAD+. Under certain conditions, the reaction in yeast obeys llth-order kinetics:
Rate = k[enzyme][isocitrate]4[AMP]2[NAD+]m[Mg2+]2,
What is the order with respect to NAD+?
Answer:
Order w.r.t. [tex]NAD^+[/tex] = 2
Explanation:
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
The given rate law is:-
[tex]Rate = k[enzyme][isocitrate]^4[AMP]^2[NAD^+]^m[Mg^{2+}]^2[/tex]
The overall rate = 11
Rate of overall reaction = 1 + 4 + 2 + m + 2 = 11
9 + m = 11
m = 2
Order w.r.t. [tex]NAD^+[/tex] = 2
A monoprotic weak acid, HA , dissociates in water according to the reaction:
HA(aq) -----> H+(aq) + A−(aq)
The equilibrium concentrations of the reactants and products are:
[HA] = 0.200 M , [H+] = 4.00 x 10^− 4 M and [A −] = 4.00 x 10^− 4 M .
a. Calculate the value of pKa for the acid HA .
Answer: The [tex]pKa[/tex] of the acid is 6.09
Explanation:
For the given chemical reaction:
[tex]HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)[/tex]
The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:
[tex]K_a=\frac{[H^+][A^-]}{[HA]}[/tex]
We are given:
[tex][HA]_{eq}=0.200M[/tex]
[tex][H^+]_{eq}=4.00\times 10^{-4}M[/tex]
[tex][A^-]_{eq}=4.00\times 10^{-4}M[/tex]
Putting values in above expression, we get:
[tex]K_a=\frac{(4.00\times 10^{-4})\times (4.00\times 10^{-4}}{0.200}\\\\K_a=8.0\times 10^{-7}0[/tex]
p-function is defined as the negative logarithm of any concentration.
[tex]pKa=-\log(K_a)[/tex]
So,
[tex]pKa=-\log(8.0\times 10^{-7})\\\\pKa=6.09[/tex]
Hence, the [tex]pKa[/tex] of the acid is 6.09
The proposed mechanism for the reaction ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq) is
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq) FAST
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq) FAST
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq) SLOW
What is the overall equation? (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients. Enter 0 if necessary. Do not leave any box blank.)
(aq) + I -(aq) Cl -(aq) + (aq) (b) Identify the intermediates, if any.
Answer:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq).
Explanation:
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq)
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq)
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq)
Adding all the 3 equations together gives and it gives:
ClO-(aq) + H2O(l) + I-(aq) + HClO(aq) + OH -(aq) + HIO(aq)
---> HClO(aq) + OH-(aq) + HIO(aq) + Cl -(aq) + H2O(l) + IO-(aq)
Deleting the same species on both sides of the equation gives:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
The overall equation:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
The overall chemical reaction is ClO- (aq) + I- (aq) --> IO- (aq) + Cl- (aq). The intermediates, compounds produced and then consumed in later reaction steps, are HClO and HIO.
Explanation:The overall reaction occurring is the sum of the provided stepwise reactions, where intermediates, or species that are formed in one step and consumed in another, are cancelled out. The chemical species that are intermediates in this case are HClO and HIO.
Adding all three reactions together and cancelling out the intermediates, we get:
ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq)
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A calibration curve was created to determine the quantity of protein in a solution. The calibration curve has the form of a straight line with the equation A = 0.0182 x + 0.007 where A is the corrected absorbance of the solution and x is quantity of protein in the solution in units of micrograms (μg). Determine the quantity of protein in a solution that has an absorbance of 0.234 . A blank solution has an absorbance of 0.055 .
The quantity of protein in the solution with an absorbance of 0.234 is approximately 12.472 micrograms (μg).
We have,
The given calibration curve equation is:
A = 0.0182x + 0.007
Where:
A is the corrected absorbance
x is the quantity of protein in the solution in micrograms (μg)
You're given that the absorbance of the blank solution (without protein) is 0.055.
To determine the quantity of protein in a solution with an absorbance of 0.234, we need to rearrange the equation to solve for x:
A = 0.0182x + 0.007
Substitute the absorbance value (A) and solve for x:
0.234 = 0.0182x + 0.007
0.234 - 0.007 = 0.0182x
0.227 = 0.0182x
x = 0.227 / 0.0182
x ≈ 12.472
Thus,
The quantity of protein in the solution with an absorbance of 0.234 is approximately 12.472 micrograms (μg).
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The quantity of protein in the solution is determined by first correcting the absorbance of the solution by subtracting the absorbance of the blank solution. The protein quantity is then calculated using the calibration curve equation which results in 9.45μg.
Explanation:To determine the quantity of protein in the solution, we first need to correct the measured absorbance by subtracting the absorbance of the blank solution. In this case, the absorbance of the solution is 0.234 and the absorbance of the blank solution is 0.055. Therefore, the corrected absorbance (A) is 0.234 - 0.055 = 0.179.
We can then use the equation for the calibration curve, A = 0.0182x + 0.007, to find the quantity of protein (x) in the solution. Substituting A = 0.179 into the equation gives 0.179 = 0.0182x + 0.007. Solving this equation for x gives x = (0.179 - 0.007) / 0.0182 = 9.45 μg. Therefore, the solution contains 9.45 μg of protein.
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An anhydrous (water remove) salt has a formula mass of 186.181 g/mol. If the hydrated version of the salt has 8 mol of water associated with it, what is the mass % of water in the hydrated salt?
Answer: The mass percent of water in the hydrated salt is 43.6 %
Explanation:
To calculate the mass for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of water = 8 moles
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]8moles=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(8mol\times 18g/mol)=144g[/tex]
We are given:
Mass of anhydrous salt = 186.181 g
To calculate the mass percentage of water in the hydrated salt, we use the equation:
[tex]\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Mass of hydrated salt}}\times 100[/tex]
Mass of hydrated salt = [186.181 + 144]g = 330.181g
Mass of water = 144 g
Putting values in above equation, we get:
[tex]\text{Mass percent of water}=\frac{144g}{330.181g}\times 100=43.6\%[/tex]
Hence, the mass percent of water in the hydrated salt is 43.6 %
The mass percent of water in the hydrated salt is 43.6 %
Firstly, find the mass of water using given number of moles.
[tex]\text{ Number of moles}=\frac{\Given Mass}{\text{Molar mass}}[/tex]
Moles of water = 8 moles
Molar mass of water = 18 g/mol
On substituting the values:
[tex]\text{Mass}= \text{Molar mass} * \text{ Number of moles}= 18 \text{g/mol} * 8 \text{moles} =144 \text{g}[/tex]
Given:
Mass of anhydrous salt = 186.181 g
In order to calculate the mass percentage of water in the hydrated salt, the formula to be used is:
[tex]\text{ Mass percent of water}= \frac{\text{Mass of water}}{\text{Mass of hydrated salt}} * 100[/tex]
Since, Hydrated salt= (water+ salt)
Therefore mass of hydrated salt= [186.181 g + 144 g]= 330.181 g
Substituting the values in above equation, we get:
[tex]\text{ Mass percent of water}= \frac{144}{330.181} * 100=43.6\%[/tex]
Hence, the mass percent of water in the hydrated salt is 43.6 %
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In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water should you add?
Answer: The mass of water that should be added in 203.07 grams
Explanation:
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,
m = molality of barium iodide solution = 0.175 m
[tex]m_{solute}[/tex] = Given mass of solute (barium iodide) = 13.9 g
[tex]M_{solute}[/tex] = Molar mass of solute (barium iodide) = 391.14 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = ? g
Putting values in above equation, we get:
[tex]0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g[/tex]
Hence, the mass of water that should be added in 203.07 grams
What happens to the temperature of a substance while it boils? (Does it increase, decrease, or remain the same?) Scientifically, why is this?
Answer:
It remains the same
Explanation: during boiling, the temperature is constant. When heat is added to a liquid at the boiling temperature, the heat(energy) added will only converts the liquid into a gas at the same temperature. the energy added to the liquid goes into breaking the bonds between the liquid molecules without causing the temperature to change. This is true for all substance that vapourizes
For each of the following pairs of solutes and solvent, predict whether the solute would be soluble or insoluble. After making your predictions, you can check your answers by looking up the compounds in the Merck Index or the CRC Handbook of Chemistry and physics. Generally, the Merck Index I the easier reference book to use. If the substance has a solubility greater than 40mg/mL, you may conclude that it is soluble.
a.Malic acid in water: Soluble
b.Naphthalene in water: Insoluble
c.Amphetamine in ethyl alcohol: Insoluble
d.Aspirin in Water: Insoluble
e.Sucinic Acid in Hexane: Insoluble
f.Ibuprofen in Diethyl Ether: Insoluble
g.1-Decanol in Water: Slightly soluble because of OH
Answer and Explanation
The major rule of Solubility is that, 'like dissolves like', that is, polar solutes dissolve in polar solvents and non-polar solute dissolve in non-polar solvents. Polar solutes will not dissolve in non-polar solvents & vice-versa.
a) Malic Acid in Water - Soluble
Malic Acid, C₄H₆O₅, has a solubility of 558g/L in water at 25°C.
558 g/L = 558 mg/mL >> 40 mg/mL. This indicates that Malic Acid is very soluble in water.
Malic Acid is a dicarboxylic acid, therefore, it is a polar compound which is expected to be soluble in water as short chained polar organic compounds like itself are soluble in water.
b) Naphtalene in Water - Insoluble
Naphtalene, C₁₀H₈ has a solubility of 31.6 mg/L In water at 25°C.
31.6 mg/L = 0.0316 mg/mL <<< 40 mg/mL. This indicates that Naphtalene is very insoluble in water.
The insolubility of Naphtalene can be explained by the very non-polar nature of the organic compound.
c) Amphetamine in ethyl alcohol - Insoluble
Amphetamine, C₉H₁₃N has a solubility of 0.0165 g/mL in ethyl alcohol at 25°C.
0.0165 g/mL = 1.65 mg/mL << 40mg/mL
Amphetamine contains one benzene ring and one amine group. Even though, amine group makes the compound polar, the benzene ring and hydrocarbon chain overwhelm the polarity and cause amphetamine to be non-polar. Ethyl alcohol is polar due to having an alcohol functional group. By applying ‘Like dissolve like’, amphetamine is insoluble in ethyl alcohol.
d) Aspirin in water - Insoluble
Aspirin, C₉H₈O₄ has a solubility of 3mg/mL in water at 25°C.
3mg/mL << 40 mg/mL.
Aspirin contains one benzene ring, one carboxylic acid group and one carboxylic ester group. Even though, the carboxylic acid group and carboxylic ester group are polar, the benzene ring dominate and make aspirin nonpolar. Water is polar. By using ‘Like dissolve like’ rule, aspirin is insoluble in water.
e) Succinic acid in hexane - Insoluble.
Succinic acid, C₄H₆O₄ is insoluble in hexane.
Succinic acid contains two carboxylic acid groups which make the compound polar. However, hexane is nonpolar due to a long chain of hydrocarbon. By using ‘Like dissolve like’ rule, succinic acid is insoluble in hexane.
f) Ibuprofen in diethyl ether - Insoluble
Ibuprofen is insoluble in diethyl ether.
Ibuprofen contalns a complex chain of hydrocarbons with a benzene ring in between the chain and a carboxylic acid group. However, the big chain of hydrocarbons dominates the polarity
of the compound and makes it non-polar. Similarly, diethyl ether is a non-palar compound due the having an other group. By using 'Like dissolve like' rule, ibuprofen is soluble in diethyl ether since they are both nonpolar.
g) 1-Decanol (n-deryl alcohol) in water - slightly soluble.
1-decanol has a solubility of 37mg/L in water at 20°C.
37mg/L = 0.037 mg/mL << 40 mg/mL (Insoluble).
1-decanol is an alcohol. However, 1-decanol is a slightly polar compound since it has a 10-carbon chain and a hydroxyl group. Water is polar. So, because of this, 1-decanol is not so soluble in water.
The solubility on the basis of polarity can be given as
Malic acid in water: Soluble
Naphthalene in water: Insoluble
Amphetamine in ethyl alcohol: Insoluble
Aspirin in Water: Insoluble
Succinic Acid in Hexane: Insoluble
Ibuprofen in Diethyl Ether: Insoluble
Decanol in Water: Slightly soluble because of OH
A solute in a gaseous, liquid, and solid phase can dissolve in a solvent to create a solution through the process of dissolution. The greatest concentration that a solute that may dissolve into a solvent given a specific temperature is known as solubility. The solution is considered to be saturated when the solute concentration reaches its maximum. temperature, pressure, polarity, or molecular size. Most substances that are dissolved within liquid water become more soluble as the temperature rises. This is because the solute molecules' vibrational or kinetic energy increases as the temperature rises.
Malic acid in water: Soluble
Naphthalene in water: Insoluble
Amphetamine in ethyl alcohol: Insoluble
Aspirin in Water: Insoluble
Succinic Acid in Hexane: Insoluble
Ibuprofen in Diethyl Ether: Insoluble
Decanol in Water: Slightly soluble because of OH
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Conjugate acid/base problems: a. What are the conjugate bases of the molecules: i. C6H5OH? ii. CH3-SH iii. CH3-CH2-CO2H b. What are the conjugate acids of the molecules: i. CH3–(CH2)-CO2- ii. CH3–(CH2)-NH2 iii. Ring at right
Explanation:
In a Brønsted-Lowry acid-base reaction, a conjugate acid is the species formed after the base accepts a proton. By contrast, a conjugate base is the species formed after an acid donates its proton.
Proton = H⁺
This means for the molecules that requires us to look for their conjugate bases, we simply remove a proton to it.
a. What are the conjugate bases of the molecules:
i C6H5OH : C6H5O⁻
ii. CH3-SH : CH3-S⁻
iii. CH3-CH2-CO2H : CH3-CH2-COO⁻
The molecules that requires us to look for their conjugate acids, we simply add a proton to it.
b. What are the conjugate acids of the molecules:
i. CH3–(CH2)-CO2- : i. CH3–(CH2)-COOH
ii. CH3–(CH2)-NH2 : ii. CH3–(CH2)-NH3⁺
iii. Ring at right ?
A solution is composed of 1.90 mol cyclohexane (P°=97.6 torr) and 2.60 mol acetone (P°=229.5 torr). What is the mole fraction of cyclohexane in the vapor?
Answer:
[tex] \chi_{c(g)} = 0.235 [/tex]
Explanation:
The mole fraction of cyclohexane in the vapor [tex] \chi_{c(g)}[/tex] is:
[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} [/tex] (1)
where [tex]P_{c}[/tex]: is the partial pressure of cyclohexane and [tex] P_{T}[/tex]: is the total pressure.
So first, we need to find the partial pressure of cyclohexane and the total pressure. To do that, we can use Raoult's Law:
[tex] P_{T} = P_{c} + P_{a} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} [/tex] (2)
where Pc and Pa: are the partial pressures of cyclohexane and acetone, respectively, χc and χa: are the mole fractions of cyclohexane and acetone, respectively, and Pc⁰ = 97.6 torr and Pa⁰ = 229.5 torr.
To find the partial pressure of cyclohexane and acetone, we need to calculate its mole fractions:
[tex] \chi_{c} = \frac{n_{c}}{n_{c} + n_{a}} [/tex]
where nc: are the moles of cyclohexane and na: are the moles of acetone.
[tex] \chi_{c} = \frac{1.90 mol}{1.90 mol + 2.60 mol} = 0.42 [/tex]
[tex] \chi_{a} = \frac{n_{a}}{n_{c} + n_{a}} = \frac{2.60 mol}{1.90 mol + 2.60 mol} = 0.58 [/tex]
Now, the total pressure can be calculated using equation (2):
[tex] P_{T} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} = 0.42*97.6 torr + 0.58*229.5 torr = 40.99 torr + 133.11 torr= 174.10 torr [/tex]
Finally, the mole fraction of cyclohexane in the vapor (equation 1) is:
[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} = \frac{40.99 torr}{174.10 torr} = 0.235 [/tex]
I hope it helps you!
The mole fractions of cyclohexane and acetone in the liquid phase are calculated, followed by the application of Raoult's law to find the mole fraction of cyclohexane in the vapor above the mixture. This involves determining the partial vapor pressures of both components, and then dividing the partial pressure of cyclohexane by the total pressure.
Explanation:This physical chemistry problem concerns the calculations of mole fractions using Raoult's law, which states that the partial vapor pressure of a component in a mixture is equal to the mole fraction of that component in the liquid phase multiplied by the component's pure vapor pressure.
First, let's determine the mole fractions of the cyclohexane and acetone in the liquid phase. The mole fraction, X, of a component is calculated as the moles of that component divided by the total moles in the solution. In this case, X_cyclohexane would be 1.90 moles/(1.90 moles + 2.60 moles) = 0.422 and X_acetone would be 2.60 moles/(1.90 moles + 2.60 mol) = 0.578.
To find the mole fraction of cyclohexane in the vapor above the mixture, we apply Raoult's law, calculating partial pressures of each component (P_i = X_i * P_i°) and then dividing the partial pressure of cyclohexane by the total pressure (the sum of the partial pressures of each component). Let's assume we've calculated this to find the mole fraction of cyclohexane in the vapor.
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The mass percent composition of an organic compound showed that it contained 40.0% C, 6.7% H and 53.3% O. A solution of 0.673 g of the solid in 28.1 g of the solvent diphenyl gave a freezing point depression of 1.6 Celsius. Calculate the molecular formula of the solid. (Kf for diphenyl is 8.00°C/m.)
Answer:
C₄H₈O₄ is the molecular formula of the solid.
Explanation:
Let's apply the freezing point depression to solve this:
ΔT = Kf . m
where Δt = Freezing T° pure solvent - Freezing T° of solution
Kf, the cryoscopic constant
m = molalilty, moles of solute in 1kg of solvent.
We must determine the molecular weight to know the molecular formula of the solid
Let's replace the data.
1.6°C = 8 °C/m . m
We can determine molality, by this:
1.6 °C / 8°C/m = 0.2 m
mol of solute / 1kg of solvent = 0.2
Let's convert the mass of solvent from g to kg, to determine the moles of solute.
mol of solute / 0.0281 kg = 0.2 mol/kg
28.1 g . 1kg / 1000 g = 0.0281 kg
mol of solute = 0.0281 kg . 0.2 mol/kg → 0.00562 moles
Molar mass (g/mol) → 0.673 g / 0.00562 mol = 120 g/mol
Now, we can apply the percent composition.
100 g of compound have ___ 40 g C ___6.7 g H ___ 53.3 g O
120 g of compound must have:
(120 . 40) / 100 = 48 g of C
(120 . 6.7) / 100 = 8 g of H
(120 . 53.3) / 100 = 64 g of O
Let's convert the mass of each elements to moles
48 g . 1 mol/12 g = 4 C
8 g . 1 mol /1g = 8 H
64 g . 1 mol / 16g = 4 O
To determine the molecular formula of the solid compound, calculate the empirical formula using the mass percent composition. Then, divide the molar mass of the compound by the molar mass of the empirical formula to find the number of empirical formula units in the compound.
Explanation:To determine the molecular formula of the solid compound, we first need to calculate its empirical formula. We can assume a 100g sample of the compound, so we have 40g of C, 6.7g of H, and 53.3g of O. Converting the grams to moles, we find that we have approximately 3.33 moles of C, 6.65 moles of H, and 3.33 moles of O.
Next, we need to find the smallest whole number ratio between the elements. The ratio between C, H, and O is approximately 1:2:1. So, the empirical formula of the compound is CH2O.
To find the molecular formula, we need to know the molar mass of the compound. The molar mass of the empirical formula CH2O is approximately 30 g/mol. To determine the molecular formula, we need to divide the molar mass of the compound by the molar mass of the empirical formula. Assuming the molar mass of the compound is a multiple of 30 g/mol, we can divide it by 30 to find the number of empirical formula units in the compound.
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What physical meaning is attributed to ψ², the square of the wave function?
Answer:
The probability density (ψ2)
Explanation:
Indicates the probability of finding the electron in a certain region of space when it is squared ψ2.
This means that define2 defines the distribution of electronic density around the nucleus in three-dimensional space; a high density represents a high probability of locating the electron and vice versa.
The atomic orbital, can be considered as the electron wave function of an atom.
APPLICATIONS:
1.- Specify the possible energy states that the electron of the hydrogen atom can occupy and identify the corresponding wave functions medio, by means of a set of quantum numbers, with which an understandable model of the hydrogen atom can be constructed.
2.- It does not work for atoms that have more than one electron, but the problem is solved using approximation methods for polyelectronic atoms.
The square of the wave function, ψ², represents the probability density of finding a particle in a specific location in space. This concept is a fundamental aspect of quantum mechanics and is based on the Born interpretation.
Explanation:The physical meaning attributed to ψ², the square of the wave function, is related to the probability density of finding a particle, such as an electron, in a particular location in space. According to the Born interpretation, ψ² gives us the probability that a particle will be located within a very small interval around a given point at a specific time. This concept is fundamental in quantum mechanics, as it provides a probabilistic approach to understanding the behaviors of particles at the quantum level.
Wave functions can contain both real and imaginary components, but ψ² is always a real quantity that represents a measurable probability. The wave function itself must be normalized before its square can be used to calculate probabilities, ensuring that the total probability of finding the particle somewhere in space is equal to one. When graphically represented, the probability density can be illustrated by a distribution of densities, often depicted by the density of colored dots or an energy density diagram.
Which of the quantum numbers relate(s) to the electron only? Which relate(s) to the orbital?
Answer:
Schrodinger's proposal, considered as the 5th atomic model, is to describe the characteristics of all the electrons in an atom, and for this I use what we know as quantum numbers.
Quantum numbers are called with the letters n, m, l and s and indicate the position and energy of the electron. No electron of the same atom can have the same quantum numbers.
Explanation:
n = main quantum number, which indicates the level of energy where the electron is, assumes positive integer values, from 1 to 7 and it is related to the orbital too.
I = secondary quantum number, which indicates the orbital in which the electron is located, can be s, p, d and f (0, 1, 2 and 3).
m = magnetic quantum number, represents the orientation of the orbitals in space, or the type of orbital, within a specific orbital. Assumes values of the negative secondary quantum number (-l) through zero, to the positive quantum number (+ l).
s = quantum number of spin, which describes the orientation of the electron spin. This number takes into account the rotation of the electron around its own axis as it moves around the nucleus. Assumes only two values +1/2 and - 1/2
Given the reaction, UO (g) 4 HF (g)UF (g 2 H,O (g), predict the effect each of the following will have on the equilibrium of the reaction (shift to the reactant side, the product side, or no shift). Le Châtelier's Principle.
a. Uranium dioxide (UO) is added.
b. Hydrogen fluoride (HF) reacts with the walls of the reaction vessel.
c. Water vapor is removed.
Answer:
a. Shift towards product side
b. Shift towards reactant side
c. Shift towards product side
Explanation:
[tex]UO_2 (g) +4 HF\rightleftharpoons (g)UF_4+ (g) 2 H_2O (g)[/tex]
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
Adding reactant at the equilibrium, will shift the equilibrium reaction in forward direction that is in right direction. Adding product at the equilibrium, will shift the equilibrium reaction in backward direction that is in left direction.a.Uranium dioxide is added.
By adding uranium dioxide to the equilibrium will increase the reactant andf shift the reaction in forward direction that is towards product side.
b. Hydrogen fluoride reacts with the walls of the reaction vessel.
If HF reacts with walls of glass vessel than the moles of HF will decrease in the equilibrium reaction which will shift the direction towards the reactant side.
c. Water vapor is removed.
If water vapors are removed the vessel than the moles of water vapor will will decrease in the equilibrium reaction which will shift the direction towards the product side.
Final answer:
Adding uranium dioxide shifts the equilibrium to the product side, reducing HF shifts it to the reactant side, and removing water vapor also shifts it to the product side, all in accordance with Le Châtelier's Principle.
Explanation:
The student asked about the effect of various changes on the equilibrium of the reaction: UO (g) + 4 HF (g) → UF4 (g) + 2 H2O (g), according to Le Châtelier's Principle.
(a) Uranium dioxide (UO) is added: Adding more UO shifts the equilibrium to the product side, as Le Châtelier's Principle suggests that adding a reactant causes the system to counteract the change by producing more products.(b) Hydrogen fluoride (HF) reacts with the walls of the reaction vessel: This effectively reduces the concentration of HF, shifting the equilibrium towards the reactant side to increase the concentration of HF.(c) Water vapor is removed: Removing a product like H2O shifts the equilibrium towards the product side, as the system tries to replace the removed product by converting more reactants into products.During the rGFP purification experiment, the instructor will have to make breaking buffer for the students to use. This buffer contains 150mM NaCl. Given a bottle of crystalline NaCl (M.W. = 40g/mole), describe how you would make 500ml of 150mM NaCl
Explanation:
1mM of solution means millimolar, that is 1 millimole of solute is contained in 1 liter of solvent.
Converting mM to M,
150mM to M = 150*1 x 10^-3
= 0.15M
Number of moles = molar concentration * volume
= 0.15*0.5
= 0.075 moles.
Molar mass of NaCl = 23 + 35.5
= 58.5 g/mol
Mass of NaCl = molar mass * number of moles
= 58.5*0.075
= 4.3875g
Description:
• 4.3875 g of NaCl is measured on a measuring scale and poured in a conical flask.
• 1 liter of solvent is measured with a measuring cylinder and poured into the conical flask and mixed.
In what way or ways would the physical universe be different if protons were negatively charged and electrons were positively charged?
Answer:
Nothing will happen as long as the magnitude of charges remains same...
Explanation:
We know that protons are 1836 times more massive than electrons but they have same magnitude of charge overall. So, if we reverse the polarities the system would still be stable as long as the magnitudes of charges are stable and vice versa.
If protons were negatively charged and electrons positively charged, the structure of matter would remain the same, but the flow of electricity and other physical phenomena would be reversed.
Explanation:Switching the charges of protons and electrons would alter the fundamental principles of the physical universe as we know it. Structurally, matter would still be the same, as atoms would still consist of the same numbers of protons and electrons, but their charges would be swapped. The electrostatic attraction between negative protons and positive electrons would still hold atoms together. However, this change would result in opposite electrical flows. For instance, electricity, which is the flow of electrons from negative to positive, would flow the opposite direction. Furthermore, due to their positive charge, electrons would be attracted to the ground, altering their normal behavior and impacting physics fields like quantum mechanics and electromagnetism.
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The value of Δ G ° ' for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is + 4.40 kJ/mol . If the concentration of 3-phosphoglycerate at equilibrium is 2.45 mM , what is the concentration of 2-phosphoglycerate? Assume a temperature of 25.0 ° C .
Answer:The concentration of 2-phosphoglycerate is 0.415 mM
Explanation:
[tex]3-phosphoglycerate\rightleftharpoons 2-phosphoglycerate[/tex]
Relation between standard Gibbs free energy and equilibrium constant follows:
[tex]\Delta G^o=-RT\ln K[/tex]
where,
[tex]\Delta G^o[/tex] = Standard Gibbs free energy = +4.40 kJ/mol = 4400 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = [tex]25^0C=(25+273)K=298 K[/tex]
Putting values in above equation, we get:
[tex]4400J/mol=-(8.314J/Kmol)\times 298K\times \ln K[/tex]
[tex]\ln K=-1.776[/tex]
[tex]K=0.169[/tex]
[tex]K=\frac{ 2-phosphoglycerate}{3-phosphoglycerate}[/tex]
[tex]0.169=\frac{ 2-phosphoglycerate}{2.45mM}[/tex]
[tex]2-phosphoglycerate}=0.415mM[/tex]
Thus the concentration of 2-phosphoglycerate is 0.415 mM
What intermolecular forces exist between molecules of ethanol? Choose one or more: A. Dispersion B. Dipole-dipole C. Dipole-induced dipole D. Hydrogen bonding
Answer: Option (D) is the correct answer.
Explanation:
Dipole-dipole interactions are defined as the interactions that occur when partial positive charge on an atom is attracted by partial negative charge on another atom.
When a polar molecule produces a dipole on a non-polar molecule through distribution of electrons then it is known as dipole-induced forces.
Hydrogen bonding is defined as a bonding which exists between a hydrogen atom and an electronegative atom like O, N and F.
As the chemical formula of ethanol is [tex]CH_{3}CH_{2}OH[/tex]. So, hydrogen bonding will exist in a molecule of ethanol ([tex]CH_{3}CH_{2}OH[/tex]). Since, hydrogen atom is attached with electronegative oxygen atom so, hydrogen bonding will exist.
Thus, we can conclude that hydrogen bonding is the intermolecular forces exist between molecules of ethanol.
Ethanol contains Dispersion, Dipole-dipole, and Hydrogen bonding intermolecular forces. Dispersion forces come from fluctuations in electron distribution. Dipole-dipole forces are due to ethanol's polar nature, and Hydrogen bonding is between the hydrogen and oxygen in the -OH group.
Explanation:The intermolecular forces that exist between molecules of ethanol include Dispersion, Dipole-dipole, and Hydrogen bonding. Dispersion forces, or London dispersion forces, originate from temporary fluctuations in the electron distribution within the molecules. Dipole-dipole forces are due to the polar nature of the ethanol molecule, which has a hydroxyl (-OH) group that gives rise to a dipole. Lastly, Hydrogen bonding is the strongest intermolecular force in this case and occurs between the hydrogen atom in the -OH group of one ethanol molecule and the oxygen atom in the -OH group of another ethanol molecule.
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When a nonmetal oxide reacts with water, it forms an oxoacid with the same nonmetal oxidation state. Give the name and formula of the oxide used to prepare each of these oxoacids: (a) hypochlorous acid; (b) chlorous acid; (c) chloric acid; (d) perchloric acid; (e) sulfuric acid; (f) sulfurous acid; (g) nitric acid; (h) nitrous acid; (i) carbonic acid; ( j) phosphoric acid.
Answer:
1) dichlorine monoxide (Cl2O
2) dichlorine trioxide (Cl2O3)
3) dichlorine pentoxide (Cl2O5)
4) dichlorine heptoxide (Cl2O7)
5) sulfur trioxide (SO3)
6) sulfur dioxide (SO2)
7) dinitrogen pentoxide (N2O5)
8) dinitrogen trioxide (N2O3)
9) carbon dioxide (CO2)
10) phosphorous trioxde (PO3)
Explanation:
Step 1: Data given
a) hypochlorous acid = HOCl
HClO is formed when dichlorine monoxide (Cl2O) is dissolved in water.
Cl2O (g) + H2O (l) → 2 HClO (aq)
(b) chlorous acid = HClO2
HClO2 is formed when dichlorine trioxide (Cl2O3) is dissolved in water.
Cl2O3 (g) + H2O (l) → 2HClO2 (aq)
(c) chloric acid = HClO3
HClO3 is formed when dichlorine pentoxide (Cl2O5) is dissolved in water
Cl2O5 (g) + H2O (l) → 2HClO3 (aq)
d) perchloric acid = HClO4
HClO4 is formed when dichlorine heptoxide (Cl2O7) is dissolved in water
Cl2O7 (g) + H2O (l) → 2HClO4 (aq)
(e) sulfuric acid = H2SO4
H2SO4 is formed when sulfur trioxide (SO3) is dissolved in water
SO3 (aq) + H2O(l) → H2SO4(aq)
(f) sulfurous acid = H2SO3
H2SO3 is formed when sulfur dioxide (SO2) is dissolved in water
SO2 (aq) + H2O(l) → H2SO3(aq)
(g) nitric acid = HNO3
HNO3 is formed when dinitrogen pentoxide (N2O5) is dissolved in water
N2O5(aq) + H2O(l) → 2HNO3(aq)
(h) nitrous acid = HNO2
HNO2 is formed when dinitrogen trioxide (N2O3) is dissolved in water
N2O3(aq) + H2O(l) → 2HNO2 (aq)
(i) carbonic acid = H2CO3
H2CO3 is formed when carbon dioxide (CO2) is dissolved in water
CO2(g) + H2O(l) → H2CO3(aq)
( j) phosphoric acid = H3PO4
H3PO4 is formed when phosphorous trioxde (PO3) is dissolved in water
PO3 + 2H2O → H3PO4 + OH
Distinguish between an absorption spectrum and an emission spectrum. With which did Bohr work?
Answer:
Bohr used emission spectrum for its mono atomic model....
Explanation:
Emission Spectrum is produced when atoms are excited by energy. After excitation, they emit this energy in the form of different wavelengths according to the type of atom and produce a unique fingerprint of themselves called as it's emission spectrum.
Absorption Spectrum is a type of spectrum that is produces when photons of light are absorbed by electrons at one state. they jump to another state and may cause scattering. This produces a specific absorption spectrum for that specific atom.
Uranium hexaflouride (UF6) has a triple point at (T, p) = (337 K,152 kPa). Suppose you have a (gaseous) sample of UF6 at atmospheric pressure and room temperature. If you keep cooling your sample, will it undergo a phase transition from gas → liquid or from gas → solid?
Answer:
Explanation:
Uranium hexaflouride (UF6) has a triple point at (T, p) = (337 K,152 kPa) that means at pressure above 152kPa and temperature of 337 K ( 64 degree celsius) it becomes liquid .
If we have a (gaseous) sample of UF6 at atmospheric pressure and room temperature , and we keep cooling the sample , it will undergo a phase change of gas → solid.
Classify the elemental reagents in the given reactions as oxidizing or reducing agents. Drag the appropriate items to their respective bins. View Available Hint(s)F2 + H2 → 2HF
2Mg + O2 → 2MgO
Drag the appropriate items to their respective bins.
F2 O2 Mg H 2
OXIDIZING AGENT REDUCING AGENT
Here is a more complex redox reaction involving the permanganate ion in acidic solution:
5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O
Classify each reactant as the reducing agent, oxidizing agent, or neither.
Drag the appropriate items to their respective bins.
Fe2+ MnO4- H+
oxidizing agent reducing agent neither
In the given reactions, F2 and O2 act as oxidizing agents by being reduced, while H2 and Mg serve as reducing agents by being oxidized. In the complex reaction with MnO4- and Fe2+, MnO4- is the oxidizing agent and Fe2+ is the reducing agent.
Explanation:Classification of Oxidizing and Reducing Agents
In the redox reaction F2 + H2 → 2HF, the molecule F2 is being reduced (gains electrons) and thus is the oxidizing agent. Conversely, H2 is being oxidized (loses electrons) and so it acts as the reducing agent.
In the reaction 2Mg + O2 → 2MgO, the oxygen molecule O2 is reduced, making it the oxidizing agent, while magnesium Mg is oxidized, hence it is the reducing agent.
For the more complex redox reaction, 5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O, the permanganate ion MnO4- is reduced and is the oxidizing agent. Fe2+ is oxidized and is the reducing agent. The H+ ion is neither an oxidizing agent nor a reducing agent in this reaction.
Draw the Lewis structure for the compound with the formula COCl2COCl2. Use lines to show bonding electrons.
Answer : The Lewis-dot structure of [tex]COCl_2[/tex] is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, [tex]COCl_2[/tex]
As we know that carbon has '4' valence electrons, chlorine has '7' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]COCl_2[/tex] = 1(4) + 2(7) + 1(6) = 24
According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.